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Chapter 21
The Electric Field 1: Discrete Charge Distributions
Conceptual Problems
*1 ••
Similarities:
Differences:
The force between charges and
masses varies as 1/r2.
There are positive and negative charges but
only positive masses.
The force is directly proportional to
the product of the charges or
masses.
Like charges repel; like masses attract.
The gravitational constant G is many orders
of magnitude smaller than the Coulomb
constant k.
2 •
Determine the Concept No. In order to charge a body by induction, it must have charges
that are free to move about on the body. An insulator does not have such charges.
3 ••
Determine the Concept During this sequence of events, negative charges are attracted
from ground to the rectangular metal plate B. When S is opened, these charges are trapped
on B and remain there when the charged body is removed. Hence B is negatively charged
and correct. is )(c
4 ••
(a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere
and disconnect the sphere from ground; then remove the insulating rod. The sphere will
be negatively charged.
(b) Bring the insulating rod in contact with the metal sphere; some of the positive charge
on the rod will be transferred to the metal sphere.
(c) Yes. First charge one metal sphere negatively by induction as in (a). Then use that
negatively charged sphere to charge the second metal sphere positively by induction.
1
Chapter 21
2
*5 ••
Determine the Concept Because the spheres are conductors, there are free electrons on
them that will reposition themselves when the positively charged rod is brought nearby.
(a) On the sphere near the positively
charged rod, the induced charge is negative
and near the rod. On the other sphere, the
net charge is positive and on the side far
from the rod. This is shown in the diagram.
(b) When the spheres are separated and far
apart and the rod has been removed, the
induced charges are distributed uniformly
over each sphere. The charge distributions
are shown in the diagram.
6 •
Determine the Concept The forces acting
on +q are shown in the diagram. The force
acting on +q due to −Q is along the line
joining them and directed toward −Q. The
force acting on +q due to +Q is along the
line joining them and directed away from
+Q.
Because charges +Q and −Q are equal in magnitude, the forces due to these charges are
equal and their sum (the net force on +q) will be to the right and so correct. is )(e Note
that the vertical components of these forces add up to zero.
*7 •
Determine the Concept The acceleration of the positive charge is given by
.0 EFa
rrr
m
q
m
== Because q0 and m are both positive, the acceleration is in the same
direction as the electric field. correct. is )(d
*8 •
Determine the Concept E
r
is zero wherever the net force acting on a test charge is
zero. At the center of the square the two positive charges alone would produce a net
electric field of zero, and the two negative charges alone would also produce a net
electric field of zero. Thus, the net force acting on a test charge at the midpoint of the
The Electric Field 1: Discrete Charge Distributions
3
square will be zero. correct. is )(b
9 ••
(a) The zero net force acting on Q could be the consequence of equal collinear charges
being equidistant from and on opposite sides of Q.
(b) The charges described in (a) could be either positive or negative and the net force on
Q would still be zero.
(c) Suppose Q is positive. Imagine a negative charge situated to its right and a larger
positive charge on the same line and the right of the negative charge. Such an arrangement
of charges, with the distances properly chosen, would result in a net force of zero acting
on Q.
(d) Because none of the above are correct, correct. is )(d
10 •
Determine the Concept We can use the
rules for drawing electric field lines to
draw the electric field lines for this system.
In the sketch to the right we’ve assigned 2
field lines to each charge q.
*11 •
Determine the Concept We can use the
rules for drawing electric field lines to
draw the electric field lines for this system.
In the field-line sketch to the right we’ve
assigned 2 field lines to each charge q.
Chapter 21
4
*12 •
Determine the Concept We can use the
rules for drawing electric field lines to
draw the electric field lines for this system.
In the field-line sketch to the right we’ve
assigned 7 field lines to each charge q.
13 •
Determine the Concept A positive charge will induce a charge of the opposite sign on
the near surface of the nearby neutral conductor. The positive charge and the induced
charge on the neutral conductor, being of opposite sign, will always attract one another.
correct. is )(a
*14 •
Determine the Concept Electric field lines around an electric dipole originate at the
positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy
this requirement. correct. is )(d
*15 ••
Determine the Concept Because θ ≠ 0, a dipole in a uniform electric field will
experience a restoring torque whose magnitude is θsinxpE . Hence it will oscillate
about its equilibrium orientation, θ = 0. If θ << 1, sinθ ≈ θ, and the motion will be simple
harmonic motion. Because the field is nonuniform and is larger in the x direction, the
force acting on the positive charge of the dipole (in the direction of increasing x) will be
greater than the force acting on the negative charge of the dipole (in the direction of
decreasing x) and thus there will be a net electric force on the dipole in the direction of
increasing x. Hence, the dipole will accelerate in the x direction as it oscillates about
θ = 0.
16 ••
(a) False. The direction of the field is toward a negative charge.
(b) True.
(c) False. Electric field lines diverge from any point in space occupied by a positive
charge.
(d) True
The Electric Field 1: Discrete Charge Distributions
5
(e) True
17 ••
Determine the Concept The diagram
shows the metal balls before they are
placed in the water. In this situation, the net
electric field at the location of the sphere
on the left is due only to the charge –q on
the sphere on the right. If the metal balls
are placed in water, the water molecules
around each ball tend to align themselves
with the electric field. This is shown for
the ball on the right with charge –q.
(a) The net electric field
r
E net that produces a force on the ball on the left is the
field
r
E due to the charge –q on the ball on the right plus the field due to the layer
of positive charge that surrounds the ball on the right. This layer of positive
charge is due to the aligning of the water molecules in the electric field, and the
amount of positive charge in the layer surrounding the ball on the left will be less
than +q. Thus, Enet < E. Because Enet < E, the force on the ball on the left is less
than it would be if the balls had not been placed in water. Hence, the force will
decrease when the balls are placed in the water.
(b) When a third uncharged metal ball is
placed between the first two, the net
electric field at the location of the sphere
on the right is the field due to the charge +q
on the sphere on the left, plus the field due
to the charge –Q and +Q on the sphere in
the middle. This electric field is directed to
the right.
The field due to the charge
–Q and +Q on the sphere in the middle at the location of the
sphere on the right is to the right. It follows that the net electric field due to the charge
+q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in
the middle is to the right and has a greater magnitude than the field due only to the charge
+q on the sphere on the left. Hence, the force on either sphere will increase if a third
uncharged metal ball is placed between them.
Remarks: The reduction of an electric field by the alignment of dipole moments
with the field is discussed in further detail in Chapter 24.
Chapter 21
6
*18 ••
Determine the Concept Yes. A positively charged ball will induce a dipole on the metal
ball, and if the two are in close proximity, the net force can be attractive.
*19 ••
Determine the Concept Assume that the wand has a negative charge. When the charged
wand is brought near the tinfoil, the side nearer the wand becomes positively charged by
induction, and so it swings toward the wand. When it touches the wand, some of the
negative charge is transferred to the foil, which, as a result, acquires a net negative charge
and is now repelled by the wand.
Estimation and Approximation
20 ••
Picture the Problem Because it is both very small and repulsive, we can ignore the
gravitational force between the spheres. It is also true that we are given no information
about the masses of these spheres. We can find the largest possible value of Q by
equating the electrostatic force between the charged spheres and the maximum force the
cable can withstand.
Using Coulomb’s law, express the
electrostatic force between the two
charged spheres:
2
2
l
kQF =
Express the tensile strength Stensile of
steel in terms of the maximum force
Fmax in the cable and the cross-
sectional area of the cable and solve
for F:
A
FS maxtensile = ⇒ tensilemax ASF =
Equate these forces to obtain:
tensile2
2
ASkQ =l
Solve for Q:
k
ASQ tensilel=
Substitute numerical values and evaluate Q:
( ) ( )( ) mC95.2
C/mN1099.8
N/m102.5m105.1m1 229
2824
=⋅×
××=
−
Q
21 ••
Picture the Problem We can use Coulomb’s law to express the net force acting on the
copper cube in terms of the unbalanced charge resulting from the assumed migration of
half the charges to opposite sides of the cube. We can, in turn, find the unbalanced charge
Qunbalanced from the number of copper atoms N and the number of electrons per atom.
The Electric Field 1: Discrete Charge Distributions
7
(a) Using Coulomb’s law, express
the net force acting on the copper
rod due to the imbalance in the
positive and negative charges:
2
2
unbalanced
r
kQF = (1)
Relate the number of copper atoms
N to the mass m of the rod, the
molar mass M of copper, and
Avogadro’s number NA:
M
V
M
m
N
N rodCu
A
ρ==
Solve for N to obtain:
A
rodCu N
M
VN ρ=
Substitute numerical values and evaluate N:
( )( ) ( )( )
atoms10461.8
kg/mol1054.63
atoms/mol1002.6m104m105.0kg/m1093.8
22
3
2322233
×=
×
××××= −
−−
N
Because each atom has 29 electrons
and protons, we can express
Qunbalanced as:
( )( )eNQ 721unbalanced 1029 −=
Substitute numerical values and evaluate Qunbalanced:
( )( )( )( ) C10963.110461.8C106.11029 22219721unbalanced −−− ×=××=Q
Substitute for Qunbalanced in equation (1) to obtain:
( )( )
( ) N1046.3m01.0
C10963.1C/mN1099.8 10
2
22229
×=×⋅×=
−
F
(b) Using Coulomb’s law, express
the maximum force of repulsion
Fmax in terms of the maximum
possible charge Qmax:
2
2
max
max r
kQF =
Solve for Qmax:
k
FrQ max
2
max =
Express Fmax in terms of the tensile
strength Stensile of copper:
ASF tensilemax =
where A is the cross sectional area of the
cube.
Chapter 21
8
Substitute to obtain:
k
ASrQ tensile
2
max =
Substitute numerical values and evaluate Qmax:
( ) ( )( ) C1060.1
C/mN1099.8
m10N/m103.2m01.0 5
229
24282
max
−
−
×=⋅×
×=Q
Because : maxunbalanced 2QQ = ( )
C0.32
C1060.12 5unbalanced
µ=
×= −Q
Remarks: A net charge of −32 µC means an excess of 2.00×1014 electrons, so the net
imbalance as a percentage of the total number of charges is 4.06×10−11 = 4×10−9 %.
22 •••
Picture the Problem We can use the definition of electric field to express E in terms of
the work done on the ionizing electrons and the distance they travel λ between collisions.
We can use the ideal-gas law to relate the number density of molecules in the gas ρ and
the scattering cross-section σ to the mean free path and, hence, to the electric field.
(a) Apply conservation of energy to
relate the work done on the
electrons by the electric field to the
change in their kinetic energy:
sFKW ∆=∆=
From the definition of electric field
we have:
qEF =
Substitute for F and ∆s to obtain: λqEW = , where the mean free path λ is
the distance traveled by the electrons
between ionizing collisions with nitrogen
atoms.
Referring to pages 545-546 for a
discussion on the mean-free path,
use its definition to relate λ to the
scattering cross-section σ and the
number density for air molecules n:
nσλ
1=
Substitute for λ and solve for E to
obtain: q
nWE σ=
Use the ideal-gas law to obtain:
kT
P
V
Nn ==
The Electric Field 1: Discrete Charge Distributions
9
Substitute for n to obtain:
qkT
PWE σ= (1)
Substitute numerical values and evaluate E:
( )( )( )( )( )( )( ) N/C1041.2K300J/K1038.1C106.1 J/eV106.1eV1N/m10m10 62319
1925219
×=××
×= −−
−−
E
(b) From equation (1) we see that: PE ∝ and 1−∝ TE
i.e., E increases linearly with pressure and
varies inversely with temperature.
*23 ••
Picture the Problem We can use Coulomb’s law to express the charge on the rod in
terms of the force exerted on it by the soda can and its distance from the can. We can
apply Newton’s 2nd law in rotational form to the can to relate its acceleration to the
electric force exerted on it by the rod. Combining these equations will yield an expression
for Q as a function of the mass of the can, its distance from the rod, and its acceleration.
Use Coulomb’s law to relate the
force on the rod to its charge Q and
distance r from the soda can:
2
2
r
kQF =
Solve for Q to obtain:
k
FrQ
2
= (1)
Apply to the
can:
ατ I=∑ mass ofcenter
αIFR =
Because the can rolls without
slipping, we know that its linear
acceleration a and angular
acceleration α are related according
to:
R
a=α
where R is the radius of the soda can.
Because the empty can is a hollow
cylinder:
2MRI =
where M is the mass of the can.
Substitute for I and α and solve for
F to obtain:
Ma
R
aMRF == 2
2
Substitute for F in equation (1):
k
MarQ
2
=
Chapter 21
10
Substitute numerical values and
evaluate Q: ( ) ( )( )
nC141
C/mN1099.8
m/s1kg018.0m1.0
229
22
=
⋅×=Q
24 ••
Picture the Problem Because the nucleus is in equilibrium, the binding force must be
equal to the electrostatic force of repulsion between the protons.
Apply 0=∑Fr to a proton:
0ticelectrostabinding =− FF
Solve for Fbinding: ticelectrostabinding FF =
Using Coulomb’s law, substitute for
Felectrostatic:
2
2
binding r
kqF =
Substitute numerical values and evaluate Felectrostatic:
( )( )( ) N230m10 C106.1C/mN1099.8
215
219229
binding =×⋅×= −
−
F
Electric Charge
25 •
Picture the Problem The charge acquired by the plastic rod is an integral number of
electronic charges, i.e., q = ne(−e).
Relate the charge acquired by the
plastic rod to the number of
electrons transferred from the wool
shirt:
( )enq −= e
Solve for and evaluate ne: 12
19e 1000.5C101.6
C8.0 ×=×−
−=−= −
µ
e
qn
26 •
Picture the Problem One faraday = NAe. We can use this definition to find the number of
coulombs in a faraday.
Use the definition of a faraday to calculate the number of coulombs in a faraday:
( )( ) C1063.9C/electron106.1electrons1002.6faraday1 41923A ×=××== −eN
The Electric Field 1: Discrete Charge Distributions
11
*27 •
Picture the Problem We can find the number of coulombs of positive charge there are in
1 kg of carbon from , where nenQ C6= C is the number of atoms in 1 kg of carbon and
the factor of 6 is present to account for the presence of 6 protons in each atom. We can
find the number of atoms in 1kg of carbon by setting up a proportion relating Avogadro’s
number, the mass of carbon, and the molecular mass of carbon to nC.
Express the positive charge in terms
of the electronic charge, the number
of protons per atom, and the number
of atoms in 1 kg of carbon:
enQ C6=
Using a proportion, relate the
number of atoms in 1 kg of carbon
nC, to Avogadro’s number and the
molecular mass M of carbon:
M
m
N
n C
A
C = ⇒
M
mNn CAC =
Substitute to obtain:
M
emNQ CA6=
Substitute numerical values and evaluate Q:
( )( )( ) C1082.4
kg/mol012.0
C101.6kg1atoms/mol106.026 71923 ×=××=
−
Q
Coulomb’s Law
28 •
Picture the Problem We can find the forces the two charges exert on each by applying
Coulomb’s law and Newton’s 3rd law. Note that because the vector pointing from
q
ir ˆˆ 2,1 =
1 to q2 is in the positive x direction.
(a) Use Coulomb’s law to express
the force that q1 exerts on q2:
2,12
2,1
21
2,1 rˆF r
qkq=r
Substitute numerical values and evaluate 2,1F
r
:
( )( )( )
( ) ( )iiF ˆmN0.24m3
µC6µC4/CmN108.99
2
229
2,1 =⋅×=
r
Chapter 21
12
(b) Because these are action-and-
reaction forces, we can apply
Newton’s 3rd law to obtain:
( )iFF ˆmN0.242,11,2 −=−= rr
(c) If q2 is −6.0 µC:
( )( )( )
( ) ( )iiF ˆmN0.24ˆm3
µC6µC4/CmN108.99
2
229
2,1 −=−⋅×=
r
and
( )iFF ˆmN0.242,11,2 =−= rr
29 •
Picture the Problem q2 exerts an attractive force 1,2F
r
on q1 and q3 a repulsive force 1,3F
r
.
We can find the net force on q1 by adding these forces.
Express the net force acting on q1:
1,31,21 FFF
rrr +=
Express the force that q2 exerts on
q1:
iF ˆ2
1,2
21
1,2 r
qqk=r
Express the force that q3 exerts on
q1:
( )iF ˆ2
1,3
31
1,3 −= r
qqkr
Substitute and simplify to obtain:
i
iiF
ˆ
ˆˆ
2
1,3
3
2
1,2
2
1
2
1,3
31
2
1,2
21
1
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
−=
r
q
r
q
qk
r
qqk
r
qqkr
Substitute numerical values and evaluate 1F
r
:
( )( ) ( ) ( ) ( )iiF ˆN1050.1ˆm6 C6m3 C4C6/CmN1099.8 2222291 −×=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⋅×= µµµr
The Electric Field 1: Discrete Charge Distributions
13
30 ••
Picture the Problem The configuration of
the charges and the forces on the fourth
charge are shown in the figure … as is a
coordinate system. From the figure it is
evident that the net force on q4 is along the
diagonal of the square and directed away
from q3. We can apply Coulomb’s law to
express 4,1F
r
, 4,2F
r
and 4,3F
r
and then add
them to find the net force on q4.
Express the net force acting on q4:
4,34,24,14 FFFF
rrrr ++=
Express the force that q1 exerts on
q4:
jF ˆ2
4,1
41
4,1 r
qkq=r
Substitute numerical values and evaluate 4,1F
r
:
( )( ) ( ) ( )jjF ˆN1024.3ˆm05.0 nC3nC3/CmN1099.8 522294,1 −×=⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×=r
Express the force that q2 exerts on q4:
iF ˆ2
4,2
42
4,2 r
qkq=r
Substitute numerical values and evaluate 4,2F
r
:
( )( ) ( ) ( )iiF ˆN1024.3ˆm05.0 nC3nC3/CmN1099.8 522294,2 −×=⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×=r
Express the force that q3 exerts on q4:
4,324,3
43
4,3 rˆF r
qkq=r , where is a unit vector
pointing from q
4,3ˆr
3 to q4.
Express 4,3r
r
in terms of 1,3r
r
and 4,1r
r
:
( ) ( ji
rrr
ˆm05.0ˆm05.0
4,11,34,3
+= )
+= rrr
Chapter 21
14
Convert to : 4,3r
r
4,3ˆr
( ) ( )
( ) ( )
ji
ji
r
r
r
ˆ707.0ˆ707.0
m05.0m05.0
ˆm05.0ˆm05.0ˆ
22
4,3
4,3
4,3
+=
+
+== r
r
Substitute numerical values and evaluate 4,3F
r
:
( )( ) ( ) ( )
( ) ( )ji
jiF
ˆN1014.1ˆN1014.1
ˆ707.0ˆ707.0
m205.0
nC3nC3/CmN1099.8
55
2
229
4,3
−− ×−×−=
+⎟⎟⎠
⎞
⎜⎜⎝
⎛−⋅×=r
Substitute and simplify to find 4F
r
:
( ) ( ) ( ) ( )
( ) ( )ji
jiijF
ˆN1010.2ˆN1010.2
ˆN1014.1ˆN1014.1ˆN1024.3ˆN1024.3
55
5555
4
−−
−−−−
×+×=
×−×−×+×=r
31 ••
Picture the Problem The configuration of the charges and the forces on q3 are shown in
the figure … as is a coordinate system. From the geometry of the charge distribution it is
evident that the net force on the 2 µC charge is in the negative y direction. We can apply
Coulomb’s law to express 3,1F
r
and 3,2F
r
and then add them to find the net force on q3.
The net force acting on q3 is given by: 3,23,13 FFF
rrr +=
The Electric Field 1: Discrete Charge Distributions
15
Express the force that q1 exerts on
q3:
jiF ˆsinˆcos3,1 θθ FF −=
r
where
( )( )(
( ) ( )
)
N3.12
m0.08m0.03
C2C5C/mN1099.8
22
229
2
31
=
+
⋅×=
=
µµ
r
qkqF
and
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 6.20
cm8
cm3tan 1θ
Express the force that q2 exerts on
q3:
jiF ˆsinˆcos3,2 θθ FF −−=
r
Substitute for 3,1F
r
and 3,2F
r
and
simplify to obtain:
j
j
ijiF
ˆsin2
ˆsin
ˆcosˆsinˆcos3
θ
θ
θθθ
F
F
FFF
−=
−
−−=r
Substitute numerical values and
evaluate 3F
r
:
( )
( ) j
jF
ˆN66.8
ˆ6.20sinN3.1223
−=
°−=r
*32 ••
Picture the Problem The positions of the
charges are shown in the diagram. It is
apparent that the electron must be located
along the line joining the two charges.
Moreover, because it is negatively charged,
it must be closer to the −2.5 µC than to the
6.0 µC charge, as is indicated in the figure.
We can find the x and y coordinates of the
electron’s position by equating the two
electrostatic forces acting on it and solving
for its distance from the origin.
We can use similar triangles to express this
radial distance in terms of the x and y
coordinates of the electron.
Express the condition that must be ee FF ,2,1 =
Chapter 21
16
satisfied if the electron is to be in
equilibrium:
Express the magnitude of the force
that q1 exerts on the electron:
( )21,1 m25.1+= r ekqF e
Express the magnitude of the force
that q2 exerts on the electron:
2
2
,2 r
eqk
F e =
Substitute and simplify to obtain:
( ) 2221 m25.1 rqr q =+
Substitute for q1 and q2 and
simplify:
( ) ( )
0m25.1
m2361.2m4.1 122
=+
+− −− rr
Solve for r to obtain:
m0.4386
and
m036.2
−=
=
r
r
Because r < 0 is unphysical, we’ll consider
only the positive root.
Use the similar triangles in the
diagram to establish the proportion
involving the y coordinate of
the
electron:
m1.12
m2.036
m5.0
=ey
Solve for ye: m909.0=ey
Use the similar triangles in the
diagram to establish the proportion
involving the x coordinate of the
electron:
m1.12
m2.036
m1
=ex
Solve for xe: m82.1=ex
The coordinates of the electron’s
position are:
( ) ( )m0.909m,1.82, −−=ee yx
The Electric Field 1: Discrete Charge Distributions
17
*33 ••
Picture the Problem Let q1 represent the
charge at the origin, q2 the charge at (0, 0.1
m), and q3 the charge at
(0.2 m, 0). The diagram shows the forces
acting on each of the charges. Note the
action-and-reaction pairs. We can apply
Coulomb’s law and the principle of
superposition of forces to find the net
force acting on each of the charges.
Express the net force acting on q1: 1,31,21 FFF
rrr +=
Express the force that q2 exerts on q1:
1,23
1,2
12
1,2
1,2
2
1,2
12
1,22
1,2
12
1,2 ˆ r
r
rF r
rr
r
qkq
rr
qkq
r
qkq ===
Substitute numerical values and evaluate 1,2F
r
:
( )( ) ( )( ) ( ) ( jjF ˆN80.1ˆm1.0m1.0 C1C2/CmN1099.8 32291,2 =− )−⋅×= µµ
r
Express the force that q3 exerts on q1:
1,331,3
13
1,3 rF
rr
r
qkq=
Substitute numerical values and evaluate 1,3F
r
:
( )( ) ( )( ) ( ) ( iiF ˆN899.0ˆm2.0m2.0 C1C4/CmN1099.8 32291,3 =− )−⋅×= µµ
r
Substitute to find : 1F
r ( ) ( ) jiF ˆN80.1ˆN899.01 +=r
Express the net force acting on q2:
( ) jF
FF
FFF
ˆN80.12,3
1,22,3
2,12,32
−=
−=
+=
r
rr
rrr
because 2,1F
r
and 1,2F
r
are action-and-reaction
forces.
Chapter 21
18
Express the force that q3 exerts on q2:
( ) ([ ]ji
rF
ˆm1.0ˆm2.03
2,3
23
2,33
2,3
23
2,3
+−=
=
r
qkq
r
qkq r
)
r
Substitute numerical values and evaluate 2,3F
r
:
( )( ) ( )( ) ( ) ( )[ ]
( ) ( ) ji
jiF
ˆN640.0ˆN28.1
ˆm1.0ˆm2.0
m224.0
C2C4/CmN1099.8 3
229
2,3
+−=
+−⋅×= µµr
Find the net force acting on q2:
( ) ( ) ( ) ( )
( ) ( ) ji
jjijFF
ˆN16.1ˆN28.1
ˆN80.1ˆN640.0ˆN28.1ˆN80.12,32
−−=
−+−=−= rr
Noting that 3,1F
r
and 1,3F
r
are an action-and-reaction pair, as are 3,2F
r
and 2,3F
r
,
express the net force acting on q3:
( ) ( ) ( )[ ]
( ) ( ) ji
jiiFFFFF
ˆN640.0ˆN381.0
ˆN640.0ˆN28.1ˆN899.02,31,33,23,13
−=
+−−−=−−=+= rrrrr
34 ••
Picture the Problem Let q1 represent the
charge at the origin and q3 the charge
initially at (8 cm, 0) and later at (17.75
cm, 0). The diagram shows the forces
acting on q3 at (8 cm, 0). We can apply
Coulomb’s law and the principle of
superposition of forces to find the net
force acting on each of the charges.
Express the net force on q2 when it
is at (8 cm, 0):
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
+=
+=
3,23
3,2
2
3,13
3,1
1
3
3,23
3,2
32
3,13
3,1
31
3,23,12 0,cm8
rr
rr
FFF
rr
rr
rrr
r
Q
r
qkq
r
qkQ
r
qkq
The Electric Field 1: Discrete Charge Distributions
19
Substitute numerical values to obtain:
( )
( )( ) ( ) ( ) ( ) ( ) ⎥⎦
⎤+⎢⎣
⎡⋅×
=−
ii
i
ˆm04.0
m04.0
ˆm08.0
m0.08
C5C2/CmN1099.8
ˆN7.19
3
2
3
229 Qµµ
Solve for and evaluate Q2: C00.32 µ−=Q
Remarks: An alternative solution is to equate the electrostatic forces acting on q2 when it is
at (17.75 cm, 0).
35 ••
Picture the Problem By considering the symmetry of the array of charges we can see that
the y component of the force on q is zero. We can apply Coulomb’s law and the principle
of superposition of forces to find the net force acting on q.
Express the net force acting on q: qQqxQq ,45ataxis,on 2 °+= FFF
rrr
Express the force on q due to the
charge Q on the x axis:
iF ˆ2axis,on R
kqQ
qxQ =
r
Express the net force on q due to the
charges at 45°:
i
iF
ˆ
2
2
ˆ45cos22
2
2,45at
R
kqQ
R
kqQ
qQ
=
°=°
r
Substitute to obtain:
i
iiF
ˆ
2
21
ˆ
2
2ˆ
2
22
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
+=
R
kqQ
R
kqQ
R
kqQ
q
r
36 •••
Picture the P oblem Let the Hr + ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0,
0), and H3 at ( )0,23,2 aa . The N−3 ion, q4 in our notation, is then at ( )32,32,2 aaa where a =1.64×10−10 m. To simplify our calculations we’ll set
N1056.8 922 −×== Cake . We can apply Coulomb’s law and the principle of
superposition of forces to find the net force acting on each ion.
Chapter 21
20
Express the net force acting on q1:
1,41,31,21 FFFF
rrrr ++=
Find 1,2F
r
:
( ) iirF ˆˆˆ 1,22
1,2
21
1,2 CCr
qkq −=−==r
Find 1,3F
r
:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+⎟⎠
⎞⎜⎝
⎛ −
=
=
ji
ji
rF
ˆ
2
3ˆ
2
1
ˆ
2
30ˆ
2
0
ˆ 1,32
1,3
13
1,3
C
a
aa
C
r
qkqr
Noting that the magnitude of q4 is three times that of the other charges and that it is
negative, express 1,4F
r
:
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+⎟⎠
⎞⎜⎝
⎛ −+⎟⎠
⎞⎜⎝
⎛ −
−==
kji
kji
kji
rF
ˆ
3
2ˆ
32
1ˆ
2
13
ˆ
3
2ˆ
32
ˆ
2
3
3
2
322
ˆ
3
20ˆ
32
0ˆ
2
0
3ˆ3
222
1,41,4
C
a
aaa
C
aaa
aaa
CC
r
The Electric Field 1: Discrete Charge Distributions
21
Substitute to find : 1F
r
k
kji
jiiF
ˆ6
ˆ
3
2ˆ
32
1ˆ
2
13
ˆ
2
3ˆ
2
1ˆ
1
C
C
CC
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛+
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−−=r
From symmetry considerations:
kFFF ˆ6132 C===
rrr
Express the condition that molecule is
in equilibrium:
04321 =+++ FFFF
rrrr
Solve for and evaluate : 4F
r ( )
k
FFFFF
ˆ63
3 13214
C−=
−=++−= rrrrr
The Electric Field
*37 •
Picture the Problem Let q represent the charge at the origin and use Coulomb’s law for
E
r
due to a point charge to find the electric field at x = 6 m and −10 m.
(a) Express the electric field at a point
P located a distance x from a charge
q:
( ) P,02 rˆE x
kqx =r
Evaluate this expression for
x = 6 m:
( ) ( )( )( )
( )i
iE
ˆN/C999
ˆ
m6
C4/CmN1099.8m6 2
229
=
⋅×= µr
(b) Evaluate E
r
at x = −10 m:
( ) ( )( )( ) ( ) ( )iiE ˆN/C360ˆm10 C4/CmN1099.8m10 2
229
−=−⋅×=− µr
(c) The following graph was plotted using a spreadsheet program:
Chapter 21
22
-500
-250
0
250
500
-2 -1 0 1 2
x (m)
E x
(N
/C
)
*38 •
Picture the Problem Let q represent the charges of +4 µC and use Coulomb’s law for
E
r
due to a point charge and the principle of superposition for fields to find the electric
field at the locations specified.
Noting that q1 = q2, use Coulomb’s law and the principle of superposition to express the
electric field due to the given charges at a point P a distance x from the origin:
( ) ( ) ( ) ( ) ( )
( ) ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−+⋅=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−+=−+=+=
P,2P,2
2
P,2P,21P,2
2
P,2
1
21
212121
ˆ
m8
1ˆ1/CmkN36
ˆ
m8
1ˆ1ˆ
m8
ˆ
qq
qqqqqq
xx
xx
kq
x
kq
x
kqxxx
rr
rrrrEEE
rrr
(a) Apply this equation to the point at x = −2 m:
( ) ( ) ( ) ( ) ( ) ( ) ( )iiiE ˆkN/C36.9ˆm101ˆm2 1/CmkN36m2 222 −=⎥⎦
⎤−+⎢⎣
⎡ −⋅=−r
(b) Evaluate E
r
at x = 2 m:
( ) ( )
( ) ( ) ( ) ( ) ( )iiiE ˆkN/C00.8ˆm6 1ˆm2 1/CmkN36m2 222 =⎥⎦⎤−+⎢⎣⎡⋅=
r
The Electric Field 1: Discrete Charge Distributions
23
(c) Evaluate E
r
at x = 6 m:
( ) ( ) ( ) ( ) ( ) ( ) ( )iiiE ˆkN/C00.8ˆm2 1ˆm6 1/CmkN36m6 222 −=⎥⎦⎤−+⎢⎣⎡⋅=
r
(d) Evaluate E
r
at x = 10 m:
( ) ( ) ( ) ( ) ( ) ( ) ( )iiiE ˆkN/C35.9ˆm2 1ˆm101/CmkN36m10 222 =⎥⎦⎤+⎢⎣⎡⋅=
r
(e) From symmetry considerations: ( ) 0m4 =E
(f) The following graph was plotted using a spreadsheet program:
-100
-50
0
50
100
-4 0 4 8
x (m)
E x
(
kN
m
2 /C
)
12
39 •
Picture the Problem We can find the electric field at the origin from its definition and
the force on a charge placed there from EF
rr
q= . We can apply Coulomb’s law to find
the value of the charge placed at y = 3 cm.
(a) Apply the definition of electric
field to obtain:
( ) ( ) jjFE ˆkN/C400
nC2
ˆN108 4
0
=×==
−
q
rr
(b) Express and evaluate the force
on a charged body in an electric
field:
( )( )
( ) j
jEF
ˆmN60.1
ˆkN/C400nC4
−=
−== rr q
Chapter 21
24
(c) Apply Coulomb’s law to obtain:
( )
( ) ( ) ( ) jj ˆmN60.1ˆm0.03 nC4 2 −=−−kq
Solve for and evaluate q: ( )( )( )( )
nC40.0
nC4/CmN1099.8
m0.03mN60.1
229
2
−=
⋅×−=q
40 •
Picture the Problem We can compare the electric and gravitational forces acting on an
electron by expressing their ratio. We can equate these forces to find the charge that
would have to be placed on a penny in order to balance the earth’s gravitational force on
it.
(a) Express the magnitude of the
electric force acting on the electron:
eEFe =
Express the magnitude of the
gravitational force acting on the
electron:
gmF eg =
Express the ratio of these forces to
obtain:
mg
eE
F
F
g
e =
Substitute numerical values and
evaluate Fe/Fg:
( )( )( )( )
12
231
19
1069.2
m/s9.81kg109.11
N/C150C101.6
×=
×
×= −
−
g
e
F
F
or ( ) ge FF 121069.2 ×= , i.e., the electric
force is greater by a factor of 2.69×1012.
(b) Equate the electric and
gravitational forces acting on the
penny and solve for q to obtain:
E
mgq =
Substitute numerical values and
evaluate q:
( )( )
C1096.1
N/C150
m/s9.81kg103
4
23
−
−
×=
×=q
The Electric Field 1: Discrete Charge Distributions
25
41 ••
Picture the Problem The diagram shows
the locations of the charges q1 and q2 and
the point on the x axis at which we are to
find .E
r
From symmetry considerations we
can conclude that the y component of E
r
at
any point on the x axis is zero. We can use
Coulomb’s law for the electric field due to
point charges to find the field at any point
on the x axis and to find the force
on a charge q
EF
rr
q=
0 placed on the x axis at
x = 4 cm.
(a) Letting q = q1 = q2, express the x-
component of the electric field due
to one charge as a function of the
distance r from either charge to the
point of interest:
iE ˆcos2 θr
kq
x =
r
Express for both charges: xE
r
iE ˆcos2 2 θr
kq
x =
r
Substitute for cosθ and r, substitute numerical values, and evaluate to obtain:
( ) ( )( )( )
( ) ( )[ ]
( )i
iiiE
ˆkN/C4.53
ˆ
m0.04m0.03
m0.04nC6/CmN108.992ˆm04.02ˆm04.02 2322
229
32
=
+
⋅×===
r
kq
rr
kq
x
r
(b) Apply to find the force
on a charge q
EF
rr
q=
0 placed on the x axis at
x = 4 cm:
( )( )
( )i
iF
ˆN0.69
ˆkN/C4.53nC2
µ=
=r
*42 ••
Picture the Problem If the electric field at x = 0 is zero, both its x and y components
must be zero. The only way this condition can be satisfied with the point charges of +5.0
µC and −8.0 µC are on the x axis is if the point charge of +6.0 µC is also on the x axis.
Let the subscripts 5, −8, and 6 identify the point charges and their fields. We can use
Coulomb’s law for E
r
due to a point charge and the principle of superposition for fields
to determine where the +6.0 µC charge should be located so that the electric field at x = 0
is zero.
Chapter 21
26
Express the electric field at x = 0 in
terms of the fields due to the charges
of +5.0 µC, −8.0 µC, and +6.0 µC:
( )
0
0 C6C8C5
=
++= − µµµ EEEE
rrrr
Substitute for each of the fields to
obtain:
0ˆˆˆ 82
8
8
62
6
6
52
5
5 =++ −
−
− rrr
r
kq
r
kq
r
kq
or ( ) ( ) 0ˆˆˆ 2
8
8
2
6
6
2
5
5 =−+−+
−
− iii
r
kq
r
kq
r
kq
Divide out the unit vector to
obtain:
iˆ
02
8
8
2
6
6
2
5
5 =−−
−
−
r
q
r
q
r
q
Substitute numerical values to
obtain: ( ) ( ) 0cm4
86
cm3
5
22
6
2 =−−− r
Solve for r6: cm38.26 =r
43 ••
Picture the Problem The diagram shows the electric field vectors at the point of interest
P due to the two charges. We can use Coulomb’s law for E
r
due to point charges and the
superposition principle for electric fields to find PE
r
. We can apply EF
rr
q= to find the
force on an electron at (−1 m, 0).
(a) Express the electric field at
(−1 m, 0) due to the charges q1 and q2:
21P EEE
rrr +=
The Electric Field 1: Discrete Charge Distributions
27
Evaluate : 1E
r
( )( )
( ) ( )
( ) ( )
( ) ( )
( )( )
( ) ( ) ji
ji
jirE
ˆkN/C575.0ˆkN/C44.1
ˆ371.0ˆ928.0N/C1055.1
m2m5
ˆm2ˆm5
m2m5
C5/CmN1099.8ˆ
3
2222
229
P1,2
P1,
1
1
−+=
+−×−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+−
+
−⋅×== µ
r
kqr
Evaluate : 2E
r
( )( )
( ) ( )
( ) ( )
( ) ( )
( )( )
( ) ( ) ji
ji
jirE
ˆkN/C54.9ˆkN/C54.9
ˆ707.0ˆ707.0N/C105.13
m2m2
ˆm2ˆm2
m2m2
C12/CmN1099.8ˆ
3
2222
229
P2,2
P2,
2
2
−+−=
−−×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−+−
+
⋅×== µ
r
kqr
Substitute for and and simplify to find 1E
r
2E
r
PE
r
:
( ) ( ) ( ) ( )
( ) ( ) ji
jijiE
ˆkN/C1.10ˆkN/C10.8
ˆkN/C54.9ˆkN/C54.9ˆkN/C575.0ˆkN/C44.1P
−+−=
−+−+−+=r
The magnitude of is: PE
r
( ) ( )
kN/C9.12
kN/C10.1kN/C8.10 22P
=
−+−=E
The direction of is: PE
r
°=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−= −
231
kN/C8.10
kN/C10.1tan 1Eθ
Note that the angle returned by your
calculator for ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−−
kN/C8.10
kN/C10.1tan 1 is the
reference angle and must be increased by
180° to yield θE.
(b) Express and evaluate the force on an electron at point P:
Chapter 21
28
( ) ( ) ( )[ ]
( ) ( )ji
jiEF
ˆN10.621ˆN10.301
ˆkN/C1.10ˆkN/C10.8C10602.1
1515
19
P
−−
−
×+×=
−+−×−== rr q
Find the magnitude of F
r
: ( ) ( )
N1008.2
N1062.1N1030.1
15
215215
−
−−
×=
×+×=F
Find the direction of F
r
: °=⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
×= −
−
− 3.51
N101.3
N101.62tan 15
15
1
Fθ
44 ••
Picture the Problem The diagram shows the locations of the charges q1 and q2 and the
point on the x axis at which we are to find E
r
. From symmetry considerations we can
conclude that the y component of E
r
at any point on the x axis is zero. We can use
Coulomb’s law for the electric field due to point charges to find the field at any point on
the x axis. We can establish the results called for in parts (b) and (c) by factoring the
radicand and using the approximation 11 ≈+α whenever α << 1.
(a) Express the x-component of the
electric field due to the charges at y
= a and y = −a as a function of the
distance r from either charge to
point P:
iE ˆcos2 2 θr
kq
x =
r
Substitute for cosθ and r to obtain:
( )
( ) i
iiiE
ˆ2
ˆ2ˆ2ˆ2
2322
232232
ax
kqx
ax
kqx
r
kqx
r
x
r
kq
x
+=
+===
r
The Electric Field 1: Discrete Charge Distributions
29
and
( ) 23222 ax kqxEx +=
(b) For a,x << x2 + a2 ≈ a2, so:
( ) 3232 22 akqxakqxEx =≈
For a,x >> x2 + a2 ≈ x2, so:
( ) 2232 22 xkqxkqxEx =≈
(c)
.2by given be wouldfield Its .2 magnitude of
charge single a be appear to wouldby separated charges the, For
2x
kqEq
aax
x =
>>
Factor the radicand to obtain: 23
2
2
2 12
−
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
x
axkqxEx
For a << x:
11 2
2
≈+
x
a
and
[ ] 2232 22 xkqxkqxEx == −
*45 ••
Picture the Problem The diagram shows the electric field vectors at the point of interest
P due to the two charges. We can use Coulomb’s law for E
r
due to point charges and the
superposition principle for electric fields to find PE
r
. We can apply EF
rr
q= to find the
force on a proton at (−3 m, 1 m).
Chapter 21
30
(a) Express the electric field at
(−3 m, 1 m) due to the charges q1 and
q2:
21P EEE
rrr +=
Evaluate : 1E
r
( )( )
( ) ( )
( ) ( )
( ) ( )
( )( ) ( ) ( jiji
jirE
ˆkN/C544.0ˆkN/C908.0ˆ514.0ˆ857.0kN/C06.1
m3m5
ˆm3ˆm5
m3m5
C4/CmN1099.8ˆ
2222
229
P1,2
1.P
1
1
−+=+−−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+−
+
−⋅×== µ
r
kqr
)
Evaluate 2E
r
:
( )( )
( ) ( )
( ) ( )
( ) ( )
( )( ) ( ) ( jiji
jirE
ˆkN/C01.1ˆkN/C01.2ˆ447.0ˆ894.0kN/C25.2
m2m4
ˆm2ˆm4
m2m4
C5/CmN1099.8ˆ
2222
229
P2,2
P2,
2
2
−+−=−−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−+−
+
⋅×== µ
r
kqr
)
Substitute and simplify to find PE
r
:
( ) ( ) ( ) ( )
( ) ( ) ji
jijiE
ˆkN/C55.1ˆkN/C10.1
ˆkN/C01.1ˆkN/C01.2ˆkN/C544.0ˆkN/C908.0P
−+−=
−+−+−+=r
The magnitude of is: PE
r
( ) ( )
kN/C90.1
kN/C55.1kN/C10.1 22P
=
+=E
The Electric Field 1: Discrete Charge Distributions
31
The direction of is: PE
r
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−= − 235
kN/C10.1
kN/C55.1tan 1Eθ
Note that the angle returned by your
calculator for ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−−
kN/C10.1
kN/C55.1tan 1 is the
reference angle and must be increased by
180° to yield θE.
(b) Express and evaluate the force on a proton at point P:
( ) ( ) ( )[ ]
( ) ( )ji
jiEF
ˆN1048.2ˆN10.761
ˆkN/C55.1ˆkN/C10.1C106.1
1616
19
P
−−
−
×−+×−=
−+−×== rr q
The magnitude of F
r
is:
( ) ( ) N1004.3N1048.2N10.761 16216216 −−− ×=×−+×−=F
The direction of F
r
is: °=⎟⎟⎠
⎞
⎜⎜⎝
⎛
×−
×−= −
−
− 235
N1076.1
N1048.2tan 16
16
1
Fθ
where, as noted above, the angle returned
by your calculator for
⎟⎟⎠
⎞
⎜⎜⎝
⎛
×−
×−
−
−
−
N1076.1
N1048.2tan 16
16
1 is the reference
angle and must be increased by 180° to
yield θE.
46 ••
Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due
to equal positive charges located at (0, a) and (0,−a), is given by ( ) .2 2322 −+= axkqxEx We can identify the locations at which Ex has it greatest values
by setting dEx/dx equal to zero.
(a) Evaluate
dx
dEx :
Chapter 21
32
( )[ ] ( )[ ]
( ) ( )
( ) ( ) ( )
( )[ ( ) ]232225222
23222522
23222322
23222322
32
2
2
32
2
22
−−
−−
−−
−−
+++−=
⎥⎦
⎤⎢⎣
⎡ +++⎟⎠
⎞⎜⎝
⎛−=
⎥⎦
⎤⎢⎣
⎡ +++=
+=+=
axaxxkq
axxaxxkq
axax
dx
dxkq
axx
dx
dkqaxkqx
dx
d
dx
dEx
Set this derivative equal to zero:
( ) ( ) 03 232225222 =+++− −− axaxx
Solve for x to obtain:
2
ax ±=
(b) The following graph was plotted using a spreadsheet program:
2kq = 1 and a = 1
-0.4
-0.2
0.0
0.2
0.4
-10 -5 0 5 10
x
E x
47 •••
Picture the Problem We can determine the stability of the equilibrium in Part (a) and
Part (b) by considering the forces the equal charges q at y = +a and y = −a exert on the
test charge when it is given a small displacement along either the x or y axis. The
application of Coulomb’s law in Part (c) will lead to the magnitude and sign of the charge
that must be placed at the origin in order that a net force of zero is experienced by each of
the three charges.
(a) Because Ex is in the x direction, a positive test charge that is displaced from
The Electric Field 1: Discrete Charge Distributions
33
(0, 0) in either the +x direction or the −x direction will experience a force pointing away
from the origin and accelerate in the direction of the force.
axis. the
alongnt displaceme small afor unstable is (0,0)at mequilibriu thely,Consequent
x
If the positive test charge is displaced in the direction of increasing y (the positive y
direction), the charge at y = +a will exert a greater force than the charge at
y = −a, and the net force is then in the −y direction; i.e., it is a restoring force. Similarly,
if the positive test charge is displaced in the direction of decreasing y (the negative y
direction), the charge at y = −a will exert a greater force than the charge at y = −a, and the
net force is then in the −y direction; i.e., it is a restoring force.
axis. the
alongnt displaceme small afor stable is (0,0)at mequilibriu thely,Consequent
y
(b)
axis. thealong ntsdisplacemefor unstable and
axis thealong ntsdisplacemefor (0,0)at stable is mequilibriu thecharge,
testnegative afor that,finds one ),(Part in as arguments same theFollowing
y
x
a
(c) Express the net force acting on the
charge at y = +a: ( ) 02 2
2
2
0
at =+=∑ += akqakqqF ayq
Solve for q0 to obtain: 0410 qq −=
Remarks: In Part (c), we could just as well have expressed the net force acting on
the charge at y = −a. Due to the symmetric distribution of the charges at y = −a and y
= +a, summing the forces acting on q0 at the origin does not lead to a relationship
between q0 and q.
*48 •••
Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due to
equal positive charges located at (0, a) and (0,−a), is given by ( ) .2 2322 −+= axkqxEx We can use k'mT π2= to express the period of the motion in
terms of the restoring constant k′.
(a) Express the force acting on the on
the bead when its displacement from
the origin is x:
( ) 2322
22
ax
xkqqEF xx +−=−=
Chapter 21
34
Factor a2 from the denominator to
obtain:
23
2
2
2
2
1
2
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−=
a
xa
xkqFx
For x << a:
x
a
kqFx 3
22−=
i.e., the bead experiences a linear restoring
force.
(b) Express the period of a simple
harmonic oscillator:
k'
mT π2=
Obtain k′ from our result in part (a):
3
22
a
kqk' =
Substitute to obtain:
2
3
3
2 2
2
2
2
kq
ma
a
kq
mT ππ ==
Motion of Point Charges in Electric Fields
49 •
Picture the Problem We can use Newton’s 2nd law of motion to find the acceleration of
the electron in the uniform electric field and constant-acceleration equations to find the
time required for it to reach a speed of 0.01c and the distance it travels while acquiring
this speed.
(a) Use data found at the back of
your text to compute e/m for an
electron:
C/kg1076.1
kg109.11
C106.1
11
31
19
×=
×
×= −
−
em
e
(b) Apply Newton’s 2nd law to relate
the acceleration of the electron to
the electric field:
ee m
eE
m
Fa == net
Substitute numerical values and
evaluate a:
( )( )
213
31
19
m/s1076.1
kg109.11
N/C100C101.6
×=
×
×= −
−
a
The Electric Field 1: Discrete Charge Distributions
35
field. electric
theopposite iselectron an of
onaccelerati theofdirection The
(c) Using the definition of
acceleration, relate the time required
for an electron to reach 0.01c to its
acceleration:
a
c
a
vt 01.0==∆
Substitute numerical values and
evaluate ∆t:
( ) s170.0
m/s101.76
m/s1030.01
213
8
µ=×
×=∆t
(d) Find the distance the electron
travels from its average speed and
the elapsed time:
( )[ ]( )
cm5.25
s170.0m/s10301.00 821
av
=
×+=
∆=∆
µ
tvx
*50 •
Picture the Problem We can use Newton’s 2nd law of motion to find the acceleration of
the proton in the uniform electric field and constant-acceleration equations to find the
time required for it to reach a speed of 0.01c and the distance it travels while acquiring
this speed.
(a) Use data found at the back of
your text to compute e/m for an
electron:
C/kg1058.9
kg1067.1
C106.1
7
27
19
×=
×
×= −
−
pm
e
Apply Newton’s 2nd law to relate the
acceleration of the electron to the
electric field:
pp m
eE
m
Fa == net
Substitute numerical values and
evaluate a:
( )( )
29
72
19
m/s1058.9
kg1067.1
N/C100C101.6
×=
×
×= −
−
a
field. electric the
ofdirection in the isproton a of
onaccelerati theofdirection The
Chapter 21
36
(b) Using the definition of
acceleration, relate the time required
for an electron to reach 0.01c to its
acceleration:
a
c
a
vt 01.0==∆
Substitute numerical values and
evaluate ∆t:
( ) s313
m/s1058.9
m/s1030.01
29
8
µ=×
×=∆t
51 •
Picture the Problem The electric force acting on the electron is opposite the direction of
the electric field. We can apply Newton’s 2nd law to find the electron’s acceleration and
use constant acceleration equations to find how long it takes the electron to travel a given
distance and its deflection during this interval of time.
(a) Use Newton’s 2nd law to relate the
acceleration of the electron first to the
net force acting on it and then the
electric field in which it finds itself:
ee m
e
m
EFa
rr
r −== net
Substitute numerical values and
evaluate a : r ( )
( )j
ja
ˆm/s1003.7
ˆN/C400
kg109.11
C101.6
213
31
19
×−=
×
×−= −
−r
(b) Relate the time to travel a given
distance in the x direction to the
electron’s speed in the x direction:
ns0.50
m/s102
m0.1
6 =×=
∆=∆
xv
xt
(c) Using a constant-acceleration
equation, relate the displacement of
the electron to its acceleration and
the elapsed time:
( )
( )( )
( ) j
j
ay
ˆcm79.8
ˆns50m/s1003.7 221321
2
2
1
−=
×−=
∆=∆ tyrr
i.e., the electron is deflected 8.79 cm
downward.
52 ••
Picture the Problem Because the electric field is uniform, the acceleration of the
electron will be constant and we can apply Newton’s 2nd law to find its acceleration and
use a constant-acceleration equation to find its speed as it leaves the region in which
there is a uniform electric field.
Using a constant-acceleration xavv ∆+= 2202
The Electric Field 1: Discrete Charge Distributions
37
equation, relate the speed of the
electron as it leaves the region of the
electric field to its acceleration and
distance of travel:
or, because v0 = 0,
xav ∆= 2
Apply Newton’s 2nd law to express
the acceleration of the electron in
terms of the electric field:
ee m
eE
m
Fa == net
Substitute to obtain:
em
xeEv ∆= 2
Substitute numerical values and evaluate v: ( )( )( ) m/s1075.3
kg109.11
m0.05N/C108C101.62 7
31
419
×=×
××= −
−
v
Remarks: Because this result is approximately 13% of the speed of light, it is only
an approximation.
53 ••
Picture the Problem We can apply the work-kinetic energy theorem to relate the change
in the object’s kinetic energy to the net force acting on it. We can express the net force
acting on the charged body in terms of its charge and the electric field.
Using the work-kinetic energy
theorem, express the kinetic energy
of the object in terms of the net force
acting on it and its displacement:
xFKW ∆=∆= net
Relate the net force acting on the
charged object to the electric field:
QEF =net
Substitute to obtain: xQEKKK ∆=−=∆ if
or, because Ki = 0,
xQEK ∆=f
Solve for Q:
xE
KQ ∆=
f
Chapter 21
38
Substitute numerical values and
evaluate Q: ( )( ) C800m0.50N/C300
J0.12 µ==Q
*54 ••
Picture the Problem We can use constant-acceleration equations to express the x and y
coordinates of the particle in terms of the parameter t and Newton’s 2nd law to express
the constant acceleration in terms of the electric field. Eliminating the parameter will
yield an equation for y as a function of x, q, and m that we can solve for Ey.
Express the x and y coordinates of
the particle as functions of time:
( )tvx θcos=
and ( ) 221sin tatvy y−= θ
Apply Newton’s 2nd law to relate the
acceleration of the particle to the net
force acting on it:
m
qE
m
F
a yy == ynet,
Substitute in the y-coordinate
equation to obtain:
( ) 2
2
sin t
m
qE
tvy y−= θ
Eliminate the parameter t between
the two equations to obtain:
( ) 222 cos2tan xmv
qE
xy y θθ −=
Set y = 0 and solve for Ey:
qx
mvEy
θ2sin2=
Substitute the non-particle specific
data to obtain:
( )
( )
( )
q
m
q
mEy
214
26
m/s1064.5
m015.0
70sinm/s103
×=
°×=
(a) Substitute for the mass and
charge of an electron and evaluate
Ey:
( )
kN/C3.21
C101.6
kg109.11m/s105.64 19
31
214
=
×
××= −
−
yE
(b) Substitute for the mass and
charge of a proton and evaluate Ey:
( )
MN/C89.5
C101.6
kg1067.1m/s105.64 19
72
214
=
×
××= −
−
yE
The Electric Field 1: Discrete Charge Distributions
39
55 ••
Picture the Problem We can use constant-acceleration equations to express the x and y
coordinates of the electron in terms of the parameter t and Newton’s 2nd law to express
the constant acceleration in terms of the electric field. Eliminating the parameter will
yield an equation for y as a function of x, q, and m. We can decide whether the electron
will strike the upper plate by finding the maximum value of its y coordinate. Should we
find that it does not strike the upper plate, we can determine where it strikes the lower
plate by setting y(x) = 0.
Express the x and y coordinates of the
electron as functions of time:
( )tvx θcos0=
and ( ) 2210 sin tatvy y−= θ
Apply Newton’s 2nd law to relate the
acceleration of the electron to the net
force acting on it:
e
y
e
y
y m
eE
m
F
a == net,
Substitute in the y-coordinate equation
to obtain:
( ) 20 2sin tm
eE
tvy
e
y−= θ
Eliminate the parameter t between the
two equations to obtain:
( ) ( ) 222
0 cos2
tan x
vm
eE
xxy
e
y
θθ −= (1)
To find ymax, set dy/dx = 0 for
extrema:
extremafor0
cos
tan 22
0
=
−= x'
vm
eE
dx
dy
e
y
θθ
Solve for x′ to obtain:
y
e
eE
vmx'
2
2sin20 θ= (See remark below.)
Substitute x′ in y(x) and simplify to
obtain ymax:
y
e
eE
vmy
2
sin220
max
θ=
Substitute numerical values and evaluate ymax:
( )( )( )( ) cm02.1N/C103.5C101.62 54sinm/s105kg109.11 319
22631
max =××
°××= −
−
y
and, because the plates are separated by 2 cm, the electron does not strike the upper
plate.
Chapter 21
40
To determine where the electron will
strike the lower plate, set
y = 0 in equation (1) and solve for x to
obtain:
y
e
eE
vmx θ2sin
2
0=
Substitute numerical values and evaluate x:
( )( )( )( ) cm07.4N/C105.3C106.1 90sinm/s105kg1011.9 319
2631
=××
°××= −
−
x
Remarks: x′ is an extremum, i.e., either a maximum or a minimum. To show that it
is a maximum we need to show that d2y/dx2, evaluated at x′, is negative. A simple
alternative is to use your graphing calculator to show that the graph of y(x) is a
maximum at x′. Yet another alternative is to recognize that, because equation (1) is
quadratic and the coefficient of x2 is negative, its graph is a parabola that opens
downward.
56 ••
Picture the Problem The trajectory of the electron while it is in the electric field is
parabolic (its acceleration is downward and constant) and its trajectory, once it is out of
the electric field is, if we ignore the small gravitational force acting on it, linear. We can
use constant-acceleration equations and Newton’s 2nd law to express the electron’s x and
y coordinates parametrically and then eliminate the parameter t to express y(x). We can
find the angle with the horizontal at which the electron leaves the electric field from the x
and y components of its velocity and its total vertical deflection by summing its
deflections over the first 4 cm and the final 12 cm of its flight.
(a) Using a constant-acceleration
equation, express the x and y
coordinates of the electron as
functions of time:
( ) tvtx 0=
and ( ) 221,0 tatvty yy +=
Because v0,y = 0:
( ) tvtx 0= (1)
and ( ) 221 taty y=
Using Newton’s 2nd law, relate the
acceleration of the electron to the
electric field:
e
y
e
y m
eE
m
Fa
−== net
The Electric Field 1: Discrete Charge Distributions
41
Substitute to obtain:
( ) 2
2
t
m
eE
ty
e
y−= (2)
Eliminate the parameter t between
equations (1) and (2) to obtain:
( ) 222
0 42
x
K
eE
x
vm
eE
xy y
e
y −=−=
Substitute numerical values and evaluate y(4 cm):
( ) ( )( )( )( ) mm40.6J1024 m0.04N/C102C101.6m04.0 16
2419
−=×
××−= −
−
y
(b) Express the horizontal and vertical
components of the electron’s speed as
it leaves the electric field:
θcos0vvx =
and
θsin0vvy =
Divide the second of these equations
by the first to obtain:
0
11 tantan
v
v
v
v y
x
y −− ==θ
Using a constant-acceleration
equation, express vy as a function of
the electron’s acceleration and its
time in the electric field:
tavv yyy += ,0
or, because v0,y = 0
0
net,
v
x
m
eE
t
m
F
tav
e
y
e
y
yy −===
Substitute to obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−= −−
K
xeE
vm
xeE y
e
y
2
tantan 12
0
1θ
Substitute numerical values and evaluate θ :
( )( )( )( ) °−=⎥⎦
⎤⎢⎣
⎡
×
××−= −
−
− 7.17
J1022
m0.04N/C102C101.6tan 16
419
1θ
(c) Express the total vertical
displacement of the electron:
cm12cm4total yyy +=
Relate the horizontal and vertical
distances traveled to the screen to
the horizontal and vertical
components of its velocity:
tvx x∆=
and
tvy y∆=
Chapter 21
42
Eliminate ∆t from these equations to
obtain:
( )xx
v
v
y
x
y θtan==
Substitute numerical values and
evaluate y:
( )[ ]( ) cm83.3m12.07.17tan −=°−=y
Substitute for y4 cm and y12 cm and
evaluate ytotal: cm47.4
cm83.3cm640.0total
−=
−−=y
i.e., the electron will strike the fluorescent
screen 4.47 cm below the horizontal axis.
57 •
Picture the Problem We can use its definition to find the dipole moment of this pair of
charges.
(a) Apply the definition of electric
dipole moment to obtain:
Lp
rr q=
and
( )( ) mC1000.8m4pC2 18 ⋅×== −µp
(b) If we assume that the dipole is
oriented as shown to the right, then
is to the right; pointing from the
negative charge toward the positive
charge.
pr
*58 •
Picture the Problem The torque on an electric dipole in an electric field is given by
and the potential energy of the dipole by Epτ
rrr ×= .Ep rr ⋅−=U
Using its definition, express the
torque on a dipole moment in a
uniform electric field:
Epτ
rrr ×=
and
θτ sinpE=
where θ is the angle between the electric
dipole moment and the electric field.
(a) Evaluate τ for θ = 0°: 00sin =°= pEτ
The Electric Field 1: Discrete Charge Distributions
43
(b) Evaluate τ for θ = 90°: ( )( )
mN1020.3
90sinN/C100.4nm5.0
24
4
⋅×=
°×⋅=
−
eτ
(c) Evaluate τ for θ = 30°: ( )( )
mN1060.1
30sinN/C100.4nm5.0
24
4
⋅×=
°×⋅=
−
eτ
(d) Using its definition, express the
potential energy of a dipole in an
electric field:
θcospEU −=⋅−= Ep rr
Evaluate U for θ = 0°: ( )( )
J1020.3
0cosN/C100.4nm5.0
24
4
−×−=
°×⋅−= eU
Evaluate U for θ = 90°: ( )( )
0
90cosN/C100.4nm5.0 4
=
°×⋅−= eU
Evaluate U for θ = 30°: ( )( )
J1077.2
30cosN/C100.4nm5.0
24
4
−×−=
°×⋅−= eU
*59 ••
Picture the Problem We can combine the dimension of an electric field with the
dimension of an electric dipole moment to prove that, in any direction, the dimension of
the far field is proportional to [ ]31 L and, hence, the electric field far from the dipole falls
off as 1/r3.
Express the dimension of an electric
field:
[ ] [ ][ ]2L
kQE =
Express the dimension an electric
dipole moment:
[ ] [ ][ ]LQp =
Write the dimension of charge in
terms of the dimension of an electric
dipole moment:
[ ] [ ][ ]L
pQ =
Substitute to obtain:
[ ] [ ][ ][ ] [ ]
[ ][ ]
[ ]32 L
pk
LL
pkE ==
This shows that the field E due to a dipole
Chapter 21
44
p falls off as 1/r3.
60 ••
Picture the Problem We can use its definition to find the molecule’s dipole moment.
From the symmetry of the system, it is evident that the x component of the dipole
moment is zero.
Using its definition, express the
molecule’s dipole moment:
jip ˆˆ yx pp +=r
From symmetry considerations we
have:
0=xp
The y component of the molecule’s
dipole moment is:
( )( )
mC1086.1
nm0.058C101.62
2
29
19
⋅×=
×=
==
−
−
eLqLpy
Substitute to obtain: ( )jp ˆmC1086.1 29 ⋅×= −r
61 ••
Picture the Problem We can express the net force on the dipole as the sum of the
forces acting on the two charges that constitute the dipole and simplify this expression
to show that We can show that, under the given conditions, is also
given by
.ˆnet iF Cp=
r
netF
r
( ) iˆpdxdEx by differentiating the dipole’s potential energy function with
respect to x.
(a) Express the net force acting on
the dipole:
qq +− += FFF
rrr
net
Apply Coulomb’s law to express the
forces on the two charges:
( )iEF ˆ1 axqCqq −−=−=− rr
and
( )iEF ˆ1 axqCqq +=+=+ rr
Substitute to obtain: ( ) ( )
ii
iiF
ˆˆ2
ˆˆ
11net
CpaqC
axqCaxqC
==
++−−=r
where p = 2aq.
The Electric Field 1: Discrete Charge Distributions
45
(b) Express the net force acting on
the dipole as the spatial derivative of
U:
[ ]
i
iiF
ˆ
ˆˆ
net
dx
dEp
Ep
dx
d
dx
dU
x
x
xx
=
−−=−=r
62 •••
Picture the Problem We can express the force exerted
on the dipole by the electric field
as −dU/dr and the potential energy of the dipole as −pE. Because the field is due to a
point charge, we can use Coulomb’s law to express E. In the second part of the problem,
we can use Newton’s 3rd law to show that the magnitude of the electric field of the dipole
along the line of the dipole a distance r away is approximately 2kp/r3.
(a) Express the force exerted by the
electric field of the point charge on
the dipole:
rF ˆ
dr
dU−=r
where is a unit radial vector pointing
from Q toward the dipole.
rˆ
Express the potential energy of the
dipole in the electric field:
2r
kQppEU −=−=
Substitute to obtain: rrF ˆ2ˆ 32 r
kQp
r
kQp
dr
d −=⎥⎦
⎤⎢⎣
⎡−−=r
(b) Using Newton’s 3rd law, express
the force that the dipole exerts on
the charge Q at the origin:
FF
rr −=Qon or rr ˆˆon FF Q −=
and
FF Q =on
Express in terms of the field in
which Q finds itself:
QFon
QEF Q =on
Substitute to obtain:
3
2
r
kQpQE = ⇒ 32r
kpE =
General Problems
*63 •
Picture the Problem We can equate the gravitational force and the electric force acting
on a proton to find the mass of the proton under the given condition.
(a) Express the condition that must
be satisfied if the net force on the
eg FF =
Chapter 21
46
proton is zero:
Use Newton’s law of gravity and
Coulomb’s law to substitute for Fg
and Fe:
2
2
2
2
r
ke
r
Gm =
Solve for m to obtain:
G
kem =
Substitute numerical values and evaluate m:
( ) kg1086.1
kg/mN1067.6
C/mN1099.8C106.1 92211
229
19 −
−
− ×=⋅×
⋅××=m
(b) Express the ratio of Fe and Fg:
2
p
2
2
2
p
2
2
Gm
ke
r
Gm
r
ke
=
Substitute numerical values to obtain:
( )( )( )( ) 362272211
219229
2
p
2
1024.1
kg1067.1kg/mN1067.6
C106.1C/mN1099.8 ×=×⋅×
×⋅×= −−
−
Gm
ke
64 ••
Picture the Problem The locations of the charges q1, q2 and q2 and the points at which
we are calculate the field are shown in the diagram. From the diagram it is evident that
E
r
along the axis has no y component. We can use Coulomb’s law for E
r
due to a point
charge and the superposition principle to find E
r
at points P1 and P2. Examining the
distribution of the charges we can see that there are two points where E = 0. One is
between q2 and q3 and the other is to the left of q1.
The Electric Field 1: Discrete Charge Distributions
47
Using Coulomb’s law, express the
electric field at P1 due to the three
charges:
i
iii
EEEE
ˆ
ˆˆˆ
2
,3
3
2
,2
2
2
,1
1
2
,3
3
2
,2
2
2
,1
1
111
111
3211
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ++=
++=
++=
PPP
PPP
qqqP
r
q
r
q
r
qk
r
kq
r
kq
r
kq
rrrr
Substitute numerical values and evaluate
1P
E
r
:
( ) ( ) ( ) ( )
( ) i
iE
ˆN/C1014.1
ˆ
cm2
C5
cm3
C3
cm4
C5/CmN1099.8
8
222
229
1
×=
⎥⎦
⎤⎢⎣
⎡ ++−⋅×= µµµP
r
Express the electric field at P2:
i
EEEE
ˆ
2
,3
3
2
,2
2
2
,1
1
222
3212
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ++=
++=
PPP
qqqP
r
q
r
q
r
qk
rrrr
Substitute numerical values and evaluate
2P
E
r
:
( ) ( ) ( ) ( )
( ) i
iE
ˆN/C1074.1
ˆ
cm14
C5
cm15
C3
cm16
C5/CmN1099.8
6
222
229
2
×=
⎥⎦
⎤+⎢⎣
⎡ +−⋅×= µµµP
r
Letting x represent the x coordinate
of a point where the magnitude of
the electric field is zero, express the
condition that E = 0 for the point
between x = 0 and x = 1 cm:
02
,3
3
2
,2
2
2
,1
1 =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ++=
PPP
P r
q
r
q
r
qkE
or
( ) ( ) 0-cm1
C5C3
cm1
C5
222 =−++
−
xxx
µµµ
Solve this equation to obtain: cm417.0=x
For x < −1 cm, let y = −x to obtain:
( ) ( ) 0cm1
C5C3
cm1
C5
222 =+−−− yyy
µµµ
Solve this equation to obtain: cm95.6=x and cm95.6−=y
Chapter 21
48
65 ••
Picture the Problem The locations of the charges q1, q2 and q2 and the point P2 at which
we are calculate the field are shown in the diagram. The electric field on the x axis due to
the dipole is given by 3dipole 2 xkpE
rr = where ip ˆ2 1aq=r . We can use Coulomb’s law
for E
r
due to a point charge and the superposition principle to find E
r
at point P2.
Express the electric field at P2 as the
sum of the field due to the dipole and
the point charge q2:
( )
i
ii
ii
EEE
ˆ4
ˆˆ22
ˆˆ2
2
1
2
2
2
3
1
2
2
3
dipole 22
⎥⎦
⎤⎢⎣
⎡ +=
+=
+=
+=
q
x
aq
x
k
x
kq
x
aqk
x
kq
x
kp
qP
rrr
where a = 1 cm.
Substitute numerical values and evaluate
2P
E
r
:
( ) ( )( ) ( )iiE ˆN/C1073.1ˆC3cm15 cm1C54m1015 C/mN1099.8 622
229
2
×=⎥⎦
⎤⎢⎣
⎡ +×
⋅×= − µ
µ
P
r
64. Problem ofwith that agreement excellent in isresult this,than
greatermuch not is i.e., interest, ofpoint the todistance theof 10%
thanmore is dipole theof charges two theof separation theWhile
a
x
*66 ••
Picture the Problem We can find the percentage of the free charge that would have to
be removed by finding the ratio of the number of free electrons ne to be removed to give
the penny a charge of 15 µC to the number of free electrons in the penny. Because we’re
assuming the pennies to be point charges, we can use Coulomb’s law to find the force of
repulsion between them.
(a) Express the fraction f of the free
charge to be removed as the quotient
of the number of electrons to be
removed and the number of free
N
nf e=
The Electric Field 1: Discrete Charge Distributions
49
electrons:
Relate N to Avogadro’s number, the
mass of the copper penny, and the
molecular mass of copper:
M
m
N
N =
A
⇒
M
mNN A=
Relate ne to the free charge Q to be
removed from the penny:
[ ]enQ −= e ⇒ e
Qn −=e
A
A
meN
QM
M
mN
e
Q
f −=−=
Substitute numerical values and evaluate f:
( )( )
( )( )( ) %1029.31029.3mol106.02C101.6g3 g/mol5.63C15 7912319 −−−− ×=×=××−−= µf
(b) Use Coulomb’s law to express the
force of repulsion between the two
pennies:
( )
2
2
e
2
2
r
enk
r
kqF ==
Substitute numerical values and evaluate F:
( )( ) ( )
( ) N4.32m25.0
C106.11038.9/CmN1099.8
2
219213229
=××⋅×=
−
F
67 ••
Picture the Problem Knowing the total charge of the two charges, we can use
Coulomb’s law to find the two combinations of charge that will satisfy the condition that
both are positive and hence repel each other. If just one charge is positive, then there is
just one distribution of charge that will satisfy the conditions that the force is attractive
and the sum of the two charges is 6 µC.
(a) Use Coulomb’s law to express
the repulsive force each charge
exerts on the other:
2
2,1
21
r
qkqF =
Express q2 in terms of the total
charge and q1:
12 qQq −=
Chapter 21
50
Substitute to obtain:
( )
2
2,1
11
r
qQkqF −=
Substitute numerical values to obtain:
( ) ( )[ ]
( )2
2
11
229
m3
C6/CmN108.99mN8 qq −⋅×= µ
Simplify to obtain:
( ) ( ) 0C01.8C6 2121 =+−+ µµ qq
Solve to obtain:
C01.2andC99.3 21 µµ == qq
or
C99.3andC01.2 21 µµ == qq
(b) Use Coulomb’s law to express
the attractive force each charge
exerts on the other:
2
2,1
21
r
qkqF −=
Proceed as in (a) to obtain: ( ) ( ) 0C01.8C6 2121 =−−+ µµ qq
Solve
to obtain:
C12.1andC12.7 21 µµ −== qq
68 ••
Picture the Problem The electrostatic forces between the charges are responsible for the
tensions in the strings. We’ll assume that these are point charges and apply Coulomb’s
law and the principle of the superposition of forces to find the tension in each string.
Use Coulomb’s law to express the
net force on the charge +q:
qq FFT 421 +=
Substitute and simplify to obtain: ( ) ( )
( ) 2
2
221
3
2
42
d
kq
d
qkq
d
qkqT =+=
Use Coulomb’s law to express the
net force on the charge +4q:
qq FFT 22 +=
Substitute and simplify to obtain: ( )( ) ( )
( ) 2
2
222
9
2
442
d
kq
d
qkq
d
qqkT =+=
The Electric Field 1: Discrete Charge Distributions
51
*69 ••
Picture the Problem We can use Coulomb’s law to express the force exerted on one
charge by the other and then set the derivative of this expression equal to zero to find the
distribution of the charge that maximizes this force.
Using Coulomb’s law, express the
force that either charge exerts on the
other:
2
21
D
qkqF =
Express q2 in terms of Q and q1: 12 qQq −=
Substitute to obtain: ( )
2
11
D
qQkqF −=
Differentiate F with respect to q1
and set this derivative equal to zero
for extreme values:
( )[ ]
( )[ ]
extremafor0
1 112
11
1
2
1
=
−+−=
−=
qQq
D
k
qQq
dq
d
D
k
dq
dF
Solve for q1 to obtain: Qq 211 =
and
QqQq 2112 =−=
To determine whether a maximum
or a minimum exists at Qq 211 = ,
differentiate F a second time and
evaluate this derivative at Qq 211 = :
[ ]
( )
. oftly independen 0
2
2
1
2
1
1
22
1
2
q
D
k
qQ
dq
d
D
k
dq
Fd
<
−=
−=
. maximizes 2121 FQqq ==∴
*70 ••
Picture the Problem We can apply Coulomb’s law and the superposition of forces to
relate the net force acting on the charge q = −2 µC to x. Because Q divides out of our
equation when F(x) = 0, we’ll substitute the data given for
x = 8.0 cm.
Using Coulomb’s law, express the net
force on q as a function of x:
( ) ( )( )22 cm12
4
x
Qkq
x
kqQxF −+−=
Chapter 21
52
Simplify to obtain: ( )
( ) Qxxkq
xF ⎥⎦
⎤⎢⎣
⎡
−+−= 22 cm12
41
Solve for Q:
( )
( ) ⎥⎦
⎤⎢⎣
⎡
−+−
=
22 cm12
41
xx
kq
xFQ
Evaluate Q for x = 8 cm:
( )( ) ( ) ( )
C00.3
cm4
4
cm8
1C2/CmN1099.8
N4.126
22
229
µ
µ
=
⎥⎦
⎤⎢⎣
⎡ +−⋅×
=Q
71 ••
Picture the Problem Knowing the total charge of the two charges, we can use
Coulomb’s law to find the two combinations of charge that will satisfy the condition that
both are positive and hence repel each other. If the spheres attract each other, then there
is just one distribution of charge that will satisfy the conditions that the force is attractive
and the sum of the two charges is 200 µC.
(a) Use Coulomb’s law to express
the repulsive force each charge
exerts on the other:
2
2,1
21
r
qkqF =
Express q2 in terms of the total
charge and q1:
12 qQq −=
Substitute to obtain:
( )
2
2,1
11
r
qQkqF −=
Substitute numerical values to obtain:
( ) ( )[ ]
( )2
2
11
229
m6.0
C200/CmN108.99N80 qq −⋅×= µ
Simplify to obtain the quadratic equation:
( ) ( 0mC1020.3mC2.0 23121 =×+−+ −qq )
Solve to obtain:
C183andC5.17 21 µµ == qq
or
The Electric Field 1: Discrete Charge Distributions
53
C5.17andC183 21 µµ == qq
(b) Use Coulomb’s law to express
the attractive force each charge
exerts on the other:
2
2,1
21
r
qkqF −=
Proceed as in (a) to obtain: ( ) ( 0mC1020.3mC2.0 23121 =×−−+ −qq )
Solve to obtain:
C215andC0.15 21 µµ =−= qq
72 ••
Picture the Problem Choose the
coordinate system shown in the diagram
and let Ug = 0 where y = 0. We’ll let our
system include the ball and the earth. Then
the work done on the ball by the electric
field will change the energy of the system.
The diagram summarizes what we know
about the motion of the ball. We can use
the work-energy theorem to our system to
relate the work done by the electric field to
the change in its energy.
Using the work-energy theorem,
relate the work done by the electric
field to the change in the energy of
the system:
g,1g,212
gfieldelectric
UUKK
UKW
−+−=
∆+∆=
or, because K1 = Ug,2 = 0,
g,12fieldelectric UKW −=
Substitute for Welectric field, K2, and
Ug,0 and simplify: ( ) mghmghghm
mghmvqEh
=−=
−=
2
2
1
2
12
1
2
Solve for m:
g
qEm =
73 ••
Picture the Problem We can use Coulomb’s law, the definition of torque, and the
condition for rotational equilibrium to find the electrostatic force between the two
charged bodies, the torque this force produces about an axis through the center of the
Chapter 21
54
meter stick, and the mass required to maintain equilibrium when it is located either 25 cm
to the right or to the left of the mid-point of the rigid stick.
(a) Using Coulomb’s law, express the
electric force between the two charges:
2
21
d
qkqF =
Substitute numerical values and evaluate F:
( )( )
( ) N225.0m1.0
C105C/mN1099.8
2
27229
=×⋅×=
−
F
(b) Apply the definition of torque to
obtain:
lF=τ
Substitute numerical values and
evaluate τ:
( )( )
ckwisecounterclo m,N113.0
m5.0N225.0
⋅=
=τ
(c) Apply 0stickmeter theofcenter =∑τ
to the meterstick:
0=− 'mglτ
Solve for m:
'g
m l
τ=
Substitute numerical values and
evaluate m: ( )( ) kg0461.0m25.0m/s81.9 N113.0 2 ==m
(d) Apply 0stickmeter theofcenter =∑τ
to the meterstick:
0=+− 'mglτ
Substitute for τ: 0=+− 'mgF ll
Substitute for F:
02
21 =+− 'mg
d
'qkq l
where q′ is the required charge.
Solve for q2′ to obtain:
l
l
1
2
2 kq
'mgdq =
Substitute numerical values and evaluate q2′:
( ) ( )( )( )( )( )( ) C1003.5m5.0C105C/mN108.99 m25.0m/s81.9kg0461.0m1.0 77229
22
2
−
− ×=×⋅×='q
The Electric Field 1: Discrete Charge Distributions
55
ic
of forces to express the field at the
rigin and use this equation to solve for Q.
xpress the electric field at the origin due to the point charges Q:
74 ••
Picture the Problem Let the numeral 1 refer to the charge in the 1st quadrant and the
numeral 2 to the charge in the 4th quadrant. We can use Coulomb’s law for the electr
field due to a point charge and the superposition
o
E
( )
( ) ( )[ ] ( ) ( )[ ] ( )
i
ijiji
rrEEE
ˆ
ˆm8ˆm2ˆm4ˆm2ˆm4
ˆˆ0,0
333
0,22
0,2
0,12
0,1
21
xE=
r
kQ
r
kQ
r
kQ
r
kQ
r
kQ
−=+−+−+−=
+=+= rrr
r is the distance from each charge to the origin and where
( )
3
m8
r
kQEx −= .
Express r in terms of the coordinates
, y) of the point charges: (x
22 yxr +=
Substitute to obtain:
( )( ) 2322 m8 yx kQEx +−=
Solve for Q to obtain: ( )
( )m8
2322
k
yxEQ x +=
merical values and
evaluate Q:
Substitute nu ( ) ( ) ( )[ ]( )( )
C97.4
m8/CmN108.99
m2m4kN/C4
229
2322
µ−=
⋅×
+−=Q
75 ••
Picture the Problem Let the numeral 1 denote one of the spheres and the numeral 2 the
other. Knowing the total charge Q on the two spheres, we can use Coulomb’s law to fin
the charge on each of them. A second application of Coulomb’s law when the spheres
d
arry the same charge and are 0.60 m apart will yield the force each exerts on the other.
ss
ach charge
xerts on the other:
c
(a) Use Coulomb’s law to expre
the repulsive force e
e
2
2,1
21
r
qkqF =
Chapter 21
56
q2 in terms of the total charge
nd q1:
Express
a
12 qQq −=
Substitute to obtain:
( )
2
2,1
11
r
qQkqF −=
ubstitute numerical values to obtain:
S
( ) ( )[ ]
( )2
2
11
229
m6.
C200/CmN108.99N201 qq −⋅×= µ
implify to obtain the quadratic equation:
0
S
( ) ( ) 0C4810C200 2121 =+−+ µµ qq
Solve to obtain:
C172andC0.28 21q µµ == q
or
C0.28andC172 21 µµ == qq
ss
arge
when
1 = q2 = 100 µC:
(b) Use Coulomb’s law to expre
the repulsive force each ch
exerts on the other
q
2
2,1
21
r
qkqF =
Substitute numerical values and evaluate F:
( )( )( ) N250m6.0 C100/CmN108.99 2
2
229 =⋅×= µF
76 ••
Picture the Problem Let the numeral 1 denote one of the spheres and the numeral 2 the
other. Knowing the total charge Q on the two spheres, we can use Coulomb’s law to fin
the charge on each of them. A second application of Coulomb’s law when the spheres
d
arry the same charge and are 0.60 m apart will yield the force each exerts on the other.
s
ach charge
xerts on the other:
c
(a) Use Coulomb’s law to expres
the attractive force e
e
2
2,1
21
r
qkqF −=
rms of the total
harge and q1:
Express q2 in te
c
12 qQq −=
The Electric Field 1: Discrete Charge Distributions
57
Substitute to obtain:
( )
2
2,1
11
r
qQkqF −−=
Substitute numerical values to obtain:
( ) ( )[ ]
( )2
2
11
229
m6.0
C200/CmN108.99N201 qq −⋅×−= µ
Simplify to obtain the quadratic equation:
( ) ( ) 0C4810C200 2121 =−−+ µµ qq
Solve to obtain:
C222andC7.21 21 µµ =−= qq
or
C7.21andC222 21 µµ −== qq
(b) Use Coulomb’s law to express
the repulsive force each charge
exerts on the other when
q1 = q2 = 100 µC:
2
2,1
21
r
qkqF =
Substitute numerical values and evaluate F:
( )( )( ) N250m6.0 C100/CmN108.99 2
2
229 =⋅×= µF
77 ••
Picture the Problem The charge configuration is shown in the diagram as are the
approximate locations, labeled x1 and x2, where the electric field is zero. We can
determine the charge Q by using Coulomb’s law and the superposition of forces to
express the net force acting on q2. In part (b), by inspection, the points where
E = 0 must be between the −3 µC and +4 µC charges. We can use Coulomb’s law for the
field due to point charges and the superposition of electric fields to determine the
coordinates x1 and x2.
Chapter 21
58
(a) Use Coulomb’s law to express the
force on the 4.0-µC charge:
( )
ii
ii
FFF
ˆˆ
ˆˆ
22
2,
2
2,1
1
2
2
2,
2
2
2,1
21
2,2,12
F
r
Q
r
qkq
r
kQq
r
qkq
Q
Q
Q
=⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=
−+=
+= rrr
Solve for Q:
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=
2
2
2
2,1
12
2, kq
F
r
qrQ Q
Substitute numerical values and evaluate Q:
( ) ( ) ( )( ) C2.97C4/CmN108.99 N240m0.2 C3m12.0 22922 µµµ −=⎥⎦
⎤
⋅×−⎢⎣
⎡ −=Q
(b) Use Coulomb’s law for electric fields and the superposition of fields to determine the
coordinate x at which E = 0:
( ) ( ) 0ˆˆm2.0ˆm32.0 21222 =+−−−−= iiiE x
kq
x
kq
x
kQr
or
( ) ( ) 0m2.0m32.0 21222 =+−−−− x
q
x
q
x
Q
Substitute numerical values to obtain:
( ) ( ) 0
C3
m2.0
C4
m32.0
C2.97
222 =−+−−−
−−
xxx
µµµ
and
( ) ( ) 0
3
m2.0
4
m32.0
2.97
222 =−−−− xxx
Solve (preferably using a graphing
calculator!) this equation to obtain:
m0508.01 =x and m169.02 =x
The Electric Field 1: Discrete Charge Distributions
59
*78 ••
Picture the Problem Each sphere is in
static equilibrium under the influence of
the tensionT
r
, the gravitational force gF
r
,
and the electric force . We can use
Coulomb’s law to relate the electric force
to the charge on each sphere and their
separation and the conditions for static
equilibrium to relate these forces to the
charge on each sphere.
EF
r
(a) Apply the conditions for static
equilibrium to the charged sphere:
0sinsin 2
2
E =−=−=∑ θθ TrkqTFFx
and
∑ =−= 0cos mgTFy θ
Eliminate T between these equations
to obtain: 2
2
tan
mgr
kq=θ
Solve for q:
k
mgrq θtan=
Referring to the figure, relate the
separation of the spheres r to the
length of the pendulum L:
θsin2Lr =
Substitute to obtain:
k
mgLq θθ tansin2=
(b) Evaluate q for m = 10 g, L = 50 cm, and θ = 10°:
( ) ( )( ) C241.0
/CmN1099.8
10tanm/s81.9kg01.010sinm5.02 229
2
µ=⋅×
°°=q
Chapter 21
60
79 ••
Picture the Problem Each sphere is in
static equilibrium under the influence of
the tensionT
r
, the gravitational force gF
r
,
and the electric force . We can use
Coulomb’s law to relate the electric force
to the charge on each sphere and their
separation and the conditions for static
equilibrium to relate these forces to the
charge on each sphere.
EF
r
(a)Apply the conditions for static
equilibrium to the charged sphere:
0sinsin 2
2
E =−=−=∑ θθ TrkqTFFx
and
0cos =−=∑ mgTFy θ
Eliminate T between these equations
to obtain: 2
2
tan
mgr
kq=θ
Referring to the figure for Problem
80, relate the separation of the spheres
r to the length of the pendulum L:
θsin2Lr =
Substitute to obtain:
θθ 22
2
sin4
tan
mgL
kq=
or
2
2
2
4
tansin
mgL
kq=θθ (1)
Substitute numerical values and evaluate : θθ tansin 2
( )( )
( )( )( ) 322
2229
2 1073.5
m1.5m/s9.81kg0.014
C75.0/CmN1099.8tansin −×=⋅×= µθθ
Because : 1tansin 2 <<θθ
θθθ ≈≈ tansin
and
33 1073.5 −×≈θ
Solve for θ to obtain:
°== 3.10rad179.0θ
The Electric Field 1: Discrete Charge Distributions
61
(b) Evaluate equation (1) with replacing q2 with q1q2:
( )( )( )
( )( )( ) 3322
229
2 1009.5
m1.5m/s9.81kg0.014
C1C5.0/CmN1099.8tansin θµµθθ ≈×=⋅×= −
Solve for θ to obtain:
°== 86.9rad172.0θ
80 ••
Picture the Problem Let the origin be at
the lower left-hand corner and designate
the charges as shown in the diagram. We
can apply Coulomb’s law for point charges
to find the forces exerted on q1 by q2, q3,
and q4 and superimpose these forces to find
the net force exerted on q1. In part (b),
we’ll use Coulomb’s law for the electric
field due to a point charge and the
superposition of fields to find the electric
field at point P(0, L/2).
(a) Using the superposition of forces,
express the net force exerted on q1:
1,41,31,21 FFFF
rrrr ++=
Apply Coulomb’s law to express 1,2F
r
:
( ) ( ) jj
rrF
ˆˆ
ˆ
2
2
3
1,23
1,2
12
1,22
1,2
12
1,2
L
kqL
L
qqk
r
qkq
r
qkq
=−−=
== rr
Apply Coulomb’s law to express 1,4F
r
:
( ) ( ) ii
rrF
ˆˆ
ˆ
2
2
3
1,43
1,4
14
1,42
1,4
14
1,4
L
kqL
L
qqk
r
qkq
r
qkq
=−−=
== rr
Apply Coulomb’s law to express 1,3F
r
:
( )
( )ji
ji
rrF
ˆˆ
2
ˆˆ
2
ˆ
223
2
323
2
1,33
1,3
13
1,32
1,3
13
1,3
+−=
−−=
==
L
kq
LL
L
kq
r
qkq
r
qkq rr
Chapter 21
62
Substitute and simplify to obtain:
( )
( ) ( )
( )ji
jiji
ijijF
ˆˆ
22
11
ˆˆ
2
ˆˆ
ˆˆˆ
2
ˆ
2
2
223
2
2
2
2
2
223
2
2
2
1
+⎟⎠
⎞⎜⎝
⎛ −=
+−+=
++−=
L
kq
L
kq
L
kq
L
kq
L
kq
L
kqr
(b) Using superposition of fields,
express the resultant field at point P:
4321 EEEEE
rrrrr +++=P (1)
Use Coulomb’s law to express 1E
r
:
jj
jrE
ˆ4ˆ
2
2
ˆ
2
ˆ
23
3
,1
,12
,1
1
1
L
kqL
L
kq
L
r
kq
r
kq
P
P
P
=⎟⎠
⎞⎜⎝
⎛
⎟⎠
⎞⎜⎝
⎛
=
⎟⎠
⎞⎜⎝
⎛==r
Use Coulomb’s law to express 2E
r
:
( )
jj
jrE
ˆ4ˆ
2
2
ˆ
2
ˆ
23
3
,2
,22
,2
2
2
L
kqL
L
kq
L
r
qk
r
kq
P
P
P
=⎟⎠
⎞⎜⎝
⎛−
⎟⎠
⎞⎜⎝
⎛
−=
⎟⎠
⎞⎜⎝
⎛−==r
Use Coulomb’s law to express 3E
r
:
⎟⎠
⎞⎜⎝
⎛ −−=
⎟⎠
⎞⎜⎝
⎛ −−==
ji
jirE
ˆ
2
1ˆ
5
8
ˆ
2
ˆˆ
223
3
,3
,32
,3
3
3
L
kq
LL
r
kq
r
kq
P
P
P
r
Use Coulomb’s law to express 4E
r
:
( )
⎟⎠
⎞⎜⎝
⎛ −=
⎟⎠
⎞⎜⎝
⎛ −−==
ji
jirE
ˆ
2
1ˆ
5
8
ˆ
2
ˆˆ
223
3
,4
,32
,4
4
4
L
kq
LL
r
qk
r
kq
P
P
P
r
Substitute in equation (1) and simplify to obtain:
jjijijjE ˆ
25
518ˆ
2
1ˆ
5
8ˆ
2
1ˆ
5
8ˆ4ˆ4
222322322 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=⎟⎠
⎞⎜⎝
⎛ −+⎟⎠
⎞⎜⎝
⎛ −−++=
L
kq
L
kq
L
kq
L
kq
L
kq
P
r
81 ••
The Electric Field 1: Discrete Charge Distributions
63
Picture the Problem We can apply Newton’s 2nd law in rotational form to obtain the
differential equation of motion of the dipole and then use the small angle approximation
sinθ ≈ θ to show that the dipole experiences a linear restoring torque and, hence, will
experience simple harmonic motion.
Apply ∑ to the dipole: = ατ I
2
2
sin
dt
dIpE θθ =−
where τ is negative because acts in such a
direction as to decrease θ.
For small values of θ, sinθ ≈ θ
and: 2
2
dt
dIpE θθ =−
Express the moment of inertia of the
dipole:
2
2
1 maI =
Relate the dipole moment of the
dipole to its charge and the charge
separation:
qap =
Substitute to obtain: θθ qaE
dt
dma −=2
2
2
2
1
or
θθ
ma
qE
dt
d 2
2
2
−=
the differential equation for a simple
harmonic oscillator with angular frequency
maqE2=ω .
Express the period of a simple
harmonic oscillator: ω
π2=T
Substitute to obtain:
qE
maT
2
2π=
82 ••
Picture the Problem We can apply conservation of energy and the definition of the
potential energy of a dipole in an electric field to relate q to the kinetic energy of the
dumbbell when it is aligned with the field.
Chapter 21
64
Using conservation of energy, relate
the initial potential energy of the
dumbbell to its kinetic energy when
it is momentarily aligned with the
electric field:
0=∆+∆ UK
or, because Ki = 0,
0=∆+ UK
where K is the kinetic energy when it is
aligned with the field.
Express the change in the potential
energy of the dumbbell as it aligns
with the electric field in terms of its
dipole moment, the electric field,
and the angle through which it
rotates:
( )160cos
coscos ff
if
−°=
+−=
−=∆
qaE
pEpE
UUU
θθ
Substitute to obtain: ( ) 0160cos =−°+ qaEK
Solve for q: ( )°−= 60cos1aE
Kq
Substitute numerical values and evaluate q:
( )( )( )
C55.6
60cos1N/C600m0.3
J105 3
µ=
°−
×=
−
q
*83 ••
Picture the Problem The forces the electron and the proton exert on each other
constitute an action-and-reaction pair. Because the magnitudes of their charges are equal
and their masses are the same, we find the speed of each particle by finding the speed of
either one. We’ll apply Coulomb’s force law for point charges and Newton’s 2nd law to
relate v to e, m, k, and r.
Apply Newton’s 2nd law to the positron:
r
vm
r
ke
2
1
2
2
2
= ⇒ 2
2
2mv
r
ke =
Solve for v to obtain:
mr
kev
2
2
=
84 ••
Picture the Problem In Problem 81 it was established that the period of an electric
dipole in an electric field is given by .22 qEmaT π= We can use this result to find
the frequency of oscillation of a KBr molecule in a uniform electric field of 1000 N/C.
The Electric Field 1: Discrete Charge Distributions
65
Express the frequency of the KBr
oscillator:
ma
qEf 2
2
1
π=
Substitute numerical values and
evaluate f:
( )( )( )( )
Hz1053.4
nm0.282kg101.4
N/C1000C101.62
2
1
8
25
19
×=
×
×= −
−
πf
85 •••
Picture the Problem We can use Coulomb’s force law for point masses and the
condition for translational equilibrium to express the equilibrium position as a function
of k, q, Q, m, and g. In part (b) we’ll need to show that the displaced point charge
experiences a linear restoring force and, hence, will exhibit simple harmonic motion.
(a) Apply the condition for
translational equilibrium to the point
mass:
02
0
=−mg
y
kqQ
Solve for y0 to obtain:
mg
kqQy =0
(b) Express the restoring force that
acts on the point mass when it is
displaced a distance ∆y from its
equilibrium position:
( )
2
00
2
0
2
0
2
0
2 y
kqQ
yyy
kqQ
y
kqQ
yy
kqQF
−∆+≈
−∆+=
because ∆y << y0.
Simplify this expression further by
writing it with a common
denominator:
3
0
0
4
0
0
3
0
4
0
0
2
21
2
2
2
y
ykqQ
y
yy
ykqQy
yyy
ykqQyF
∆−≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ∆+
∆−=
∆+
∆−=
again, because ∆y << y0.
From the 1st step of our solution: mg
y
kqQ =2
0
Chapter 21
66
Substitute to obtain: y
y
mgF ∆−=
0
2
Apply Newton’s 2nd law to the
displaced point charge to obtain:
y
y
mg
dt
ydm ∆−=∆
0
2
2 2
or
02
0
2
2
=∆+∆ y
y
g
dt
yd
the differential equation of simple
harmonic motion with 02 yg=ω .
86 •••
Picture the Problem The free-body
diagram shows the Coulomb force the
positive charge Q exerts on the bead that is
constrained to move along the x axis. The x
component of this force is a restoring
force, i.e., it is directed toward the bead’s
equilibrium position. We can show that,
for x << L, this restoring force is linear
and, hence, that the bead will exhibit
simple harmonic motion about its
equilibrium position. Once we’ve obtained
the differential equation of SHM we can
relate the period of the motion to its
angular frequency.
Using Coulomb’s law for point
charges, express the force F that +Q
exerts on −q:
( )
2222 xL
kqQ
xL
QqkF +−=+
−=
Express the component of this force
along the x axis:
( ) xxL
kqQ
xL
x
xL
kqQ
xL
kqQFx
2322
2222
22 cos
+−=
++−=
+−= θ
The Electric Field 1: Discrete Charge Distributions
67
Factor L2 from the denominator of this
equation to obtain:
x
L
kqQx
L
xL
kqQFx 323
2
2
3 1
−≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−=
because x << L.
Apply ∑ to the bead to
obtain:
= xx maF
x
L
kqQ
dt
xdm 32
2
−=
or
032
2
=+ x
mL
kqQ
dt
xd
the differential equation of simple
harmonic motion with 3mLkqQ=ω .
Express the period of the motion of
the bead in terms of the angular
frequency of the motion:
kqQ
mLL
kqQ
mLT ππω
π 222
3
===
87 •••
Picture the Problem Each sphere is in
static equilibrium under the influence of
the tensionT
r
, the gravitational force gF
r
,
and the force exerted by the
electric field. We can use Coulomb’s law
to relate the electric force to the charges
on the spheres and their separation and the
conditions for static equilibrium to relate
these forces to the charge on each sphere.
CoulombF
r
EF
r
(a)Apply the conditions for static
equilibrium to the charged sphere:
0sin
sin
2
2
Coulomb
=−=
−=∑
θ
θ
T
r
kq
TFFx
and
0cos =−−=∑ qEmgTFy θ
Eliminate T between these equations
to obtain: ( ) 2
2
tan
rqEmg
kq
+=θ
Referring to the figure for Problem
78, relate the separation of the
θsin2Lr =
Chapter 21
68
spheres r to the length of the
pendulum L:
Substitute to obtain:
( ) θθ 22
2
sin4
tan
LqEmg
kq
+=
or
( ) 2
2
2
4
tansin
LqEmg
kq
+=θθ (1)
Substitute numerical values and
evaluate to obtain: θθ tansin 2
32 1025.3tansin −×=θθ
Because : 1tansin 2 <<θθ
θθθ ≈≈ tansin
and
33 1025.3 −×≈θ
Solve for θ to obtain:
°== 48.8rad148.0θ
(b) The downward electrical forces
acting on the two spheres are no
longer equal. Let the mass of the
sphere carrying the charge of 0.5 µC
be m1, and that of the sphere
carrying the charge of 1.0 µC be m2.
The free-body diagrams show the
tension, gravitational, and electrical
forces acting on each sphere.
Because we already know from part
(a) that the angles are small, we can
use the small-angle approximation
sinθ ≈ tanθ ≈θ.
Apply the conditions for static
equilibrium to the charged sphere
whose mass is m1:
( )
( )
0
sin
sinsin
sin
112
21
2
21
112
21
21
112
21
1,
=
++−≈
++−=
+−=∑
θθθ
θθθ
θ
T
L
qkq
T
LL
qkq
T
r
qkqFx
and
The Electric Field 1: Discrete Charge Distributions
69
∑ =−−= 011,11, EqgmTF yy
Apply the conditions for static
equilibrium to the charged sphere
whose mass is m2:
( )
( )
0
sin
sinsin
sin
222
21
2
21
222
21
21
222
21
2,
=
++≈
++=
−=∑
θθθ
θθθ
θ
T
L
qkq
T
LL
qkq
T
r
qkqFx
and
022,22, =−−=∑ EqgmTF yy
Express θ1 and θ 2 in terms of the
components of T and : 1
r
2T
r
y
x
T
T
,1
,1
1 =θ (1)
and
y
x
T
T
,2
,2
2 =θ (2)
Divide equation (1) by equation (2)
to obtain:
y
y
y
x
y
x
T
T
T
T
T
T
,1
,2
,2
,2
,1
,1
2
1 ==θ
θ
because the horizontal components of and 1T
r
2T
r
are equal.
Substitute for T2,y and T1,y to obtain:
Eqgm
Eqgm
11
22
2
1
+
+=θ
θ
Add equations (1) and (2) to obtain:
( ) ⎥⎦
⎤⎢⎣
⎡
++++=+=+ EqgmEqgmL
qkq
T
T
T
T
y
x
y
x
2211
2
21
2
21
,2
,2
,1
,1
21
11
θθθθ
Solve for θ1 + θ2:
3
2211
2
21
21
11 ⎥⎦
⎤⎢⎣
⎡
+++=+ EqgmEqgmL
qkqθθ
Chapter 21
70
Substitute numerical values and evaluate
1 + θ2 and θ1/θ2:
θ
°==+ 4.16rad287.021 θθ
and
34.1
2
1 =θ
θ
olve for θ1 and θ2 to obtain: °= 42.91θ and °= 98.61θ S
88 •••
Picture the Problem Each sphere is in
static equilibrium under the influence of a
tension, gravitational and Coulomb forc
Let the mass of the sphere carrying the
charge of 2.0 µC be m
e.
C
s
to
e forces to the charges on the
pheres.
1 = 0.01 kg, and that
of the sphere carrying the charge of 1.0 µ
be m2 = 0.02 kg. We can use Coulomb’
law to relate the Coulomb force to the
charge on each sphere and their separation
and the conditions for static equilibrium
relate thes
s
Apply the conditions for static equilibrium
to the charged sphere whose mass is m1:
( )
( )
0
sin
sinsin
sin
112
21
2
21
112
21
21
112
21
1,
=
++−≈
++−=
+−=∑
θθθ
θθθ
θ
T
L
qkq
T
LL
qkq
T
r
qkqFx
and
∑ =−= 01,11, gmTF yy
to the charged sphere whose mass is m2:
Apply the conditions for static equilibrium
( )
( )
0
sin
sinsin
sin
222
21
2
21
222
21
21
222
21
2,
=
++≈
++=
−=∑
θθθ
θθθ
θ
T
L
qkq
T
LL
qkq
T
r
qkqFx
The Electric Field 1: Discrete Charge Distributions
71
and
02,22, =−=∑ gmTF yy
a
ter ponents of and
Using the small-angle approximation
sinθ ≈ tanθ ≈θ, express θ1 nd θ2 in
ms of the com 1T
r
2T
r
:
y
x
T
T
,1
,1
1 =θ (1)
and
y
x
T
T
,2
,2
2 =θ (2)
ation (1) by equation (2)
obtain:
Divide equ
to
y
y
yT ,2
x
y
x
T
T
T
T
T
,1
,2
,2
,1
,1
2
1 ==θ
θ
because the horizontal components of
1T
r
and 2T
r
are equal.
Substitute for T2,y and T1,y to obtain:
1
2
2
1
m
m=θ
θ
Add equations (1) and (2) to obtain:
( ) ⎥⎦
⎤+⎢⎣
⎡
+=
+=+
gmgmL
qkq
T
T
T
T
y
x
y
x
21
2
21
2
21
,2
,2
,1
,1
21
11
θθ
θθ
Solve for θ1 + θ2:
3
21
2
21
21
11 ⎥⎦
⎤⎢⎣
⎡ +=+
gmgmL
qkqθθ
and
valuate θ1 + θ2 and θ1/θ2:
Substitute numerical values
e
°==+ 4.28rad496.021 θθ
and
2
1
2
1 =θ
θ
Solve for θ1 and θ2 to obtain: °= 47.91θ and °= 9.181θ
Remarks: While the small angle approximation is not as good here as it was in the
receding problems, the error introduced is less than 3%.
p
Chapter 21
72
89 •••
Picture the Problem We can find the effective value of the gravitational field by finding
the force on the bob due to and gr E
r
and equating this sum to the product of the mass of
the bob and . We can then solve this equation for 'gr E
r
in terms of gr , 'gr , q, and M and
use the equation for the period of a simple pendulum to find the magnitude of 'gr
Express the force on the bob due to
and gr E
r
:
'M
M
qMqM gEgEgF r
rrrrr =⎟⎠
⎞⎜⎝
⎛ +=+=
where
Egg
rrr
M
q' +=
Solve for E
r
to obtain:
( )ggE rrr −= '
q
M
Using the expression for the period
of a simple pendulum, find the
magnitude of g′:
g'
LT' π2=
and ( )
( ) 22
2
2
2
m/s4.27
s1.2
m144 === ππ
T
Lg'
Substitute numerical values and evaluate E
r
:
( ) ( )[ ] ( )jjjE ˆN/C1010.1ˆm/s81.9ˆm/s4.27
C8.0
kg105 4223 ×−=−−
×=
−
µ
r
*90 •••
Picture the Problem We can relate the force of attraction that each molecule exerts on
the other to the potential energy function of either molecule using .dxdUF −= We can
relate U to the electric field at either molecule due to the presence of the other through U
= −pE. Finally, the electric field at either molecule is given by .2 3xkpE =
Express the force of attraction
between the dipoles in terms of the
spatial derivative of the potential
energy function of p1:
dx
dUF 1−= (1)
Express the potential energy of the
dipole p1:
111 EpU −=
where E1 is the field at p1 due to p2.
The Electric Field 1: Discrete Charge Distributions
73
Express the electric field at p1 due to
p2:
3
2
1
2
x
kpE =
where x is the separation of the dipoles.
Substitute to obtain:
3
21
1
2
x
pkpU −=
Substitute in equation (1) and
differentiate with respect to x: 4
21
3
21 62
x
pkp
x
pkp
dx
dF =⎥⎦
⎤⎢⎣
⎡−−=
Evaluate F for p1 = p2 = p and
x = d to obtain: 4
26
d
kpF =
91 •••
Picture the Problem We can use Coulomb’s law for the electric field due to a point
charge and superposition
of fields to find the electric field at any point on the y axis. By
applying Newton’s 2nd law, with the charge on the ring negative, we can show that the
ring experiences a linear restoring force and, therefore, will execute simple harmonic
motion. We can find ω from the differential equation of motion and use f = ω/2π to find
the frequency of the motion.
(a) Use Coulomb’s law for the electric field due to a point charge and
superposition of fields, express the field at point P on the y axis:
( ) ( )
( ) j
jiji
rrrrEEE
ˆ2
ˆˆ
2
ˆˆ
2
ˆˆ
2322
23222322
,23
,2
,13
,1
,22
,2
2
,12
,1
1
21
ya
kQy
yL
ya
kQyL
ya
kQ
r
kQ
r
kQ
r
kq
r
kq
P
P
P
P
P
P
P
P
P
+=
⎟⎠
⎞⎜⎝
⎛ +−++⎟⎠
⎞⎜⎝
⎛ ++=
+=+=+= rrrrr
where a = L/2.
(b) Relate the force on the charged
ring to its charge and the electric
field:
( ) jEF ˆ2 2322 ya kqQyq yy +==
rr
where q must be negative if yF
r
is to be a
restoring force.
Chapter 21
74
) Apply Newton’s 2nd law to the
ring to obtain:
(c
( ) yya kqQdt ydm 23222
2 2
+−=
or
( ) yyam kqQdt yd 23222
2 2
+−=
Factor the radicand to obtain:
y
mL
kqQy
ma
kqQ 162
y
a
yma
kqQ
dt
yd
33
23
2
2
3
2
2
1
2
−=−≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−=
provided y << a = L .
Thus we have:
/2
y
mL
kqQ
dt
yd 2
32
16−=
or
016 32
2
+ y
mL
kqQ
dt
yd =
simple
harmonic motion.
cy of the simple
the differential equation of
Express the frequen
harmonic motion in terms of its
angular frequency:
π
ω
2
=f
From the differential equation
describing the motion we have: 3
2 16
mL
kqQ=ω
and
3
16
2
1
mL
kqQf π=
Substitute numerical values and evaluate f:
( )( )( )
( )( ) Hz37.9m0.24kg0.03
C2C5/CmN1099.816
2
1
3
229
=⋅×= µµπf
The Electric Field 1: Discrete Charge Distributions
75
92 •••
Picture the Problem The free body
diagram shows the forces acting on the
microsphere of mass m and having an
excess charge of q = Ne when the electric
field is downward. Under terminal-speed
conditions the sphere is in equilibrium
under the influence of the electric force eF
r
,
its weight ,m and the drag force gr .dF
r
We
can apply Newton’s 2nd law, under
terminal-speed conditions, to relate the
number of excess charges N on the sphere
to its mass and, using Stokes’ law, find its
terminal speed.
(a) Apply to the
microsphere:
∑ = yy maF ymaFmgF =−− de
or, because ay = 0,
0terminald,e =−− FmgF
Substitute for Fe, m, and Fd,terminal to
obtain:
06 t =−− rvVgqE πηρ
or, because q = Ne,
06 t
3
3
4 =−− rvgrNeE πηρπ
Solve for N to obtain:
eE
rvgrN t
3
3
4 6πηρπ +=
Substitute numerical values and
evaluate gr ρπ 334 :
( )
( )(
N1018.7
m/s81.9kg/m1005.1
m105.5
15
233
37
3
43
3
4
−
−
×=
××
×= πρπ gr
)
Substitute numerical values and
evaluate t6 rvπη :
( )( )
( )
N1016.2
m/s1016.1
m105.5sPa108.166
14
4
75
t
−
−
−−
×=
××
×⋅×= ππηrv
Substitute numerical values in
equation (1) and evaluate N: ( )( )
3
V/m106C106.1
N1016.2N1018.7
419
1415
=
××
×+×= −
−−
N
(b) With the field pointing upward,
the electric force is downward and
the application of to ∑ = yy maF
0eterminald, =−− mgFF
or
06 334t =−−− grNeErv ρππη
Chapter 21
76
the bead yields:
Solve for vt to obtain:
r
grNeEv πη
ρπ
6
3
3
4
t
+=
Substitute numerical values and evaluate vt:
( )( ) ( ) ( )( )( )( )
m/s1093.1
m105.5sPa108.16
m/s81.9kg/m1005.1m105.5V/m106C106.13
4
75
23337
3
4419
t
−
−−
−−
×=
×⋅×
××+××= π
πv
*93 •••
Picture the Problem The free body
diagram shows the forces acting on the
microsphere of mass m and having an
excess charge of q = Ne when the electric
field is downward. Under terminal-speed
conditions the sphere is in equilibrium
under the influence of the electric force eF
r
,
its weight ,m and the drag force gr .dF
r
We
can apply Newton’s 2nd law, under
terminal-speed conditions, to relate the
number of excess charges N on the sphere
to its mass and, using Stokes’ law, to its
terminal speed.
(a) Apply to the
microsphere when the electric field is
downward:
∑ = yy maF ymaFmgF =−− de
or, because ay = 0,
0terminald,e =−− FmgF
Substitute for Fe and Fd,terminal to
obtain:
06 u =−− rvmgqE πη
or, because q = Ne,
06 u =−− rvmgNeE πη
Solve for vu to obtain:
r
mgNeEv πη6u
−= (1)
With the field pointing upward, the
electric force is downward and the
application of to the
microsphere yields:
∑ = yy maF
0eterminald, =−− mgFF
or
06 d =−− mgNeErvπη
Solve for vd to obtain:
r
mgNeEv πη6d
+= (2)
The Electric Field 1: Discrete Charge Distributions
77
Add equations (1) and (2) to obtain:
r
qE
r
NeE
r
mgNeE
r
mgNeEvvv
πηπη
πη
πη
33
6
6du
==
++
−=+=
e.microspher
theof mass theknow toneedt don'you that advantage thehas This
(b) Letting ∆v represent the change
in the terminal speed of the
microsphere due to a gain (or loss)
of one electron we have:
NN vvv −=∆ +1
Noting that ∆v will be the same
whether the microsphere is moving
upward or downward, express its
terminal speed when it is moving
upward with N electronic charges on
it:
r
mgNeEvN πη6
−=
Express its terminal speed upward
when it has N + 1 electronic
charges:
( )
r
mgeENvN πη6
1
1
−+=+
Substitute and simplify to obtain: ( )
r
eE
r
mgNeE
r
mgeENvN
πη
πηπη
6
66
1
1
=
−−−+=∆ +
Substitute numerical values and
evaluate ∆v: ( )( )( )( )
m/s1015.5
m105.5mPa108.16
V/m106C106.1
5
75
419
−
−−
−
×=
×⋅×
××=∆ πv
Chapter 21
78
79
Chapter 22
The Electric Field 2: Continuous Charge
Distributions
Conceptual Problems
*1 ••
(a) False. Gauss’s law states that the net flux through any surface is given
by insideS nnet 4 kQdAE πφ == ∫ . While it is true that Gauss’s law is easiest to apply to
symmetric charge distributions, it holds for any surface.
(b) True
2 ••
Determine the Concept Gauss’s law states that the net flux through any surface is given
by insideS nnet 4 kQdAE πφ == ∫ . To use Gauss’s law the system must display some
symmetry.
3 •••
Determine the Concept The electric field is that due to all the charges, inside and
outside the surface. Gauss’s law states that the net flux through any surface is given
by insideS nnet 4 kQdAE πφ == ∫ . The lines of flux through a Gaussian surface begin on
charges on one side of the surface and terminate on charges on the other side of the
surface.
4 ••
Picture the Problem We can show that the charge inside a sphere of radius r is
proportional to r3 and that the area of a sphere is proportional to r2. Using Gauss’s law,
we can show that the field must be proportional to r3/r2 = r.
Use Gauss’s law to express the
electric field inside a spherical
charge distribution of constant
volume charge density:
A
kQE inside4π=
where 24 rA π= .
Express Qinside as a function of ρ and
r:
3
3
4
inside rVQ πρρ ==
Substitute to obtain: rk
r
rkE
3
4
4
4
2
3
3
4 πρ
π
πρπ ==
Chapter 22
80
*5
•
(a) False. Consider a spherical shell, in which there is no charge, in the vicinity of an
infinite sheet of charge. The electric field due to the infinite sheet would be non-zero
everywhere on the spherical surface.
(b) True (assuming there are no charges inside the shell).
(c) True.
(d) False. Consider a spherical conducting shell. Such a surface will have equal charges
on its inner and outer surfaces but, because their areas differ, so will their charge
densities.
6 •
Determine the Concept Yes. The electric field on a closed surface is related to the net
flux through it by Gauss’s law: 0insideS ∈== ∫ QEdAφ . If the net flux through the closed
surface is zero, the net charge inside the surface must be zero by Gauss’s law.
7 •
Determine the Concept The negative point charge at the center of the conducting shell
induces a charge +Q on the inner surface of the shell. correct. is )(a
8 •
Determine the Concept The negative point charge at the center of the conducting shell
induces a charge +Q on the inner surface of the shell. Because a conductor does not have
to be neutral, correct. is )(d
*9 ••
Determine the Concept We can apply Gauss’s law to determine the electric field for
r < R1 and r > R2. We also know that the direction of an electric field at any point is
determined by the direction of the electric force acting on a positively charged object
located at that point.
From the application of Gauss’s law
we know that the electric field in
both of these regions is not zero and
is given by:
2n r
kQE =
A positively charged object placed in either of these regions would experience an
attractive force from the charge –Q located at the center of the shell. correct. is )(b
The Electric Field 2: Continuous Charge Distributions
81
*10 ••
Determine the Concept We can decide what will happen when the conducting shell is
grounded by thinking about the distribution of charge on the shell before it is grounded
and the effect on this distribution of grounding the shell.
The negative point charge at the center of the conducting shell induces a positive charge
on the inner surface of the shell and a negative charge on the outer surface.
Grounding the shell attracts positive charge from ground; resulting in the outer surface
becoming electrically neutral. correct. is )(b
11 ••
Determine the Concept We can apply Gauss’s law to determine the electric field for r <
R1 and r > R2. We also know that the direction of an electric field at any point is
determined by the direction of the electric force acting on a positively charged object
located at that point.
From the application of Gauss’s law we know that the electric field in the region r < R1
is given by 2n r
kQE = . A positively charged object placed in the region r < R1 will
experience an attractive force from the charge –Q located at the center of the shell. With
the conducting shell grounded, the net charge enclosed by a spherical Gaussian surface
of radius r > R2 is zero and hence the electric field in this region is zero.
correct. is )(c
12 ••
Determine the Concept No. The electric field on a closed surface is related to the net
flux through it by Gauss’s law: 0insideS ∈== ∫ QEdAφ . φ can be zero without E being
zero everywhere. If the net flux through the closed surface is zero, the net charge inside
the surface must be zero by Gauss’s law.
13 ••
False. A physical quantity is discontinuous if its value on one side of a boundary differs
from that on the other. We can show that this statement is false by citing a
counterexample. Consider the field of a uniformly charged sphere. ρ is discontinuous at
the surface, E is not.
Estimation and Approximation
*14 ••
Picture the Problem We’ll assume that the total charge is spread out uniformly (charge
density = σ) in a thin layer at the bottom and top of the cloud and that the area of each
Chapter 22
82
surface of the cloud is 1 km2. We can then use the definition of surface charge density
and the expression for the electric field at the surface of a charged plane surface to
estimate the total charge of the cloud.
Express the total charge Q of a
thundercloud in terms of the surface
area A of the cloud and the charge
density σ :
AQ σ=
Express the electric field just outside
the cloud: 0∈
= σE
Solve for σ :
E0=∈σ
Substitute for σ to obtain:
EAQ 0=∈
Substitute numerical values and evaluate Q:
( )( )( ) C6.26km1V/m103mN/C1085.8 262212 =×⋅×= −Q
Remarks: This charge is in reasonably good agreement with the total charge
transferred in a lightning strike of approximately 30 C.
15 ••
Picture the Problem We’ll assume that the field is strong enough to produce a spark.
Then we know that field must be equal to the dielectric strength of air. We can then use
the relationship between the field and the charge density to estimate the latter.
Suppose the field is large enough to
produce a spark. Then:
V/m103 6×≈E
Because rubbing the balloon leaves it
with a surface charge density of +σ
and the hair with a surface charge
density of −σ, the electric field
between the balloon and the hair is:
02∈
= σE
Solve for σ : E02∈=σ
Substitute numerical values and evaluate σ :
( )( ) 2562212 C/m1031.5V/m103mN/C1085.82 −− ×=×⋅×=σ
The Electric Field 2: Continuous Charge Distributions
83
16 •
Picture the Problem For x << r, we can model the disk as an infinite plane. For
x >> r, we can approximate the ring charge by a point charge.
For x << r, express the electric field
near an infinite plane of charge:
σπkEx 2=
(a) and (b) Because Ex is
independent of x for x << r:
( )( )
N/C1003.2
C/m6.3/CmN1099.82
5
2229
×=
⋅×= µπxE
For x >> r, use Coulomb’s law for
the electric field due to a point
charge to obtain:
( ) 2
2
2 x
rk
x
kQxEx
σπ==
(c) Evaluate Ex at x = 5 m:
( ) ( )( ) ( )( ) N/C54.2m5 C/m6.3cm5.2/CmN1099.8m5 2
22229
=⋅×= µπxE
(d) Evaluate Ex at x = 5 cm:
( ) ( )( ) ( )( ) N/C1054.2m05.0 C/m6.3cm5.2/CmN1099.8cm5 42
22229
×=⋅×= µπxE
Note that this is a very poor approximation because x = 2r is not much greater than r.
Calculating E
r
From Coulomb’s Law
*17 •
Picture the Problem We can use the definition of λ to find the total charge of the line of
charge and the expression for the electric field on the axis of a finite line of charge to
evaluate Ex at the given locations along the x axis. In part (d) we can apply Coulomb’s law
for the electric field due to a point charge to approximate the electric field at x = 250 m.
(a) Use the definition of linear
charge density to express Q in terms
of λ:
( )( ) nC17.5m5nC/m3.5 ==
= LQ λ
Express the electric field on the axis
of a finite line charge:
( ) ( )Lxx
kQxEx −= 000
Chapter 22
84
(b) Substitute numerical values and
evaluate Ex at x = 6 m:
( ) ( )( )( )( )
N/C26.2
m5m6m6
nC17.5/CmN108.99m6
229
=
−
⋅×=xE
(c) Substitute numerical values and
evaluate Ex at x = 9 m:
( ) ( )( )( )( )
N/C37.4
m5m9m9
nC17.5/CmN108.99m9
229
=
−
⋅×=xE
(d) Substitute numerical values and evaluate Ex at x = 250 m:
( ) ( )( )( )( ) mN/C57.2m5m502m502 nC17.5/CmN108.99m502
229
=−
⋅×=xE
(e) Use Coulomb’s law for the
electric field due to a point charge to
obtain:
( ) 2x
kQxEx =
Substitute numerical values and evaluate Ex(250 m):
( ) ( )( )( ) mN/C52.2m250 nC17.5/CmN108.99m250 2
229
=⋅×=xE
Note that this result agrees to within 2% with the exact value obtained in (d).
18 •
Picture the Problem
Let the charge
densities on the two plates be σ1 and σ2
and denote the three regions of interest as
1, 2, and 3. Choose a coordinate system in
which the positive x direction is to the
right. We can apply the equation for
E
r
near an infinite plane of charge and the
superposition of fields to find the field in
each of the three regions.
The Electric Field 2: Continuous Charge Distributions
85
(a) Use the equation for E
r
near
an infinite plane of charge to
express the field in region 1
when σ1 = σ2 = +3 µC/m2:
i
ii
EEE
ˆ4
ˆ2ˆ2 21
1 21
σπ
σπσπ
σσ
k
kk
−=
−−=
+= rrr
Substitute numerical values and evaluate :1E
r
( )( ) ( )iiE ˆN/C1039.3ˆC/m3/CmN1099.84 522291 ×−=⋅×−= µπr
Proceed as above for region 2:
0ˆ2ˆ2
ˆ2ˆ2 212 21
=−=
−=+=
ii
iiEEE
σπσπ
σπσπσσ
kk
kk
rrr
Proceed as above for region 3:
( )( )
( )i
i
i
iiEEE
ˆN/C1039.3
ˆC/m3/CmN1099.84
ˆ4
ˆ2ˆ2
5
2229
213 21
×=
⋅×=
=
+=+=
µπ
σπ
σπσπσσ
k
kk
rrr
The electric field lines are shown
to the right:
(b) Use the equation for E
r
near
an infinite plane of charge to
express and evaluate the field in
region 1 when σ1 = +3 µC/m2 and
σ2 = −3 µC/m2:
0ˆ2ˆ2
ˆ2ˆ2 211 21
=−=
−=+=
ii
iiEEE
σπσπ
σπσπσσ
kk
kk
rrr
Proceed as above for region 2:
( )( )
( )i
i
i
iiEEE
ˆN/C1039.3
ˆC3/CmN1099.84
ˆ4
ˆ2ˆ2
5
229
211 21
×=
⋅×=
=
+=+=
µπ
σπ
σπσπσσ
k
kk
rrr
Chapter 22
86
Proceed as above for region 3:
0ˆ2ˆ2
ˆ2ˆ2 213 21
=−=
−=+=
ii
iiEEE
σπσπ
σπσπσσ
kk
kk
rrr
The electric field lines are shown to the
right:
19 •
Picture the Problem The magnitude of the electric field on the axis of a ring of charge
is given by ( ) ( ) 2322 axkQxxEx += where Q is the charge on the ring and a is the
radius of the ring. We can use this relationship to find the electric field on the x axis at
the given distances from the ring.
Express E
r
on the axis of a ring charge:
( ) ( ) 2322 ax kQxxEx +=
(a) Substitute numerical values and evaluate Ex for x = 1.2 cm:
( ) ( )( )( )( ) ( )[ ] N/C1069.4cm5.8cm2.1 cm2.1C75.2/CmN1099.8cm2.1 52322
229
×=
+
⋅×= µxE
(b) Proceed as in (a) with x = 3.6 cm:
( ) ( )( )( )( ) ( )[ ] N/C1013.1cm5.8cm6.3 cm6.3C75.2/CmN1099.8cm6.3 62322
229
×=
+
⋅×= µxE
(c) Proceed as in (a) with x = 4.0 m:
( ) ( )( )( )( ) ( )[ ] N/C1054.1cm5.8m4 m4C75.2/CmN1099.8m4 32322
229
×=
+
⋅×= µxE
(d) Using Coulomb’s law for the
electric field due to a point charge,
express Ex:
( ) 2x
kQxEx =
The Electric Field 2: Continuous Charge Distributions
87
Substitute numerical values and evaluate Ex at x = 4.0 m:
( ) ( )( )( ) N/C1055.1m4 C75.2/CmN1099.8m4 32
229
×=⋅×= µxE
ring. theis than m 4 nearer is chargepoint thebecauselarger slightly
isIt ).(Part in obtainedresult with the1% within toagreesresult This
=x
c
20 •
Picture the Problem We can use ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 Rx
xkxEx σπ , the expression for the
electric field on the axis of a disk charge, to find Ex at x = 0.04 cm and 5 m.
Express the electric field on the axis
of a disk charge:
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 Rx
xkxEx σπ
(a) Evaluate this expression for x = 0.04 cm:
( )( ) ( ) ( )
N/C1000.2
cm5.2cm0.04
cm04.01C/m6.3C/mN1099.82
5
22
2229
×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−⋅×= µπxE
This value is about 1.5% smaller than the approximate value obtained in Problem 9.
(b) Proceed as in (a) for x = 5 m:
( )( ) ( ) ( ) N/C54.2cm2.5m5
m51C/m6.3/CmN1099.82
22
2229 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−⋅×= µπxE
Note that the exact and approximate (from Problem 16) agree to within 1%.
21 •
Picture the Problem We can use the definition of λ to find the total charge of the line of
charge and the expression for the electric field on the perpendicular bisector of a finite
line of charge to evaluate Ey at the given locations along the y axis. In part (e) we can
apply Coulomb’s law for the electric field due to a point charge to approximate the
electric field at y = 4.5 m.
(a) Use the definition of linear ( )( ) nC300.0cm5nC/m6 === LQ λ
Chapter 22
88
charge density to express Q in terms
of λ:
Express the electric field on the
perpendicular bisector of a finite line
charge:
( ) ( ) 2221
2
12
yL
L
y
kyEy +
= λ
(b) Evaluate Ey at y = 4 cm:
( ) ( ) ( )( )( ) ( ) kN/C43.1m04.0m025.0
m05.0nC/m6
m04.0
/CmN1099.82cm4
22
2
1229 =
+
⋅×=yE
(c) Evaluate Ey at y = 12 cm:
( ) ( ) ( )( )( ) ( ) N/C183m12.0m025.0
m05.0nC/m6
m12.0
/CmN1099.82cm12
22
2
1229 =
+
⋅×=yE
(d) Evaluate Ey at y = 4.5 m:
( ) ( ) ( )( )( ) ( ) N/C133.0m5.4m025.0
m05.0nC/m6
m5.4
/CmN1099.82m.54
22
2
1229 =
+
⋅×=yE
(e) Using Coulomb’s law for the electric
field due to a point charge, express Ey:
( ) 2y
kQyEy =
Substitute numerical values and evaluate Ey at y = 4.5 m:
( ) ( )( )( ) N/C133.0m5.4 nC3.0/CmN1099.8m5.4 2
229
=⋅×=yE
This result agrees to three decimal places with the value calculated in Part (d).
22 •
Picture the Problem The electric field on the axis of a disk charge is given by
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 ax
xkqEx π . We can equate this expression and 021 2∈= σxE and
solve for x.
Express the electric field on the axis of a
disk charge: ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 ax
xkqEx π
The Electric Field 2: Continuous Charge Distributions
89
We’re given that: 021 2∈= σxE
Equate these expressions:
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 220
12
4 ax
xkσπε
σ
Simplify to obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 220
12
4 ax
xkσπε
σ
or, because k = 1/4πε0,
22
1
2
1
ax
x
+−=
Solve for x to obtain:
3
ax =
23 •
Picture the Problem We can use ( ) 2322 ax kQxEx += to find the electric field at the given
distances from the center of the charged ring.
(a) Evaluate Ex at x = 0.2a: ( ) ( )( )[ ]
2
2322
189.0
2.0
2.02.0
a
kQ
aa
akQaEx
=
+
=
(b) Evaluate Ex at x = 0.5a: ( ) ( )( )[ ]
2
2322
358.0
5.0
5.05.0
a
kQ
aa
akQaEx
=
+
=
(c) Evaluate Ex at x = 0.7a: ( ) ( )( )[ ]
2
2322
385.0
7.0
7.07.0
a
kQ
aa
akQaEx
=
+
=
(d) Evaluate Ex at x = a: ( ) [ ] 22322 354.0 akQaa kQaaEx =+=
Chapter 22
90
(e) Evaluate Ex at x = 2a: ( ) ( )[ ] 22322 179.02 22 akQaa kQaaEx =+=
The field along the x axis is plotted below. The x coordinates are in units of x/a and E is in
units of kQ/a2.
-0.4
-0.2
0.0
0.2
0.4
-3 -2 -1 0 1 2 3
x /a
E x
24 •
Picture the Problem We can use ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 Rx
xkEx σπ , where R is the radius of
the disk, to find the electric field on the axis of a disk charge.
Express Ex in terms of ε0:
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−∈=
22
0
22
0
1
2
1
4
2
Rx
x
Rx
xEx
σ
π
πσ
(a) Evaluate Ex at x = 0.2a: ( ) ( )
0
22
0
2
804.0
2.0
2.01
2
2.0
∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−∈=
σ
σ
aa
aaEx
The Electric Field 2: Continuous Charge Distributions
91
(b) Evaluate Ex at x = 0.5a: ( ) ( )
0
22
0
2
553.0
5.0
5.01
2
5.0
∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−∈=
σ
σ
aa
aaEx
(c) Evaluate Ex at x = 0.7a: ( ) ( )
0
22
0
2
427.0
7.0
7.01
2
7.0
∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−∈=
σ
σ
aa
aaEx
(d) Evaluate Ex at x = a: ( )
0
22
0
2
293.0
1
2
∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−∈=
σ
σ
aa
aaEx
(e) Evaluate Ex at x = 2a: ( ) ( )
0
22
0
2
106.0
2
21
2
2
∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−∈=
σ
σ
aa
aaEx
The field along the x axis is plotted below. The x coordinates are in units of x/a and E is
in units of .2 0∈σ
0.0
0.4
0.8
1.2
1.6
2.0
-3 -2 -1 0 1 2 3
x/R
E x
Chapter 22
92
*25 ••
Picture the Problem
(a) The electric field on the x axis of
a disk of radius r carrying a surface
charge density σ is given by:
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 rx
xkEx σπ
(b) The electric field due to an
infinite sheet of charge density σ is
independent of the distance from the
plane and is given by:
σπkE 2plate =
A spreadsheet solution is shown below. The formulas used to calculate the quantities in
the columns are as follows:
Cell Content/Formula Algebraic Form
B3 9.00E+09 k
B4 5.00E−10 σ
B5 0.3 r
A8 0 x0
A9 0.01 x0 + 0.01
B8 2*PI()*$B$3*$B$4*(1−A8/
(A8^2+$B$5^2)^2)^0.5) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+− 2212 rx
xkσπ
C8 2*PI()*$B$3*$B$4 σπk2
A B C
1
2
3 k= 9.00E+09 Nm^2/C^2
4 sigma= 5.00E-10 C/m^2
5 r= 0.3 m
6
7 x E(x) E plate
8 0.00 28.27 28.3
9 0.01 27.33 28.3
10 0.02 26.39 28.3
11 0.03 25.46 28.3
12 0.04 24.54 28.3
13 0.05 23.63 28.3
14 0.06 22.73 28.3
15 0.07 21.85 28.3
73 0.65 2.60 28.3
74 0.66 2.53 28.3
75 0.67 2.47 28.3
76 0.68 2.41 28.3
77 0.69 2.34 28.3
The Electric Field 2: Continuous Charge Distributions
93
78 0.70 2.29 28.3
The following graph shows E as a function of x. The electric field from an infinite sheet
with the same charge density is shown for comparison – the magnitude of the electric
fields differ by more than 10 percent for x = 0.03 m.
0
5
10
15
20
25
30
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
x (m)
E
(N
/C
E
E plate
26 ••
Picture the Problem Equation 22-10 expresses the electric field on the axis of a ring
charge as a function of distance along the axis from the center of the ring. We can show
that it has its maximum and minimum values at 2ax += and 2ax −= by setting
its first derivative equal to zero and solving the resulting equation for x. The graph of Ex
will confirm that the maximum and minimum occur at these coordinates.
Express the variation of Ex with x on
the axis of a ring charge:
( ) 2322 ax kQxEx +=
Differentiate this expression with respect to x to obtain:
( )
( ) ( )
( )
( ) ( )( ) ( )( ) ( ) ( )( )322
212222322
322
2122
2
32322
322
23222322
2322
32
ax
axxaxkQ
ax
xaxxaxkQ
ax
ax
dx
dxax
kQ
ax
x
dx
dkQ
dx
dEx
+
+−+=+
+−+=
+
+−+
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
+=
Set this expression equal to zero for
extrema and simplify:
( ) ( )( ) 03 322
212222322
=+
+−+
ax
axxax
,
( ) ( ) 03 212222322 =+−+ axxax ,
Chapter 22
94
and
03 222 =−+ xax
Solve for x to obtain:
2
ax ±=
as our candidates for maxima or minima.
A plot of E, in units of kQ/a2, versus x/a is shown to the right. This graph shows that E is
a minimum at 2ax −= and a maximum at 2ax = .
-0.4
-0.2
0.0
0.2
0.4
-3 -2 -1 0 1 2 3
x/a
E x
27 ••
Picture the Problem The line charge and
point (0, y) are shown in the diagram. Also
shown is a line element of length dx and the
field E
r
d its charge produces at (0, y). We
can find dEx from E
r
d and then integrate
from x = x1 to x = x2.
Express the x component of E
r
d :
( ) dxyx xk
dx
yx
x
yx
k
dx
yx
kdEx
2322
2222
22 sin
+−=
++−=
+−=
λ
λ
θλ
The Electric Field 2: Continuous Charge Distributions
95
Integrate from x = x1 to x2 to obtain: ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+++−−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+++−−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+−−=
+−= ∫
22
1
22
2
22
1
22
2
22
2322
11
1
2
1
2
1
yx
y
yx
y
y
k
yxyx
k
yx
k
dx
yx
xkE
x
x
x
x
x
λ
λ
λ
λ
From the diagram we see that:
22
2
2cos
yx
y
+=θ or ⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
y
x21
2 tanθ
and
22
1
1cos
yx
y
+=θ or ⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
y
x11
1 tanθ
Substitute to obtain: [ ]
[ ]12
12
coscos
coscos
θθλ
θθλ
−=
+−−=
y
k
y
kEx
28 ••
Picture the Problem The diagram shows a
segment of the ring of length ds that has a
charge dq = λds. We can express the
electric field E
r
d at the center of the ring
due to the charge dq and then integrate this
expression from θ = 0 to 2π to find the
magnitude of the field in the center of the
ring.
(a) and (b) The field E
r
d at the
center of the ring due to the charge
dq is:
ji
EEE
ˆsinˆcos θθ dEdE
ddd yx
−−=
+= rrr
(1)
The magnitude dE of the field at the
center of the ring is: 2r
kdqdE =
Chapter 22
96
Because dq = λds:
2r
dskdE λ=
The linear charge density varies with
θ according to
λ(θ) = λ0 sin θ :
2
0 sin
r
dskdE θλ=
Substitute rdθ for ds:
r
dk
r
rdkdE θθλθθλ sinsin 020 ==
Substitute for dE in equation (1) to
obtain:
j
iE
ˆsin
ˆcossin
2
0
0
r
dk
r
dkd
θθλ
θθθλ
−
−=r
Integrate E
r
d from θ = 0 to 2π:
j
j
j
iE
ˆ
ˆ0
ˆsin
ˆ2sin
2
0
0
2
0
20
2
0
0
r
k
r
k
d
r
k
d
r
k
λπ
λπ
θθλ
θθλ
π
π
−=
−=
−
−=
∫
∫r
. is magnitude its anddirection negative in the isorigin at the field The 0
r
ky λπ
29 ••
Picture the Problem The line charge and
the point whose coordinates are
(0, y) are shown in the diagram. Also
shown is a segment of the line of length dx.
The field that it produces at (0, y) is .E
r
d
We can find dEy from E
r
d and then
integrate from x = 0 to x = a to find the y
component of the electric field at a point on
the y axis.
(a) Express the magnitude of the
field E
r
d due to charge dq of the 2r
kdqdE =
The Electric Field 2: Continuous Charge Distributions
97
element of length dx:
where 222 yxr +=
Because :dxdq λ=
22 yx
dxkdE +=
λ
Express the y component of dE:
dx
yx
kdEy θλ cos22 +=
Refer to the diagram to express cosθ
in terms of x and y:
22
cos
yx
y
+=θ
Substitute for cosθ in the expression
for dEy to obtain:
( ) dxyx ykdEy 2322 += λ
Integrate from x = 0 to x = a and
simplify to obtain:
( )
22
22
0
222
0
2322
1
ya
a
y
k
yay
ak
yxy
xyk
dx
yx
ykE
a
a
y
+=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+=
+= ∫
λ
λ
λ
λ
*30 •••
Picture the Problem Consider the ring
with its axis along the z direction shown in
the diagram. Its radius is z = rcosθ and its
width is rdθ. We can use the equation for
the field on the axis of a ring charge and
then integrate to express the field at the
center of the hemispherical shell.
Express the field on the axis of the
ring charge:
( )
3
232222 cossin
r
kzdq
rr
kzdqdE
=
+= θθ
where z = rcosθ
Chapter 22
98
Express the charge dq on the ring: ( )
θθπσ
θθπσσ
dr
rdrdAdq
sin2
sin2
2=
==
Substitute to obtain: ( )
θθθσπ
θθπσθ
dk
r
drrkdE
cossin2
sin2cos
3
2
=
=
Integrate dE from θ = 0 to π/2 to obtain:
[ ] σπθσπ
θθθσπ
π
π
kk
dkE
==
= ∫
2
0
2
2
1
2
0
sin2
cossin2
Gauss’s Law
31 •
Picture the Problem The definition of electric flux is ∫ ⋅= S ˆdAnErφ . We can apply this
definition to find the electric flux through the square in its two orientations.
(a) Apply the definition of φ to find
the flux of the field when the square
is parallel to the yz plane:
( ) ( )
( )( ) /CmN0.20m1.0kN/C2
kN/C2ˆˆkN/C2
22
SS
⋅==
=⋅= ∫∫ dAdAiiφ
(b) Proceed as in (a) with °=⋅ 30cosˆˆ ni : ( )
( )
( )( )
/CmN3.17
30cosm1.0kN/C2
30coskN/C2
30coskN/C2
2
2
S
S
⋅=
°=
°=
°=
∫
∫
dA
dAφ
*32 •
Determine the Concept While the number of field lines that we choose to draw radially
outward from q is arbitrary, we must show them originating at q and, in the absence of
other charges, radially symmetric. The number of lines that we draw is, by agreement, in
proportion to the magnitude of q.
The Electric Field 2: Continuous Charge Distributions
99
(a) The sketch of the field lines and of the
sphere is shown in the diagram to the
right.
surface. spherical theenteredhave would6 lines, field 24draw chosen to weHad
sphere. enter the lines 3 , fromdrawn lines field ofnumber Given the q
(b) zero. is surface thecrossinglines ofnumber net The
(c) zero. isflux net The
33 •
Picture the Problem The field at both circular faces of the cylinder is parallel to the
outward vector normal to the surface, so the flux is just EA. There is no flux through the
curved surface because the normal to that surface is perpendicular to .E
r
The net flux
through the closed surface is related to the net charge inside by Gauss’s law.
(a) Use Gauss’s law to calculate the
flux through the right circular
surface:
( ) ( )( )
/CmN51.1
m04.0ˆˆN/C300
ˆ
2
2
rightrightright
⋅=
⋅=
⋅=
π
φ
ii
nE A
r
Apply Gauss’s law to left circular
surface:
( ) ( )( )( )
/CmN51.1
m04.0ˆˆN/C300
ˆ
2
2
leftleftleft
⋅=
−⋅−=
⋅=
π
φ
ii
nE A
r
Chapter 22
100
(b) Because the field lines are
parallel to the curved surface of the
cylinder:
0curved =φ
(c) Express and evaluate the net flux
through the entire cylindrical
surface:
/CmN02.3
0/CmN51.1/CmN51.1
2
22
curvedleftrightnet
⋅=
+⋅+⋅=
++= φφφφ
(d) Apply Gauss’s law to obtain:
insidenet 4 kQπφ =
Solve for Qinside:
k
Q π
φ
4
net
inside =
Substitute numerical values and
evaluate Qinside: ( )
C1067.2
/CmN1099.84
/CmN20.3
11
229
2
inside
−×=
⋅×
⋅= πQ
34 •
Picture the Problem We can use Gauss’s law in terms of ε0 to find the net charge inside
the box.
(a) Apply Gauss’s law in terms of
ε0 to find the net charge inside the
box:
inside
0
net
1 Q∈=φ
or
net0inside φ=∈Q
Substitute numerical values and
evaluate Qinside:
( )( )
C1031.5
/CmkN6m/NC1085.8
8
22212
inside
−
−
×=
⋅⋅×=Q
(b)
box. theinsidepresent charges negative and positive ofnumber
equalan bemay There zero. is chargenet that theconcludeonly can You
35 •
Picture the Problem We can apply Gauss’s law to find the flux of the electric field
through the surface of the sphere.
(a) Use the formula for the surface
area of a sphere to obtain:
( ) 222 m14.3m5.044 === ππrA
The Electric Field 2: Continuous Charge Distributions
101
(b) Apply Coulomb’s law to express
and evaluate E:
( ) ( )
N/C1019.7
m5.0
C2
m/NC1085.84
1
4
1
4
22212
2
0
×=
⋅×=
∈=
−
µ
π
π r
qE
(c) Apply Gauss’s law to obtain:
( )( )
/CmN1026.2
m14.3N/C1019.7
ˆ
25
24
SS
⋅×=
×=
=⋅= ∫∫ EdAdAnErφ
(d)
sphere. theinside located is charge
the whereoft independen is surface gh theflux throu The No.
(e) Because the cube encloses the sphere,
the flux through the surface of the sphere
will also be the flux through the cube:
/CmN1026.2 25cube ⋅×=φ
*36 •
Picture the Problem We’ll define the flux of the gravitational field in a manner that is
analogous to the definition of the flux of the electric field and then substitute for the
gravitational field and evaluate the integral over the closed spherical surface.
Define the gravitational flux as:
∫ ⋅= Sg ˆdAngrφ
Substitute for gr and evaluate the
integral to obtain:
( ) Gmr
r
Gm
dA
r
GmdA
r
Gm
ππ
φ
44
ˆˆ
2
2
S2S 2g
−=⎟⎠
⎞⎜⎝
⎛−=
−=⋅⎟⎠
⎞⎜⎝
⎛−= ∫∫ nr
37 ••
Picture the Problem We’ll let the square be one face of a cube whose side is 40 cm.
Then the charge is at the center of the cube and we can apply Gauss’s law in terms of ε0
to find the flux through the square.
Apply Gauss’s law to the cube to
express the net flux:
inside
0
net
1 Q∈=φ
Chapter 22
102
Express the flux through one face of
the cube: inside0
square 6
1 Q∈=φ
Substitute numerical values and
evaluate φsquare: ( )
/CmN1077.3
m/NC1085.86
C2
24
2212square
⋅×=
⋅×= −
µφ
38 ••
Picture the Problem We can treat this portion of the earth’s atmosphere as though it is a
cylinder with cross-sectional area A and height h. Because the electric flux increases with
altitude, we can conclude that there is charge inside the cylindrical region and use
Gauss’s law to find the charge and hence the charge density of the atmosphere in this
region.
The definition of volume charge
density is:
V
Q=ρ
Express the charge inside a cylinder
of base area A and height h for a
charge density ρ:
AhQ ρ=
Taking upward to be the positive
direction, apply Gauss’s law to the
charge in the cylinder:
( ) ( ) 0000 ∈−=∈−−= AEAEAEAEQ hh
where we’ve taken our zero at 250 m above
the surface of a flat earth.
Substitute to obtain: ( ) ( )
h
EE
Ah
AEAE hh 0000 ∈−=∈−=ρ
Substitute numerical values and evaluate ρ:
( )( ) 3132212 C/m1008.7
m250
m/NC1085.8N/C170N/C150 −− ×−=⋅×−=ρ
where we’ve been able to neglect the curvature of the earth because the maximum height
of 400 m is approximately 0.006% of the radius of the earth.
Spherical Symmetry
39 •
Picture the Problem To find En in these three regions we can choose Gaussian surfaces
of appropriate radii and apply Gauss’s law. On each of these surfaces, Er is constant and
The Electric Field 2: Continuous Charge Distributions
103
Gauss’s law relates Er to the total charge inside the surface.
(a) Use Gauss’s law to find the
electric field in the region r < R1:
inside
0
S n
1 QdAE ∈=∫
and
0
0
inside
1
=∈=< A
Q
E Rr
because Qinside = 0.
Apply Gauss’s law in the region
R1 < r < R2:
( ) 2120 1411 r
kq
r
qE RrR =∈=<< π
Using Gauss’s law, find the electric
field in the region r > R2:
( ) ( )2 2120 21 42 r
qqk
r
qqE Rr
+=∈
+=> π
(b) Set 0
2
=>RrE to obtain:
021 =+ qq
or
1
2
1 −=
q
q
(c) The electric field lines for the
situation in (b) with q1 positive is shown
to the right.
40 •
Picture the Problem We can use the definition of surface charge density and the formula
for the area of a sphere to find the total charge on the shell. Because the charge is
distributed uniformly over a spherical shell, we can choose a spherical Gaussian surface
and apply Gauss’s law to find the electric field as a function of the distance from the
center of the spherical shell.
(a) Using the definition of surface
charge density, relate the charge on
the
sphere to its area:
( )( )
nC407.0
m06.0nC/m94
4
22
2
=
=
==
π
πσσ rAQ
Chapter 22
104
Apply Gauss’s law to a spherical
surface of radius r that is concentric
the spherical shell to obtain:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n
24 ∈=
QErπ
Solve for En:
2
inside
2
0
inside
n
1
4 r
kQ
r
QE =∈= π
(b) Qinside a sphere whose radius is 2
cm is zero and hence:
( ) 0cm2n =E
(c) Qinside a sphere whose radius is
5.9 cm is zero and hence:
( ) 0cm9.5n =E
(d) Qinside a sphere whose radius is 6.1 cm is 0.407 nC and hence:
( ) ( )( )( ) N/C983m061.0 nC407.0/CmN1099.8cm1.6 2
229
n =⋅×=E
(e) Qinside a sphere whose radius is 10 cm is 0.407 nC and hence:
( ) ( )( )( ) N/C366m1.0 nC407.0/CmN1099.8cm10 2
229
n =⋅×=E
41 ••
Picture the Problem We can use the definition of volume charge density and the
formula for the volume of a sphere to find the total charge of the sphere. Because the
charge is distributed uniformly throughout the sphere, we can choose a spherical
Gaussian surface and apply Gauss’s law to find the electric field as a function of the
distance from the center of the sphere.
(a) Using the definition of volume
charge density, relate the charge on
the sphere to its volume:
( )( )
nC407.0
m06.0nC/m450 3334
3
3
4
=
=
==
π
πρρ rVQ
Apply Gauss’s law to a spherical
surface of radius r < R that is
concentric with the spherical shell to
obtain:
inside
0
S n
1 QdAE ∈=∫
or
The Electric Field 2: Continuous Charge Distributions
105
0
inside
n
24 ∈=
QErπ
Solve for En:
2
inside
2
0
inside
n
1
4 r
kQ
r
QE =∈= π
Because the charge distribution is
uniform, we can find the charge
inside the Gaussian surface by using
the definition of volume charge
density to establish the proportion:
V'
Q
V
Q inside=
where V′ is the volume of the Gaussian
surface.
Solve for Qinside to obtain:
3
3
inside R
rQ
V
V'QQ ==
Substitute to obtain:
( ) r
R
kQ
r
QRrE 32
0
inside
n
1
4
=∈=< π
(b) Evaluate En at r = 2 cm:
( ) ( )( )( ) ( ) N/C339m0.02m0.06 nC0.407/CmN1099.8cm2 3
229
n =⋅×=E
(c) Evaluate En at r = 5.9 cm:
( ) ( )( )( ) ( ) N/C999m0.059m0.06 nC0.407/CmN1099.8cm9.5 3
229
n =⋅×=E
Apply Gauss’s law to the Gaussian
surface with r > R:
0
inside
n
24 επ
QEr =
Solve for En to obtain:
22
inside
n r
kQ
r
kQE ==
(d) Evaluate En at r = 6.1 cm:
( ) ( )( )( ) N/C983m0.061 nC0.407/CmN1099.8cm1.6 2
229
n =⋅×=E
(e) Evaluate En at r = 10 cm:
Chapter 22
106
( ) ( )( )( ) N/C366m0.1 nC0.407/CmN1099.8cm10 2
229
n =⋅×=E
Note that, for r > R, these results are the same as those obtained for in Problem 40 for a
uniform charge distribution on a spherical shell. This agreement is a consequence of the
choices of σ and ρ so that the total charges on the two spheres is the same.
*42 ••
Determine the Concept The charges on a conducting sphere, in response to the repulsive
Coulomb forces each experiences, will separate until electrostatic equilibrium conditions
exit. The use of a wire to connect the two spheres or to ground the outer sphere will cause
additional redistribution of charge.
(a) Because the outer sphere is conducting, the field in the thin shell must vanish.
Therefore, −2Q, uniformly distributed, resides on the inner surface, and −5Q, uniformly
distributed, resides on the outer surface.
(b) Now there is no charge on the inner surface and −5Q on the outer surface of the
spherical shell. The electric field just outside the surface of the inner sphere changes from
a finite value to zero.
(c) In this case, the −5Q is drained off, leaving no charge on the outer surface and −2Q
on the inner surface. The total charge on the outer sphere is then −2Q.
43 ••
Picture the Problem By symmetry; the electric field must be radial. To find Er inside
the sphere we choose a spherical Gaussian surface of radius r < R. On this surface, Er is
constant. Gauss’s law then relates Er to the total charge inside the surface.
Apply Gauss’s law to a spherical
surface of radius r < R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er:
2
inside
2
0
inside
r
1
4 r
kQ
r
QE =∈= π
Use the definition of charge density
to relate Qinside to ρ and the volume
defined by the Gaussian surface:
3
3
4
surfaceGaussianinside rVQ ρπρ ==
The Electric Field 2: Continuous Charge Distributions
107
Substitute to obtain:
( ) kr
r
krRrE ρπρπ 342
3
3
4
r ==<
Substitute numerical values and evaluate Er at r = 0.5R = 0.05 m:
( ) ( )( )( ) N/C77.3m05.0/CmN108.99nC/m2m05.0 229334r =⋅×= πE
44 ••
Picture the Problem We can find the total charge on the sphere by expressing the charge
dq in a spherical shell and integrating this expression between r = 0 and
r = R. By symmetry, the electric fields must be radial. To find Er inside the charged
sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the
charged sphere we choose a spherical Gaussian surface of radius r > R. On each of these
surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface.
(a) Express the charge dq in a shell
of thickness dr and volume 4πr2 dr:
( )
drAr
drArrdrrdq
3
22
4
44
π
πρπ
=
==
Integrate this expression from
r = 0 to R to find the total charge on
the sphere:
[ ] 404
0
34 ARArdrrAQ R
R
πππ === ∫
(b) Apply Gauss’s law to a spherical
surface of radius r > R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er:
( )
2
0
4
2
4
2
inside
2
0
inside
r
4
1
4
r
AR
r
RkA
r
kQ
r
QRrE
∈==
=∈=>
π
π
Apply Gauss’s law to a spherical
surface of radius r < R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Chapter 22
108
Solve for Er: ( )
0
2
0
2
4
0
2
inside
r 444 ∈=∈=∈=<
Ar
r
Ar
r
QRrE π
π
π
The graph of Er versus r/R, with Er in units of A/4∈0, was plotted using a spreadsheet
program.
0.0
0.2
0.4
0.6
0.8
1.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
r/R
E r
Remarks: Note that the results for (a) and (b) agree at r = R.
45 ••
Picture the Problem We can find the total charge on the sphere by expressing the charge
dq in a spherical shell and integrating this expression between r = 0 and r = R. By
symmetry, the electric fields must be radial. To find Er inside the charged sphere we
choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere
we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is
constant. Gauss’s law then relates Er to the total charge inside the surface.
(a) Express the charge dq in a shell
of thickness dr and volume 4πr2 dr:
Brdr
dr
r
Brdrrdq
π
πρπ
4
44 22
=
==
Integrate this expression from
r = 0 to R to find the total charge on
the sphere:
[ ]
2
0
2
0
2
24
BR
BrdrrBQ R
R
π
ππ
=
=== ∫
The Electric Field 2: Continuous Charge Distributions
109
(b) Apply Gauss’s law to a spherical
surface of radius r > R
that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er:
( )
2
0
2
2
2
2
inside
2
0
inside
r
2
2
1
4
r
BR
r
BRk
r
kQ
r
QRrE
∈==
=∈=>
π
π
Apply Gauss’s law to a spherical
surface of radius r < R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er: ( )
0
0
2
2
0
2
inside
r
2
4
2
4
∈=
∈=∈=<
B
r
Br
r
QRrE π
π
π
The graph of Er versus r/R, with Er in units of B/2∈0, was plotted using a spreadsheet
program.
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 0.5 1.0 1.5 2.0 2.5 3.0
r /R
E r
Remarks: Note that our results for (a) and (b) agree at r = R.
Chapter 22
110
*46 ••
Picture the Problem We can find the total charge on the sphere by expressing the charge
dq in a spherical shell and integrating this expression between r = 0 and r = R. By
symmetry, the electric fields must be radial. To find Er inside the charged sphere we
choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere
we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is
constant. Gauss’s law then relates Er to the total charge inside the surface.
(a) Express the charge dq in a shell
of thickness dr and volume 4πr2 dr:
Cdr
dr
r
Crdrrdq
π
πρπ
4
44 2
22
=
==
Integrate this expression from
r = 0 to R to find the total charge on
the sphere:
[ ]
CR
CrdrCQ R
R
π
ππ
4
44 0
0
=
== ∫
(b) Apply Gauss’s law to a spherical
surface of radius r > R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er:
( )
2
0
2
2
inside
2
0
inside
r
4
1
4
r
CR
r
CRk
r
kQ
r
QRrE
∈==
=∈=>
π
π
Apply Gauss’s law to a spherical
surface of radius r < R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er: ( )
r
C
r
Cr
r
QRrE
0
0
2
0
2
inside
r 4
4
4
∈=
∈=∈=< π
π
π
The graph of Er versus r/R, with Er in units of RC 0/∈ , was plotted using a spreadsheet
The Electric Field 2: Continuous Charge Distributions
111
program.
0
2
4
6
8
10
0.0 0.5 1.0 1.5 2.0 2.5 3.0
r/R
E r
47 •••
Picture the Problem By symmetry, the electric fields resulting from this charge
distribution must be radial. To find Er for r < a we choose a spherical Gaussian surface of
radius r < a. To find Er for a < r < b we choose a spherical Gaussian surface of radius a <
r < b. To find Er for r > b we choose a spherical Gaussian surface of radius r > b. On
each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge
inside the surface.
(a), (b) Apply Gauss’s law to a
spherical surface of radius r that is
concentric with the nonconducting
spherical shell to obtain:
inside
0
S r
1 QdAE ∈=∫
or
0
inside
r
24 ∈=
QErπ
Solve for Er:
( ) 2inside2
0
inside
r
1
4 r
kQ
r
QrE =∈= π
Evaluate Er(r < a):
( ) 01
4 2
inside
2
0
inside
r ==∈=< r
kQ
r
QarE π
because ρ(r < a) = 0 and, therefore, Qinside =
0.
Chapter 22
112
Integrate dq from r = a to r to find
the total charge in the spherical shell
in the interval a < r < b:
( )33
3
2
inside
3
4
3
'4''4
ar
CrdrrQ
r
a
r
a
−=
⎥⎦
⎤⎢⎣
⎡== ∫
πρ
ππρ
Evaluate Er(a < r < b):
( )
( )
( )332
0
33
2
2
inside
r
3
3
4
ar
r
ar
r
k
r
kQbraE
−∈=
−=
=<<
ρ
ρπ
For r > b: ( )33inside 34 abQ −= πρ
and
( ) ( )
( )332
0
33
2r
3
3
4
ab
r
ab
r
kbrE
−∈=
−=>
ρ
ρπ
Remarks: Note that E is continuous at r = b.
Cylindrical Symmetry
48 ••
Picture the Problem From symmetry, the field in the tangential direction must vanish.
We can construct a Gaussian surface in the shape of a cylinder of radius r and length L
and apply Gauss’s law to find the electric field as a function of the distance from the
centerline of the infinitely long, uniformly charged cylindrical shell.
Apply Gauss’s law to the cylindrical
surface of radius r and length L that
is concentric with the infinitely long,
uniformly charged cylindrical shell:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
Lr
kQ
rL
QE inside
0
inside
n
2
2
=∈= π
The Electric Field 2: Continuous Charge Distributions
113
For r < R, Qinside = 0 and: ( ) 0n =< RrE
For r > R, Qinside = λL and: ( ) ( )
r
R
r
Rk
r
k
Lr
LkRrE
0
n
2222
∈=
===>
σ
σπλλ
49 ••
Picture the Problem We can use the definition of surface charge density to find the total
charge on the shell. From symmetry, the electric field in the tangential direction must
vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and
length L and apply Gauss’s law to find the electric field as a function of the distance from
the centerline of the uniformly charged cylindrical shell.
(a) Using its definition, relate the
surface charge density to the total
charge on the shell:
σπ
σ
RL
AQ
2=
=
Substitute numerical values and
evaluate Q:
( )( )( )
nC679
nC/m9m200m0.062 2
=
= πQ
(b) From Problem 48 we have, for
r = 2 cm:
( ) 0cm2 =E
(c) From Problem 48 we have, for
r = 5.9 cm:
( ) 0cm9.5 =E
(d) From Problem 48 we have, for r = 6.1 cm:
r
RE
0
r ∈=
σ
and
( ) ( )( )( )( ) kN/C00.1m0.061m/NC108.85 m0.06nC/m9cm1.6 2212
2
=⋅×= −E
(e) From Problem 48 we have, for r = 10 cm:
( ) ( )( )( )( ) N/C610m1.0m/NC108.85 m0.06nC/m9cm10 2212
2
=⋅×= −E
Chapter 22
114
50 ••
Picture the Problem From symmetry, the field tangent to the surface of the cylinder
must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r
and length L and apply Gauss’s law to find the electric field as a function of the distance
from the centerline of the infinitely long nonconducting cylinder.
Apply Gauss’s law to a cylindrical
surface of radius r and length L that
is concentric with the infinitely long
nonconducting cylinder:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
Lr
kQ
rL
QE inside
0
inside
n
2
2
== επ
Express Qinside for r < R: ( ) ( )LrVrQ 20inside πρρ ==
Substitute to obtain: ( ) ( ) r
Lr
LrkRrE
0
0
2
0
n 2
2
∈==<
ρπρ
or, because 2Rρπλ =
( ) r
R
RrE 2
0
n 2 ∈=< π
λ
Express Qinside for r > R: ( ) ( )LRVrQ 20inside πρρ ==
Substitute to obtain: ( ) ( )
r
R
Lr
LRkRrE
0
2
0
2
0
n 2
2
∈==>
ρπρ
or, because 2Rρπλ =
( )
r
RrE
0
n 2 ∈=> π
λ
51 ••
Picture the Problem We can use the definition of volume charge density to find the total
charge on the cylinder. From symmetry, the electric field tangent to the surface of the
cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of
radius
r and length L and apply Gauss’s law to find the electric field as a function of the
distance from the centerline of the uniformly charged cylinder.
The Electric Field 2: Continuous Charge Distributions
115
(a) Use the definition of volume
charge density to express the total
charge of the cylinder:
( )LRVQ 2tot πρρ ==
Substitute numerical values to
obtain:
( )( ) ( )
nC679
m200m0.06nC/m300 23tot
=
= πQ
From Problem 50, for r < R, we
have:
rEr
02∈
= ρ
(b) For r = 2 cm:
( ) ( )( )( ) N/C339m/NC108.852 m0.02nC/m300cm2 2212
3
=⋅×= −rE
(c) For r = 5.9 cm:
( ) ( )( )( ) kN/C00.1m/NC108.852 m0.059nC/m300cm9.5 2212
3
=⋅×= −rE
From Problem 50, for r > R, we have:
r
REr
0
2
2∈=
ρ
(d) For r = 6.1 cm:
( ) ( )( )( )( ) kN/C00.1m061.0m/NC108.852 m06.0nC/m300cm1.6 2212
23
=⋅×= −rE
(e) For r = 10 cm:
( ) ( )( )( )( ) N/C610m1.0m/NC108.852 m06.0nC/m300cm10 2212
23
=⋅×= −rE
Note that, given the choice of charge densities in Problems 49 and 51, the electric fields
for r > R are the same.
*52 ••
Picture the Problem From symmetry; the field tangent to the surfaces of the shells must
vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and
length L and apply Gauss’s law to find the electric field as a function of the distance from
Chapter 22
116
the centerline of the infinitely long, uniformly charged cylindrical shells.
(a) Apply Gauss’s law to the
cylindrical surface of radius r and
length L that is concentric with the
infinitely long, uniformly charged
cylindrical shell:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
Lr
kQE insiden
2= (1)
For r < R1, Qinside = 0 and: ( ) 01n =< RrE
Express Qinside for R1 < r < R2: LRAQ 1111inside 2πσσ ==
Substitute in equation (1) to obtain:
( ) ( )
r
R
Lr
LRkRrRE
0
11
11
21n
22
∈=
=<<
σ
πσ
Express Qinside for r > R2:
LRLR
AAQ
2211
2211inside
22 πσπσ
σσ
+=
+=
Substitute in equation (1) to obtain:
( ) ( )
r
RR
Lr
LRLRkRrE
0
2211
2211
2n
222
∈
+=
+=>
σσ
πσπσ
(b) Set E = 0 for r > R2 to obtain: 0
0
2211 =∈
+
r
RR σσ
or
02211 =+ RR σσ
Solve for the ratio of σ1 to σ2:
1
2
2
1
R
R−=σ
σ
The Electric Field 2: Continuous Charge Distributions
117
Because the electric field is
determined by the charge inside the
Gaussian surface, the field under
these conditions would be as given
above:
( )
r
RRrRE
0
11
21n ∈=<<
σ
(c) Assuming that σ1 is positive, the
field lines would be directed as
shown to the right.
53 ••
Picture the Problem The electric field is directed radially outward. We can construct a
Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s
law to find the electric field as a function of the distance from the centerline of the
infinitely long, uniformly charged cylindrical shell.
(a) Apply Gauss’s law to a
cylindrical surface of radius r and
length L that is concentric with the
inner conductor:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
Lr
kQE insiden
2= (1)
For r < 1.5 cm, Qinside = 0 and: ( ) 0cm5.1n =<rE
Letting R = 1.5 cm, express Qinside
for 1.5 cm < r < 4.5 cm: RL
LQ
πσ
λ
2
inside
=
=
Substitute in equation (1) to obtain:
( ) ( )
r
k
Lr
LkrE
λ
λ
2
2cm5.4cm5.1n
=
=<<
Substitute numerical values and evaluate En(1.5 cm < r < 4.5 cm):
Chapter 22
118
( ) ( ) ( ) ( )
rr
rE m/CN108nC/m6/CmN108.992cm5.4cm5.1 229n
⋅=⋅×=<<
Express Qinside for
4.5 cm < r < 6.5 cm:
0inside =Q
and
( ) 0cm5.6cm5.4n =<< rE
Letting σ2 represent the charge
density on the outer surface, express
Qinside for r > 6.5 cm:
LRAQ 2222inside 2πσσ ==
where R2 = 6.5 cm.
Substitute in equation (1) to obtain:
( ) ( )
r
R
Lr
LRkRrE
0
2222
2n
22
∈==>
σπσ
In (b) we show that σ2 = 21.2 nC/m2. Substitute numerical values to obtain:
( ) ( )( )( ) rrrE m/CN156mN/C1085.8 cm5.6nC/m2.21cm5.6 2212
2
n
⋅=⋅×=> −
(b) The surface charge densities on
the inside and the outside surfaces of
the outer conductor are given by:
1
1 2 Rπ
λσ −= and 12 σσ −=
Substitute numerical values and evaluate σ1
and σ2: ( )
2
1 nC/m2.21m045.02
nC/m6 −=−= πσ
and
2
2 nC/m2.21=σ
54 ••
Picture the Problem From symmetry considerations, we can conclude that the field
tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in
the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric
field as a function of the distance from the centerline of the infinitely long nonconducting
cylinder.
(a) Apply Gauss’s law to a
cylindrical surface of radius r and
length L that is concentric with the
infinitely long nonconducting
cylinder:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
The Electric Field 2: Continuous Charge Distributions
119
because no flux crosses them.
Solve for En:
0
inside
n 2 ∈= rL
QE π (1)
Express dQinside for ρ(r) = ar: ( ) ( )
Ldrar
drrLardVrdQ
2
inside
2
2
π
πρ
=
==
Integrate dQinside from r = 0 to R to
obtain:
3
0
3
0
2
inside
3
2
3
22
RaL
raLdrraLQ
RR
π
ππ
=
⎥⎦
⎤⎢⎣
⎡== ∫
Divide both sides of this equation
by L to obtain an expression for the
charge per unit length λ of the
cylinder:
3
2 3inside aR
L
Q πλ ==
(b) Substitute for Qinside in equation
(1) to obtain: ( ) 2
00
3
n 32
3
2
ra
Lr
raL
RrE ∈=∈=< π
π
For r > R: 3
inside 3
2 RaLQ π=
Substitute for Qinside in equation (1)
to obtain: ( )
0
3
0
3
n 32
3
2
∈=∈=> r
aR
rL
RaL
RrE π
π
55 ••
Picture the Problem From symmetry; the field tangent to the surface of the cylinder
must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r
and length L and apply Gauss’s law to find the electric field as a function of the distance
from the centerline of the infinitely long nonconducting cylinder.
(a) Apply Gauss’s law to a
cylindrical surface of radius r and
length L that is concentric with the
infinitely long nonconducting
cylinder:
inside
0
S n
1 QdAE ∈=∫
or
Chapter 22
120
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
0
inside
n 2 ∈= rL
QE π (1)
Express dQinside for ρ(r) = br2: ( ) ( )
Ldrbr
drrLbrdVrdQ
3
2
inside
2
2
π
πρ
=
==
Integrate dQinside from r = 0 to R to
obtain:
4
0
4
0
3
inside
2
4
22
RbL
rbLdrrbLQ
RR
π
ππ
=
⎥⎦
⎤⎢⎣
⎡== ∫
Divide both sides of this equation
by L to obtain an expression for the
charge per unit length λ of the
cylinder:
2
4
inside bR
L
Q πλ ==
(b) Substitute for Qinside in equation
(1) to obtain: ( ) 3
00
4
n 42
2 rb
Lr
rbL
RrE ∈=∈=< π
π
For r > R: 4
inside 2
RbLQ π=
Substitute
for Qinside in equation (1)
to obtain: ( )
0
4
0
4
n 42
2
∈=∈=> r
bR
rL
RbL
RrE π
π
56 •••
Picture the Problem From symmetry; the field tangent to the surface of the cylinder
must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r
and length L and apply Gauss’s law to find the electric field as a function of the distance
from the centerline of the infinitely long nonconducting cylindrical shell.
Apply Gauss’s law to a cylindrical
surface of radius r and length L that inside0S
n
1 QdAE ∈=∫
The Electric Field 2: Continuous Charge Distributions
121
is concentric with the infinitely
long nonconducting cylindrical
shell:
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
0
inside
n 2 ∈= rL
QE π
For r < a, Qinside = 0: ( ) 0n =< arE
Express Qinside for a < r < b: ( )22
22
inside
arL
LaLrVQ
−=
−==
ρπ
ρπρπρ
Substitute for Qinside to obtain: ( ) ( )
( )
r
ar
Lr
arLbraE
0
22
0
22
n
2
2
∈
−=
∈
−=<<
ρ
π
ρπ
Express Qinside for r > b: ( )22
22
inside
abL
LaLbVQ
−=
−==
ρπ
ρπρπρ
Substitute for Qinside to obtain: ( ) ( )
( )
r
ab
rL
abLbrE
0
22
0
22
n
2
2
∈
−=
∈
−=>
ρ
π
ρπ
57 •••
Picture the Problem We can integrate the density function over the radius of the inner
cylinder to find the charge on it and then calculate the linear charge density from its
definition. To find the electric field for all values of r we can construct a Gaussian surface
in the shape of a cylinder of radius r and length L and apply Gauss’s law to each region of
the cable to find the electric field as a function of the distance from its centerline.
(a) Find the charge Qinner on the
inner cylinder: ( )
CLRdrCL
rLdr
r
CVdrQ
R
RR
ππ
πρ
22
2
0
00
inner
==
==
∫
∫∫
Chapter 22
122
Relate this charge to the linear
charge density:
CR
L
CLR
L
Q ππλ 22innerinner ===
Substitute numerical values and
evaluate λinner:
( )( )
nC/m8.18
m0.015nC/m2002inner
=
= πλ
(b) Apply Gauss’s law to a
cylindrical surface of radius r and
length L that is concentric with the
infinitely long nonconducting
cylinder:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n2 ∈=
QrLEπ
where we’ve neglected the end areas
because no flux crosses them.
Solve for En:
0
inside
n 2 ∈= rL
QE π
Substitute to obtain, for
r < 1.5 cm:
( )
00
n 2
2cm5.1 ∈=∈=<
C
Lr
CLrrE π
π
Substitute numerical values and
evaluate En(r < 1.5 cm):
( )
kN/C22.6
m/NC108.85
nC/m200cm5.1 2212
2
n
=
⋅×=< −rE
Express Qinside for
1.5 cm < r < 4.5 cm:
CLRQ π2inside =
Substitute to obtain, for
1.5 cm < r < 4.5 cm:
( )
r
CR
rL
RLCrE
0
0
n 2
2cm5.4cm5.1
∈=
∈=<< π
π
where R = 1.5 cm.
Substitute numerical values and evaluate En(1.5 cm < r < 4.5 cm):
( ) ( )( )( ) rrrE m/CN339m/NC108.85 m0.015nC/m200cm5.4cm5.1 2212
2
n
⋅=⋅×=<< −
Because the outer cylindrical shell
is a conductor:
( ) 0cm5.6cm5.4n =<< rE
The Electric Field 2: Continuous Charge Distributions
123
For r > 6.5 cm, CLRQ π2inside =
and:
( )
r
rE m/CN339cm5.6n
⋅=>
Charge and Field at Conductor Surfaces
*58 •
Picture the Problem Because the penny is in an external electric field, it will have
charges of opposite signs induced on its faces. The induced charge σ is related to the
electric field by E = σ/ε0. Once we know σ, we can use the definition of surface charge
density to find the total charge on one face of the penny.
(a) Relate the electric field to the
charge density on each face of the
penny:
0∈
= σE
Solve for and evaluate σ: ( )( )
2
2212
0
nC/m2.14
kN/C1.6m/NC108.85
=
⋅×=
=∈
−
Eσ
(b) Use the definition of surface
charge density to obtain:
2r
Q
A
Q
πσ ==
Solve for and evaluate Q: ( )( )
pC4.45
m0.01nC/m14.2 222
=
== πσπrQ
59 •
Picture the Problem Because the metal slab is in an external electric field, it will have
charges of opposite signs induced on its faces. The induced charge σ is related to the
electric field by ./ 0∈=σE
Relate the magnitude of the electric
field to the charge density on the
metal slab:
0∈
= σE
Use its definition to express σ :
2L
Q
A
Q ==σ
Substitute to obtain:
0
2 ∈= L
QE
Chapter 22
124
Substitute numerical values and
evaluate E: ( ) ( )
kN/C9.42
m/NC108.85m0.12
nC1.2
22122
=
⋅×= −E
60 •
Picture the Problem We can apply its definition to find the surface charge density of the
nonconducting material and calculate the electric field at either of its surfaces from
σ/2∈0. When the same charge is placed on a conducting sheet, the charge will distribute
itself until half the charge is on each surface.
(a) Use its definition to find σ : ( ) 22 nC/m150m0.2
nC6 ===
A
Qσ
(b) Relate the electric field on either
side of the sheet to the density of
charge on its surfaces:
( )
kN/C47.8
m/NC108.852
nC/m150
2 2212
2
0
=
⋅×=∈= −
σE
(c) Because the slab is a conductor
the charge will distribute uniformly
on its two surfaces so that:
( ) 22 nC/m0.57m0.22
nC6
2
===
A
Qσ
(d) The electric field just outside the
surface of a conductor is given by:
kN/C47.8
m/NC108.85
nC/m75
2212
2
0
=
⋅×=∈= −
σE
61 •
Picture the Problem We can construct a Gaussian surface in the shape of a sphere of
radius r with the same center as the shell and apply Gauss’s law to find the electric field
as a function of the distance from this point. The inner and outer surfaces of the shell will
have charges induced on them by the charge q at the center of the shell.
(a) Apply Gauss’s law to a spherical
surface of radius r that is concentric
with the point charge:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n
24 ∈=
QErπ
Solve for En:
0
2
inside
n 4 ∈= r
QE π (1)
The Electric Field 2: Continuous Charge Distributions
125
For r < a, Qinside = q. Substitute in
equation (1) and simplify to obtain:
( ) 2
0
2n 4 r
kq
r
qarE =∈=< π
Because the spherical shell is a
conductor, a charge –q will be
induced on its inner surface. Hence,
for a < r < b:
0inside =Q
and
( ) 0n =<< braE
For r > b, Qinside = q. Substitute in
equation (1) and simplify to obtain:
( ) 2
0
2n 4 r
kq
r
qbrE =∈=> π
(b) The electric field lines are shown
in the diagram to the right:
(c) A charge –q is induced on the
inner surface. Use the definition of
surface charge density to obtain:
22inner 44 a
q
a
q
ππσ −=
−=
A charge q is induced on the outer
surface. Use the definition of surface
charge density to obtain:
2outer 4 b
q
πσ =
62 ••
Picture the Problem We can construct a spherical Gaussian surface at the surface of the
earth (we’ll assume the Earth is a sphere) and apply Gauss’s law to relate the electric
field to its total charge.
Apply Gauss’s law to a spherical
surface of radius RE that is
concentric with the earth:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n
2
E4 ∈=
QERπ
Solve for Qinside = Qearth to obtain:
k
ERERQ n
2
E
n
2
E0earth 4 =∈= π
Chapter 22
126
Substitute numerical values and
evaluate Qearth:
( ) ( )
C106.77
/CmN108.99
N/C150m106.37
5
229
26
earth
×=
⋅×
×=Q
*63 ••
Picture the Problem Let the inner and outer radii of the uncharged spherical conducting
shell be a and b and q represent the positive point charge at the center of the shell. The
positive point charge at the center will induce a negative charge on the inner surface of
the shell and, because the shell is uncharged, an equal positive charge will be induced on
its outer surface. To solve part (b), we can construct a Gaussian surface in the shape of a
sphere of radius r with the same center as the shell and apply Gauss’s law to find the
electric field as a function of the distance from this point. In part (c) we can use a similar
strategy with the additional charge placed on the shell.
(a) Express the charge density on
the inner surface:
A
qinner
inner =σ
Express the relationship between the
positive point charge q and the
charge induced on the inner surface
qinner:
0inner =+ qq
Substitute for qinner to obtain:
2inner 4 a
q
πσ
−=
Substitute numerical values and
evaluate σinner: ( )
2
2inner C/m553.0m6.04
C5.2 µ−=−= π
µσ
Express the charge density on the
outer surface: A
qouter
outer =σ
Because the spherical shell is
uncharged:
0innerouter =+ qq
Substitute for qouter to obtain:
2
inner
outer 4 b
q
πσ
−=
Substitute numerical values and
evaluate σouter: ( )
2
2outer C/m246.0m9.04
C5.2 µ== π
µσ
The Electric Field 2: Continuous Charge Distributions
127
(b) Apply Gauss’s law to a spherical
surface of radius r that is concentric
with the point charge:
inside
0
S n
1 QdAE ∈=∫
or
0
inside
n
24 ∈=
QErπ
Solve for En:
0
2
inside
n 4 ∈= r
QE π (1)
For r < a = 0.6 m, Qinside = q. Substitute in equation (1) and evaluate
En(r < 0.6 m) to obtain:
( ) ( )( )
( ) 224
2
229
2
0
2n
1/CmN1025.2
C5.2/CmN1099.8
4
r
rr
kq
r
qarE
⋅×=
⋅×==∈=<
µ
π
Because the spherical shell is a
conductor, a charge –q will be
induced on its inner surface. Hence,
for 0.6 m < r < 0.9 m:
0inside =Q
and
( ) 0m9.0m6.0n =<< rE
For r > 0.9 m, the net charge inside the Gaussian surface is q and:
( ) ( ) 2242n 1/CmN1025.2m9.0 rrkqrE ⋅×==>
(c) Because E = 0 in the conductor:
C5.2inner µ−=q
and
2
inner C/m553.0 µ−=σ
as before.
Express the relationship between the
charges on the inner and outer
surfaces of the spherical shell:
C5.3innerouter µ=+ qq
and
C0.6-C5.3 innerouter µµ == qq
σouter is now given by: ( ) 22outer C/m589.0m9.04
C6 µ== π
µσ
Chapter 22
128
For r < a = 0.6 m, Qinside = q and
En(r < 0.6 m) is as it was in (a):
( ) ( ) 224n 1/CmN1025.2 rarE ⋅×=<
Because the spherical shell is a
conductor, a charge –q will be
induced on its inner surface. Hence,
for 0.6 m < r < 0.9 m:
0inside =Q
and
( ) 0m9.0m6.0n =<< rE
For r > 0.9 m, the net charge inside the Gaussian surface is 6 µC and:
( ) ( )( ) ( ) 22422292n 1/CmN1039.51C6/CmN1099.8m9.0 rrrkqrE ⋅×=⋅×==> µ
64 ••
Picture the Problem From Gauss’s law we know that the electric field at the surface of
the charged sphere is given by 2RkQE = where Q is the charge on the sphere and R is
its radius. The minimum radius for dielectric breakdown corresponds to the maximum
electric field at the surface of the sphere.
Use Gauss’s law to express the
electric field at the surface of the
charged sphere:
2R
kQE =
Express the relationship between E
and R for dielectric breakdown:
2
min
max R
kQE =
Solve for Rmin:
max
min E
kQR =
Substitute numerical values and
evaluate Rmin:
( )( )
cm2.23
N/C103
C18/CmN1099.8
6
229
min
=
×
⋅×= µR
The Electric Field 2: Continuous Charge Distributions
129
65 ••
Picture the Problem We can use its
definition to find the surface charge
density just outside the face of the slab.
The electric field near the surface of the
slab is given by .0face ∈=σE We can
find the electric field on each side of the
slab by adding the fields due to the slab
and the plane of charge.
(a) Express the charge density per
face in terms of the net charge on the
slab:
2face 2L
q=σ
Substitute numerical values to
obtain: ( ) 22face C/m60.1m52
C80 µµσ ==
Express the electric field just outside
one face of the slab in terms of its
surface charge density:
0
face
slab ∈=
σE
Substitute numerical values and
evaluate Eface:
N/C101.81
m/NC108.85
C/m1.60
5
2212
2
slab
×=
⋅×= −
µE
(b) Express the total field on the side
of the slab closest to the infinite
charged plane:
rr
rr
EEE
ˆˆ
2
ˆˆ
0
face
0
plane
slabplane
slabplanenear
∈−∈=
−=
+=
σσ
EE
rrr
where rˆ is a unit vector pointing away from
the slab.
Substitute numerical values and
evaluate nearE
r
: ( )
( )
( ) r
r
rE
ˆN/C10680.0
ˆN/C1081.1
ˆ
m/NC108.852
C/m2
5
5
2212
2
near
×−=
×−
⋅×= −
µr
Chapter 22
130
Express the total field on the side of
the slab away from the infinite
charged plane:
rrE ˆˆ
2 0
face
0
plane
far ∈+∈=
σσr
Substitute numerical values and
evaluate farE
r
: ( )
( )
( ) r
r
rE
ˆN/C1094.2
ˆN/C1081.1
ˆ
m/NC108.852
C/m2
5
5
2212
2
far
×=
×+
⋅×= −
µr
The charge density on the side of the slab near the plane is:
( )( ) 252212near0near C/m602.0N/C10680.0m/NC108.85 µσ =×⋅×==∈ −E
The charge density on the far side of the slab is:
( )( ) 252212near0near C/m60.2N/C1094.2m/NC108.85 µσ =×⋅×==∈ −E
General Problems
66 ••
Determine the Concept We can determine the direction of the electric field between
spheres I and II by imagining a test charge placed between the spheres and determining
the direction of the force acting on it. We can determine the amount and sign of the
charge on each sphere by realizing that the charge on a given surface induces a charge of
the same magnitude but opposite sign on the next surface of larger radius.
(a) The charge placed on sphere III has no bearing on the electric field between spheres I
and II. The field in this region will be in the direction of the force exerted on a test charge
placed between the spheres. Because the charge at the center is negative,
center. therdpoint towa willfield the
(b) The charge on sphere I (−Q0) will induce a charge of the same magnitude but
opposite sign on sphere II: 0Q+
(c) The induction of charge +Q0 on the inner surface of sphere II will leave its outer
surface with a charge of the same magnitude but opposite sign: 0Q−
(d) The presence of charge −Q0 on the outer surface of sphere II will induce a charge of
the same magnitude but opposite sign on the inner surface of sphere III: 0Q+
The Electric Field 2: Continuous Charge Distributions
131
(e) The presence of charge +Q0 on the inner surface of sphere III will leave the outer
surface of sphere III neutral: 0
(f) A graph of E as a function of r is shown
to the right:
67 ••
Picture the Problem Because the difference between the field just to the right of the
origin right,xE and the field just to the left of the origin left,xE is the field due to the
nonuniform surface charge, we can express left,xE and the difference between right,xE and
.0∈σ
Express the electric field just to the
left of the origin in terms of right,xE
and 0∈σ :
0
right,left, ∈−=
σ
xx EE
Substitute numerical values and evaluate left,xE :
N/C1015.1
m/NC108.85
C/m3.10N/C1065.4 52212
2
5
left, ×=⋅×−×= −
µEx
Chapter 22
132
68 ••
Picture the Problem Let P denote the point of interest at (2 m, 1.5 m). The electric field
at P is the sum of the electric fields due to the infinite line charge and the point charge.
Express the resultant electric field at P:
qEEE
rrr += λ
Find the field at P due the infinite line charge:
( )( ) ( )iirE ˆkN/C74.6ˆ
m4
C/m5.1/CmN1099.82ˆ2
229
−=−⋅×== µλλ r
kr
Express the field at P due the point
charge:
rE ˆ2r
kq
q =
r
Referring to the diagram above,
determine r and rˆ :
m12.1=r
and
jir ˆ446.0ˆ893.0ˆ −=
Substitute and evaluate ( )mm,1.52qEr :
( ) ( )( )( ) ( )
( )( )
( ) ( ) ji
ji
jiE
ˆkN/C16.4ˆkN/C32.8
ˆ446.0ˆ893.0kN/C32.9
ˆ446.0ˆ893.0
m12.1
C3.1/CmN1099.8mm,1.52 2
229
−=
−=
−⋅×= µq
r
Substitute to obtain:
The Electric Field 2: Continuous Charge Distributions
133
( ) ( ) ( ) ( )
( ) ( ) ji
jiiE
ˆkN/C17.4ˆkN/C61.1
ˆkN/C17.4ˆkN/C35.8ˆkN/C74.6mm,1.52
−=
−+−=r
*69 ••
Picture the Problem If the patch is small enough, the field at the center of the patch
comes from two contributions. We can view the field in the hole as the sum of the field
from a uniform spherical shell of charge Q plus the field due to a small patch with surface
charge density equal but opposite to that of the patch cut out.
(a) Express the magnitude of the
electric field at the center of the
hole:
holeshell spherical EEE +=
Apply Gauss’s law to a spherical
gaussian surface just outside the
given sphere:
( )
00
enclosed2
shell spherical 4 ∈=∈=
QQrE π
Solve for Espherical shell to obtain:
2
0
shell spherical 4 r
QE ∈= π
The electric field due to the small
hole (small enough so that we can
treat it as a plane surface) is:
0
hole 2∈
−= σE
Substitute and simplify to obtain:
( )
2
0
2
0
2
0
0
2
0
8
424
24
r
Q
r
Q
r
Q
r
QE
∈=
∈−∈=
∈
−+∈=
π
ππ
σ
π
(b) Express the force on the patch:
qEF =
where q is the charge on the patch.
Assuming that the patch has radius
a, express the proportion between
its charge and that of the spherical
shell:
22 4 r
Q
a
q
ππ = or Qr
aq 2
2
4
=
Substitute for q and E in the
expression for F to obtain:
4
0
22
2
0
2
2
3284 r
aQ
r
QQ
r
aF ∈=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈⎟⎟⎠
⎞
⎜⎜⎝
⎛= ππ
(c) The pressure is the force
divided by the area of the patch:
4
0
2
2
2
4
0
22
32
32
r
Q
a
r
aQ
P ∈=
∈= ππ
π
Chapter 22
134
70 ••
Picture the Problem The work done by the electrostatic force in expanding the soap
bubble is ∫= .PdVW
From Problem 69:
4
0
2
2
32 r
QP ∈= π
Express W in terms of dr: ∫∫ == drrPPdVW 24π
Substitute for P and simplify:
∫∈=
m2.0
m1.0
2
0
2
8 r
drQW π
Evaluating the integral yields:
( )( )
J1002.2
m1.0
1
m2.0
1
mN/C1085.88
nC31
8
7
2212
2m0.2
m1.00
2
−
−
×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−⋅×=⎥⎦
⎤⎢⎣
⎡−∈= ππ r
QW
71 ••
Picture the Problem We can use E = kq/R2, where R is the radius of the droplet, to find
the electric field at its surface. We can find R by equating the volume of the bubble at the
moment it bursts to the volume of the resulting spherical droplet.
Express the field at the surface of
the spherical water droplet:
2R
kqE = (1)
where R is the radius of the droplet and q is
its charge.
Express the volume of the bubble
just before it pops:
trV 24π≈
where t is the thickness of the soap bubble.
Express the volume of the sphere
into which the droplet collapses:
3
3
4 RV π=
Because the volume of the droplet
and the volume of the bubble are
equal:
32
3
44 Rtr ππ =
Solve for R: 3 23 trR =
Assume a thickness t of 1 µm and
evaluate R:
( ) ( ) m1093.4m1m2.03 33 2 −×== µR
The Electric Field 2: Continuous Charge Distributions
135
Substitute numerical values in
equation (1) and evaluate E:
( )( )( )
N/C1011.1
m1093.4
nC3C/mN1099.8
6
23
229
×=
×
⋅×= −E
72 ••
Picture the Problem Let the numeral 1
refer to the infinite plane at x = −2 m and
the numeral 2 to the plane at x = 2 m and
let the letter A refer to the region to the left
of plane 1, B to the region between the
planes, and C to the region to the right of
plane 2. We can use the expression for the
electric field of in infinite plane of charge
to express the electric field due to each
plane of charge in each of the three
regions. Their sum will be the resultant
electric field in each region.
Express the resultant electric field as
the sum of the fields due to planes 1
and 2:
21 EEE
rrr += (1)
(a) Express and evaluate the field
due to plane 1 in region A:
( ) ( )
( ) ( )
( )i
i
iE
ˆkN/C198
ˆ
m/NC1085.82
C/m5.3
ˆ
2
2212
2
0
1
1
=
−⋅×
−=
−∈=
−
µ
σAr
Express and evaluate the field due to
plane 2 in region A:
( ) ( )
( ) ( )
( )i
i
iE
ˆkN/C339
ˆ
m/NC1085.82
C/m6
ˆ
2
2212
2
0
2
2
−=
−⋅×=
−∈=
−
µ
σAr
Substitute in equation (1) to obtain:
( ) ( ) ( )
( )i
iiE
ˆkN/C141
ˆkN/C339ˆkN/C198
−=
−+=Ar
Chapter 22
136
(b) Express and evaluate the field
due to plane 1 in region B:
( )
( )
( )i
i
iE
ˆkN/C198
ˆ
m/NC1085.82
C/m5.3
ˆ
2
2212
2
0
1
1
−=
⋅×
−=
∈=
−
µ
σBr
Express and evaluate the field due
to plane 2 in region B:
( ) ( )
( ) ( )
( )i
i
iE
ˆkN/C339
ˆ
m/NC1085.82
C/m6
ˆ
2
2212
2
0
2
2
−=
−⋅×=
−∈=
−
µ
σBr
Substitute in equation (1) to obtain:
( ) ( ) ( )
( )i
iiE
ˆkN/C537
ˆkN/C339ˆkN/C198
−=
−+−=Br
(c) Express and evaluate the field
due to plane 1 in region C:
( )
( )
( )i
i
iE
ˆkN/C198
ˆ
m/NC1085.82
C/m5.3
ˆ
2
2212
2
0
1
1
−=
⋅×
−=
∈=
−
µ
σCr
Express and evaluate the field due to
plane 2 in region C:
( )
( )
( )i
i
iE
ˆkN/C339
ˆ
m/NC1085.82
C/m6
ˆ
2
2212
2
0
2
2
=
⋅×=
∈=
−
µ
σCr
Substitute in equation (1) to obtain:
( ) ( ) ( )
( )i
iiE
ˆkN/C141
ˆkN/C339ˆkN/C198
=
+−=Cr
*73 ••
Picture the Problem We can find the electric fields at the three points of interest by
adding the electric fields due to the infinitely long cylindrical shell and the spherical
shell. In Problem 42 it was established that, for an infinitely long cylindrical shell of
radius R, ( ) ,0=< RrEr and ( ) .0 rRRrEr ∈=> σ We know that, for a spherical shell
of radius R, ( ) ,0=< RrEr and ( ) .202 rRRrEr ∈=> σ
The Electric Field 2: Continuous Charge Distributions
137
Express the resultant electric field as
the sum of the fields due to the
cylinder and sphere:
sphcyl EEE
rrr += (1)
(a) Express and evaluate the electric
field due to the cylindrical shell at
the origin:
( ) 00,0cyl =Er
because the origin is inside the cylindrical
shell.
Express and evaluate the electric field due to the spherical shell at the origin:
( ) ( ) ( )( )( )( ) ( ) ( )iiiE ˆkN/C339ˆm0.5m/NC1085.8 m25.0C/m12ˆ0,0 22212
22
2
0
2
sph =−⋅×
−=−∈= −
µσ
r
Rr
Substitute in equation (1) to obtain:
( ) ( )
( )i
iE
ˆkN/C339
ˆkN/C33900,0
=
+=r
or
( ) kN/C3390,0 =E
and
°= 0θ
(b) Express and evaluate the electric field due to the cylindrical shell at
(0.2 m, 0.1 m):
Chapter 22
138
( ) ( )( )( )( ) ( )iiiE ˆkN/C508ˆm0.2m/NC1085.8 m15.0C/m6ˆmm,0.10.2 2212
2
0
cyl =⋅×=∈= −
µσ
r
Rr
Express the electric field due to the
charge on the spherical shell as a
function of the distance from its
center:
( ) rE ˆ2
0
2
sph r
Rr ∈=
σr
where rˆ is a unit vector pointing from (50
cm, 0) to (20 cm, 10 cm).
Referring to the diagram shown
above, find r and rˆ :
m316.0=r
and
jir ˆ316.0ˆ949.0 +−=r
Substitute to obtain:
( ) ( )( )( )( ) ( )
( )( )
( ) ( ) ji
ji
jiE
ˆkN/C268ˆkN/C806
ˆ316.0ˆ949.0kN/C849
ˆ316.0ˆ949.0
m0.316m/NC1085.8
m25.0C/m12mm,0.10.2 22212
22
sph
−+=
+−−=
+−⋅×
−= −
µr
Substitute in equation (1) to obtain:
( ) ( ) ( ) ( )
( ) ( ) ji
jiiE
ˆkN/C268ˆkN/C1310
ˆkN/C268ˆkN/C806ˆkN/C508mm,0.10.2
−+=
−++=r
or
( ) ( ) ( ) kN/C1340kN/C268kN/C1310mm,0.10.2 22 =−+=E
and
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= − 348
kN/C1310
kN/C268tan 1θ
(c) Express and evaluate the electric field due to the cylindrical shell at
(0.5 m, 0.2 m):
( ) ( )( )( )( ) ( )iiE ˆkN/C203ˆm0.5m/NC1085.8 m15.0C/m6mm,0.20.5 2212
2
cyl =⋅×= −
µr
Express and evaluate the electric
field due to the spherical shell at
( ) 0mm,0.20.5sph =Er
The Electric Field 2: Continuous Charge Distributions
139
(0.5 m, 0.5 m):
because (0.5 m, 0.2 m) is inside the
spherical shell.
Substitute in equation (1) to obtain:
( ) ( )
( )i
iE
ˆkN/C203
0ˆkN/C203mm,0.20.5
=
+=r
or
( ) kN/C203mm,0.20.5 =E
and
°= 0θ
74 ••
Picture the Problem Let the numeral 1 refer to the plane with charge density σ 1 and the
numeral 2 to the plane with charge density σ 2. We can find the electric field at the two
points of interest by adding the electric fields due to the charge distributions of the two
infinite planes.
Express the electric field at any
point in space due to the charge
distributions on the two planes:
21 EEE
rrr += (1)
(a) Express the electric field at (6 m, 2 m) due to plane 1:
( ) ( ) ( ) jjjE ˆkN/C67.3ˆm/NC108.852 nC/m65ˆ2mm,26 2212
2
0
1
1 =⋅×=∈= −
σr
Express the electric field at (6 m, 2 m) due to plane 2:
( ) ( ) ( )rrrE ˆkN/C54.2ˆm/NC108.852 nC/m45ˆ2mm,26 2212
2
0
2
2 =⋅×=∈= −
σr
where rˆ is a unit vector pointing from plane 2 toward the point whose coordinates are (6
m, 2 m).
Refer to the diagram below to obtain:
jir ˆ30cosˆ30sinˆ °−°=
Chapter 22
140
Substitute to obtain:
( ) ( )( ) ( ) ( ) jijiE ˆkN/C20.2ˆkN/C27.1ˆ30cosˆ30sinkN/C54.2mm,262 −+=°−°=r
Substitute in equation (1) to obtain:
( ) ( ) ( ) ( )
( ) ( ) ji
jijE
ˆkN/C47.1ˆkN/C27.1
ˆkN/C20.2ˆkN/C27.1ˆkN/C67.3mm,26
+=
−++=r
(b) Note that ( ) ( )mm,26mm,56 11 EE rr = so that:
( ) ( ) ( ) jjjE ˆkN/C67.3ˆmN/C1085.82 nC/m65ˆ2m5m,6 2212
2
0
1 =⋅×=∈= −
σr
Note also that ( ) ( )mm,26mm,56 22 EE rr −= so that:
( ) ( ) ( ) jiE ˆkN/C20.2ˆkN/C27.1mm,562 +−=r
Substitute in equation (1) to obtain:
( ) ( ) ( ) ( )
( ) ( ) ji
jijE
ˆkN/C87.5ˆkN/C27.1
ˆkN/C20.2ˆkN/C27.1ˆkN/C67.3mm,56
+−=
+−+=r
75 ••
Picture the Problem Because the atom is uncharged, we know that the integral of the
electron’s charge distribution over all of space must equal its charge e. Evaluation of this
integral will lead to an expression for ρ0. In (b) we can express the resultant field at any
point as the sum of the fields due to the proton and the electron cloud.
(a) Because the atom is uncharged:
( ) ( )∫∫ ∞∞ ==
0
2
0
4 drrrdVre πρρ
Substitute for ρ(r):
∫∫ ∞ −∞ − ==
0
22
0
0
22
0 44 drerdrree
arar πρπρ
Use integral tables or integration by
parts to obtain:
4
3
0
22 adrer ar =∫
∞
−
The Electric Field 2: Continuous Charge Distributions
141
Substitute to obtain:
0
3
3
0 4
4 ρππρ aae =⎟⎟⎠
⎞
⎜⎜⎝
⎛=
Solve for ρ0:
30 a
e
πρ =
(b) The field will be the sum of the
field due to the proton and that of
the electron charge cloud:
cloud2cloudp Er
kqEEE +=+=
Express the field due to the electron
cloud: ( ) ( )2cloud r
rkQrE =
where Q(r) is the net negative charge
enclosed a distance r from the proton.
Substitute to obtain: ( ) ( )22 r
rkQ
r
kerE +=
As in (a), Q(r) is given by:
')'('4)(
0
drrrrQ
r
ρπ∫=
Integrate to find Q(r) and substitute
in the expression for E to obtain: ⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++= − 2
2
/2
2
221)(
a
r
a
re
r
kerE ar
*76 ••
Picture the Problem We will assume that the radius at which they balance is large
enough that only the third term in the expression matters. Apply a condition for
equilibrium will yield an equation that we can solve for the distance r.
Apply 0=∑F to the proton:
02 /22
2
=−− mge
a
ke ar
To solve for r, isolate the
exponential factor and take the
natural logarithm of both sides of the
equation:
⎟⎟⎠
⎞
⎜⎜⎝
⎛= 2
22ln
2 mga
kear
Substitute numerical values and evaluate r:
( )( )( )( )( ) nm16.1nm0529.0m/s81.9kg1067.1 C1060.1C/mN1099.82ln2 nm0529.0 2227
219229
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
×
×⋅×= −
−
r
Chapter 22
142
.nanometers few a within force nalgravitatio
an thesmaller th it to reduces screening force, nalgravitatio n thelarger tha
magnitude of orders 40 is force ticelectrosta unscreened h theeven thoug Thus,
Remarks: Note that the argument of the logarithm contains the ratio between the
gravitational potential energy of a mass held a distance a0 above the surface of the
earth and the electrostatic potential energy for two unscreened charges a distance a0
apart.
77 ••
Picture the Problem In parts (a) and (b) we can express the charges on each of the
elements as the product of the linear charge density of the ring and the length of the
segments. Because the lengths of the segments are the product of the angle subtended at
P and their distances from P, we can express the charges in terms of their distances from
P. By expressing the ratio of the fields due to the charges on s1 and s2 we can determine
their dependence on r1 and r2 and, hence, the resultant field at P. We can proceed
similarly in part (c) with E varying as 1/r rather than 1/r2. In part (d), with s1 and s2
representing areas, we’ll use the definition of the solid angle subtended by these areas to
relate their charges to their distances from point P.
(a) Express the charge q1 on the
element of length s1:
111 rsq λθλ ==
where θ is the angle subtended by the arcs
of length s1 and s2.
Express the charge q2 on the
element of length s2:
222 rsq λθλ ==
Divide the first of these equations
by the second to obtain:
2
1
2
1
2
1
r
r
r
r
q
q == λθ
λθ
Express the electric field at P due to
the charge associated with the
element of length s1:
1
2
1
1
2
1
1
2
1
1
1 r
k
r
rk
r
sk
r
kqE λθλθλ ====
Express the electric field at P due to
the charge associated with the
element of length s2:
2
2 r
kE λθ=
Divide the first of these equations
by the second to obtain:
1
2
2
1
2
1
r
r
r
k
r
k
E
E == λθ
λθ
The Electric Field 2: Continuous Charge Distributions
143
and, because r2 > r1,
21 EE >
(b) The two fields point away from
their segments of arc. . towardpoints field
resultant the, Because
2
21
s
EE >
(c) If E varies as 1/r: λθλθλ k
r
rk
r
sk
r
kqE ====
1
1
1
1
1
1
1
and
λθλθλ k
r
rk
r
sk
r
kqE ====
2
2
2
2
2
2
2
Therefore:
21 EE =
(d) Use the definition of the solid
angle Ω subtended by the area s1 to
obtain:
2
1
1
44 r
s
ππ =
Ω
or
2
11 rs Ω=
Express the charge q1 of the area s1:
2
111 rsq Ω== σσ
Similarly, for an element of area s2: 222 rs Ω=
and
2
22 rq Ω=σ
Express the ratio of q1 to q2 to
obtain:
2
2
2
1
2
2
2
1
2
1
r
r
r
r
q
q =Ω
Ω= σ
σ
Proceed as in (a) to obtain:
12
2
2
1
2
1
2
2
2
2
1
1
2
2
2
2
2
2
1
1
2
1 =Ω
Ω===
rr
rr
qr
qr
r
kq
r
kq
E
E
σ
σ
Chapter 22
144
Because the two fields are of equal
magnitude and oppositely directed:
0=Er
1.
2
toward
point would and at fieldstronger theproduce would then ,1/ If
s
PsrE E
r∝
78 ••
Picture the Problem We can apply the condition for translational equilibrium to the
particle and use the expression for the electric field on the axis of a ring charge to obtain
an expression for |q|/m. Doing so will lead us to the conclusion that |q|/m will be a
minimum when Ez is a maximum and so we’ll use the result from Problem 26 that
2Rz −= maximizes Ez.
(a) Apply ∑ = 0zF to the particle:
0=−mgEq z
Solve for |q|/m:
zE
g
m
q = (1)
Note that this result tells us that the
minimum value of |q|/m will be where the
field due to the ring is greatest.
Express the electric field along the z
axis due to the ring of charge:
( ) 2322 Rz
kQzEz +=
Differentiate this expression with respect to z to obtain:
( )
( ) ( )
( )
( ) ( )( ) ( )( ) ( ) ( )( )322
212222322
322
2122
2
32322
322
23222322
2322
32
Rz
RzzRzkQ
Rz
zRzzRzkQ
Rz
Rz
dx
dzRz
kQ
Rz
x
dz
dkQ
dz
dEx
+
+−+=+
+−+=
+
+−+
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
+=
The Electric Field 2: Continuous Charge Distributions
145
Set this expression equal to zero for
extrema and simplify:
( ) ( )( ) 03 322
212222322
=+
+−+
Rz
RzzRz
,
( ) ( ) 03 212222322 =+−+ RzzRz ,
and
03 222 =−+ zRz
Solve for x to obtain:
2
Rz ±=
as candidates for maxima or minima.
You can either plot a graph of Ez or
evaluate its second derivative at
these points to show that it is a
maximum at:
2
Rz −=
Substitute to obtain an expression
Ez,max:
223
2
2max, 27
2
2
2
R
kQ
RR
RkQ
Ez =
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛−
⎟⎠
⎞⎜⎝
⎛−
=
Substitute in equation (1) to obtain:
kQ
gR
m
q
2
27 2=
(b) If |q|/m is twice as great as in (a),
then the electric field should be half
its value in (a), i.e.:
( ) 2322227 Rz
kQz
R
kQ
+=
or
3
2
2
6
2
4
1
27
1
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
R
zR
z
R
Let a = z2/R2 and simplify to obtain:
01243 23 =+−+ aaa
The graph of ( ) 1243 23 +−+= aaaaf shown below was plotted using a spreadsheet
program.
Chapter 22
146
-30
-25
-20
-15
-10
-5
0
5
10
15
20
0.0 1.0 2.0 3.0 4.0
a
f(
a
)
Use your calculator or trial-and-error
methods to obtain:
04188.0=a and 596.3=a
The corresponding z values are: Rz 205.0−= and Rz 90.1−=
The condition for a stable equilibrium position is that the particle, when displaced from
its equilibrium position, experiences a restoring force, i.e. a force that acts toward the
equilibrium position. When the particle in this problem is just above its equilibrium
position the net force on it must be downward and when it is just below the equilibrium
position the net force on it must be upward. Note that the electric force is zero at the
origin, so the net force there is downward and remains downward to the first equilibrium
position as the weight force exceeds the electric force in this interval. The net force is
upward between the first and second equilibrium positions as the electric force exceeds
the weight force. The net force is downward below the second equilibrium position as the
weight force exceeds the electric force. Thus, the first (higher) equilibrium position is
stable and the second (lower) equilibrium position is unstable.
You might also find it instructive to use
your graphing calculator to plot a graph of
the electric force (the gravitational force is
constant and only shifts the graph of the
total force downward). Doing so will
produce a graph similar to the one shown
in the sketch to the right.
Note that the slope of the graph is negative on both sides of −0.205R whereas it is
positive on both sides of −1.90R; further evidence that −0.205R is a position of stable
equilibrium and −1.90R a position of unstable equilibrium.
The Electric Field 2: Continuous Charge Distributions
147
79 ••
Picture the Problem The loop with the
small gap is equivalent to a closed loop and
a charge of RQ π2l− at the gap. The field
at the center of a closed loop of uniform
line charge is zero. Thus the field is
entirely due to the charge RQ π2l− .
(a) Express the field at the center of
the loop:
gaploopcenter EEE
rrr += (1)
Relate the field at the center of the
loop to the charge in the gap:
rE ˆ2gap R
kq−=r
Use the definition of linear charge
density to relate the charge in the
gap to the length of the gap:
R
Qq
πλ 2== l
or
R
Qq π2
l=
Substitute to obtain:
rE ˆ
2 3gap R
kQ
π
lr −=
Substitute in equation (1) to obtain:
rrE ˆ
2
ˆ
2
0 33center R
kQ
R
kQ
ππ
llr −=−=
outward.radially pointsorigin at the field thepositive, is If Q
(b) From our result in (a) we see
that the magnitude of centerE
r
is: 3center 2 R
kQE π
l=
80 ••
Picture the Problem We can find the electric fields at the three points of interest,
labeled 1, 2, and 3 in the diagram, by adding the electric fields due to the charge
distributions on the nonconducting sphere and the spherical shell.
Chapter 22
148
Express the electric field due to the
nonconducting sphere and the
spherical shell at any point in space:
shellsphere EEE
rrr += (1)
(a) Because (4.5 m, 0) is inside the
spherical shell:
( ) 00,m5.4shell =Er
Apply Gauss’s law to a spherical
surface inside the nonconducting
sphere to obtain:
( ) iE ˆ
3
4
sphere rkr ρπ=
r
Evaluate ( )m5.0sphereEr :
( ) ( )( )( ) ( )iiE ˆkN/C1.94ˆm5.0C/m5/CmN10988.8
3
4m5.0 2229sphere =⋅×= µπ
r
Substitute in equation (1) to obtain: ( ) ( )
( )i
iE
ˆkN/C1.94
0ˆkN/C1.940,m5.4
=
+=r
Find the magnitude and direction of
( )0,m5.4Er :
( ) kN/C1.940,m5.4 =E
and
°= 0θ
(b) Because (4 m, 1.1m) is inside
the spherical shell:
( ) 0m1.1,m4shell =Er
The Electric Field 2: Continuous Charge Distributions
149
Evaluate ( )m1.1sphereEr :
( ) ( )( )( )( ) ( ) jjE ˆkN/C6.33ˆm1.13 m6.0C/m5/CmN1099.84m1.1 2
32229
sphere =⋅×= µπ
r
Substitute in equation (1) to obtain: ( ) ( )
( ) j
jE
ˆkN/C6.33
0ˆkN/C6.330,m5.4
=
+=r
Find the magnitude and direction of
( )m1.1,m5.4Er :
( ) kN/C6.33m1.1,m5.4 =E
and
°= 90θ
(c) Because (2 m, 3 m) outside the
spherical shell:
( ) rE ˆ2shellshell r
kQr =r
where rˆ is a unit vector pointing from
(4 m, 0) to (2 m, 3 m).
Evaluate Qshell:
( )( )
C1.27
m2.1C/m5.14 22shellshell
µ
µπσ
−=
−== AQ
Refer to the diagram below to find rˆ and r:
m61.3=r
and
jir ˆ832.0ˆ555.0ˆ +−=
Substitute and evaluate ( )mm,32shellEr :
( ) ( )( )( )
( )( )
( ) ( ) ji
ji
rE
ˆkN/C6.15ˆkN/C4.10
ˆ832.0ˆ555.0kN/C7.18
ˆ
m3.61
C1.27/CmN1099.8m61.3 2
229
shell
−+=
+−−=
−⋅×= µr
Chapter 22
150
Express the electric field due to the
charged nonconducting sphere at a
distance r from its center that is
greater than its radius:
( ) rE ˆ2spheresphere r
kQ
r =r
Find the charge on the sphere:
( )( )
C52.4
m6.0C/m5
3
4 32
spheresphere
µ
µπρ
=
== VQ
Evaluate ( )m61.3sphereEr :
( ) ( )( )( ) ( )
( )( )
( ) ( ) ji
ji
rrE
ˆkN/C59.2ˆkN/C73.1
ˆ832.0ˆ555.0kN/C12.3
ˆkN/C12.3ˆ
m61.3
C52.4/CmN1099.8mm,32 2
229
sphere
+−=
+−=
=⋅×= µr
Substitute in equation (1) to obtain:
( ) ( ) ( ) ( ) ( )
( ) ( ) ji
jijiE
ˆkN/C0.13ˆkN/C67.8
ˆkN/C59.2ˆkN/C73.1ˆkN/C6.15ˆkN/C4.10m3,m2
−+=
+−+−+=r
Find the magnitude and direction of ( )m3,m2Er :
( ) ( ) ( ) kN/C6.15kN/C0.13kN/C67.8m3,m2 22 =−+=E
and
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= − 304
kN/C67.8
kN/C0.13tan 1θ
81 ••
Picture the Problem Let the numeral 1
refer to the infinite plane whose charge
density is σ 1 and the numeral 2 to the
infinite plane whose charge density is
σ 2. We can find the electric fields at the
two points of interest by adding the electric
fields due to the charge distributions on the
infinite planes and the sphere.
The Electric Field 2: Continuous Charge Distributions
151
Express the electric field due to the
infinite planes and the sphere at any
point in space:
21sphere EEEE
rrrr ++= (1)
(a) Because (0.4 m, 0) is inside the
sphere:
( ) 00,m4.0sphere =Er
Find the field at (0.4 m, 0) due to
plane 1:
( )
( )
( ) j
j
jE
ˆkN/C169
ˆ
m/NC1085.82
C/m3
ˆ
2
0,m4.0
2212
2
0
1
1
=
⋅×=
∈=
−
µ
σr
Find the field at (0.4 m, 0) due to plane 2:
( ) ( ) ( ) ( ) ( )iiiE ˆkN/C113ˆm/NC1085.82 C/m2ˆ20,m4.0 2212
2
0
2
2 =−⋅×
−=−∈= −
µσr
Substitute in equation (1) to obtain: ( ) ( )
( )
( ) ( ) ji
i
jE
ˆkN/C169ˆkN/C113
ˆkN/C113
ˆkN/C16900,m4.0
+=
+
+=r
Find the magnitude and direction of
( )0,m4.0Er :
( ) ( ) ( )
kN/C203
kN/C169kN/C1130,m4.0 22
=
+=E
and
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 2.56
kN/C113
kN/C169tan 1θ
(b) Because (2.5 m, 0) is outside the
sphere:
( ) rE ˆ0,m4.0 2spheresphere r
kQ=r
where rˆ is a unit vector pointing from
(1 m, −0.6 m) to (2.5 m, 0).
Chapter 22
152
Evaluate Qsphere:
( )( )
C7.37
m1C/m34
4
22
2
spheresphere
µ
µπ
πσσ
−=
−=
== RAQ
Referring to the diagram above,
determine r and rˆ :
m62.1=r
and
jir ˆ371.0ˆ928.0ˆ +=
Substitute and evaluate ( )0,m5.2sphereEr :
( ) ( )( )( )
( )( )
( ) ( ) ji
ji
rE
ˆkN/C9.47ˆkN/C120
ˆ371.0ˆ928.0kN/C129
ˆ
m62.1
7.37/CmN1099.80,m5.2 2
229
sphere
−+−=
+−=
−⋅×= Cµr
Find the field at (2.5 m, 0) due to
plane 1:
( )
( )
( ) j
j
jE
ˆkN/C169
ˆ
m/NC1085.82
C/m3
ˆ
2
0,m5.2
2212
2
0
1
1
=
⋅×=
∈=
−
µ
σr
Find the field at (2.5 m, 0) due to
plane 2:
( )
( )
( )i
i
iE
ˆkN/C113
ˆ
m/NC1085.82
C/m2
ˆ
2
0,m5.2
2212
2
0
2
2
−=
⋅×
−=
∈=
−
µ
σr
Substitute in equation (1) to obtain:
( ) ( ) ( ) ( ) ( )
( ) ( ) ji
ijjiE
ˆkN/C121ˆkN/C233
ˆkN/C113ˆkN/C169ˆkN/C9.47ˆkN/C1200,m4.0
+−=
−++−+−=r
Find the magnitude and direction of ( )0,m5.2Er :
The Electric Field 2: Continuous Charge Distributions
153
( ) ( ) ( ) kN/C263kN/C121kN/C2330,m5.2 22 =+−=E
and
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛
−=
− 153
kN/C233
kN/C121tan 1θ
82 ••
Picture the Problem Let P represent the point of interest at (1.5 m, 0.5 m). We can find
the electric field at P by adding the electric fields due to the infinite plane, the infinite
line, and the sphere. Once we’ve expressed the field at P in vector form, we can find its
magnitude and direction.
Express the electric field at P:
spherelineplane EEEE
rrrr ++=
Find planeE
r
at P:
( )
( )i
i
iE
ˆkN/C113
ˆ
m/NC1085.82
C/m2
ˆ
2
2212
2
0
plane
−=
⋅×−=
∈−=
−
µ
σr
Express lineE
r
at P: rE ˆ2line r
kλ=r
Refer to the diagram to obtain:
( ) ( ) jir ˆm5.0ˆm5.0 −=r
and
( ) ( ) jir ˆ707.0ˆ707.0ˆ −=
Substitute to obtain:
( )( ) ( ) ( )[ ]
( ) ( ) ( )[ ] ( ) ( ) jiji
jiE
ˆkN/C1.72ˆkN/C1.72ˆ707.0ˆ707.0kN/C102
ˆ707.0ˆ707.0
m707.0
C/m4/CmN1099.82 229
line
−+=−=
−⋅×= µr
Letting r′ represent the distance
from the center of the sphere to P,
'kr' rE ˆ
3
4
sphere ρπ=
r
Chapter 22
154
apply Gauss’s law to a spherical
surface of radius r′ centered at
(1 m, 0) to obtain an expression
for sphereE
r
at P:
where 'rˆ is directed toward the center of
the sphere.
Refer to the diagram used above to obtain:
( ) ( ) jir ˆm5.0ˆm5.0 −−='r
and
( ) ( ) jir ˆ707.0ˆ707.0ˆ −−='
Substitute to obtain:
( )( )( ) ( ) ( )[ ]
( )( ) ( ) ( ) jiji
jiE
ˆkN/C113ˆkN/C113ˆˆkN/C113
ˆ707.0ˆ707.0C/m6m707.0/CmN1099.8
3
4 3229
sphere
−+−=+−=
+−⋅×= µπr
Substitute and evaluate E
r
:
( ) ( ) ( ) ( )
( )
( ) ( ) ji
j
ijiiE
ˆkN/C185ˆkN/C154
ˆkN/C113
ˆkN/C113ˆkN/C1.72ˆkN/C1.72ˆkN/C113
−+−=
−+
−+−++−=r
Finally, find the magnitude and direction
of E
r
:
( ) ( )
kN/C241
kN/C185kN/C154 22
=
−+−=E
and
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−= − 220
kN/C185
kN/C154tan 1θ
83 ••
Picture the Problem We can find the
period of the motion from its angular
frequency and apply Newton’s 2nd law to
relate ω to m, q, R, and the electric field
due to the infinite line charge. Because the
electric field is given by rkEr λ2= we
can express ω and, hence, T as a function
of m, q, R, and λ.
Relate the period T of the particle to
its angular frequency ω:
ω
π2=T (1)
The Electric Field 2: Continuous Charge Distributions
155
Apply Newton’s 2nd law to the
particle to obtain:
2
radial ωmRqEF r ==∑
Solve for ω:
mR
qEr=ω
Express the electric field at a
distance R from the infinite line
charge:
R
kEr
λ2=
Substitute in the expression for ω:
m
qk
RmR
qk λλω 212 2 ==
Substitute in equation (1) to obtain:
qk
mRT λπ 22=
*84 ••
Picture the Problem Starting with the equation for the electric field on the axis of ring
charge, we can factor the denominator of the expression to show that, for
x << R, Ex is proportional to x. We can use Fx = qEx to express the force acting on the
particle and apply Newton’s 2nd law to show that, for small displacements from
equilibrium, the particle will execute simple harmonic motion. Finally, we can find the
period of the motion from its angular frequency, which we can obtain from the
differential equation of motion.
(a) Express the electric field on the
axis
of the ring of charge:
( ) 2322 Rx
kQxEx +=
Factor R2 from the denominator of
Ex to obtain:
x
R
kQ
R
xR
kQx
R
xR
kQxEx
323
2
2
3
23
2
2
2
1
1
≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
provided x << R.
(b) Express the force acting on the
particle as a function of its charge
and the electric field:
x
R
kqQqEF xx 3==
Chapter 22
156
(c) Because the negatively charged
particle experiences a linear
restoring force, we know that its
motion will be simple harmonic.
Apply Newton’s 2nd law to the
negatively charged particle to
obtain:
x
R
kqQ
dt
xdm 32
2
−=
or
032
2
=+ x
mR
kqQ
dt
xd
the differential equation of simple
harmonic motion.
Relate the period T of the simple
harmonic motion to its angular
frequency ω:
ω
π2=T
From the differential equation we
have: 3
2
mR
kqQ=ω
Substitute to obtain:
kqQ
mRT
3
2π=
85 ••
Picture the Problem Starting with the equation for the electric field on the axis of a ring
charge, we can factor the denominator of the expression to show that, for x << R, Ex is
proportional to x. We can use Fx = qEx to express the force acting on the particle and
apply Newton’s 2nd law to show that, for small displacements from equilibrium, the
particle will execute simple harmonic motion. Finally, we can find the angular frequency
of the motion from the differential equation and use this expression to find its value when
the radius of the ring is doubled and all other parameters remain unchanged.
Express the electric field on the axis
of the ring of charge:
( ) 2322 Rx
kQxEx +=
Factor R2 from the denominator of
Ex to obtain:
x
R
kQ
R
xR
kQx
R
xR
kQxEx
323
2
2
3
23
2
2
2
1
1
≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
provided x << R.
The Electric Field 2: Continuous Charge Distributions
157
Express the force acting on the
particle as a function of its charge
and the electric field:
x
R
kqQqEF xx 3==
Because the negatively charged
particle experiences a linear
restoring force, we know that its
motion will be simple harmonic.
Apply Newton’s 2nd law to the
negatively charged particle to
obtain:
x
R
kqQ
dt
xdm 32
2
−=
or
032
2
=+ x
mR
kqQ
dt
xd
the differential equation of simple
harmonic motion.
The angular frequency of the simple
harmonic motion of the particle is
given by:
3mR
kqQ=ω (1)
Express the angular frequency of the
motion if the radius of the ring is
doubled:
( )32' Rm
kqQ=ω (2)
Divide equation (2) by equation (1)
to obtain:
( )
8
12'
3
3
==
mR
kqQ
Rm
kqQ
ω
ω
Solve for and evaluate ω′: rad/s7.42
8
rad/s21
8
' === ωω
86 ••
Picture the Problem Starting with the equation for the electric field on the axis of a ring
charge, we can factor the denominator of the expression to show that, for x << R, Ex is
proportional to x. We can use Fx = qEx to express the force acting on the particle and
apply Newton’s 2nd law to show that, for small displacements from equilibrium, the
particle will execute simple harmonic motion. Finally, we can find the angular frequency
of the motion from the differential equation and use this expression to find its value when
the radius of the ring is doubled while keeping the linear charge density on the ring
constant.
Express the electric field on the axis
of the ring of charge:
( ) 2322 Rx
kQxEx +=
Chapter 22
158
Factor R2 from the denominator of
Ex to obtain:
x
R
kQ
R
xR
kQx
R
xR
kQxEx
323
2
2
3
23
2
2
2
1
1
≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
provided x << R.
Express the force acting on the
particle as a function of its charge
and the electric field:
x
R
kqQqEF xx 3==
Because the negatively charged
particle experiences a linear
restoring force, we know that its
motion will be simple harmonic.
Apply Newton’s 2nd law to the
negatively charged particle to
obtain:
x
R
kqQ
dt
xdm 32
2
−=
or
032
2
=+ x
mR
kqQ
dt
xd
,
the differential equation of simple
harmonic motion.
The angular frequency of the simple
harmonic motion of the particle is
given by:
3mR
kqQ=ω (1)
Express the angular frequency of
the motion if the radius of the ring
is doubled while keeping the linear
charge density constant (i.e.,
doubling Q):
( )
( )32
2'
Rm
Qkq=ω (2)
Divide equation (2) by equation (1)
to obtain:
( )
( )
2
12
2
'
3
3
==
mR
kqQ
Rm
Qkq
ω
ω
Solve for and evaluate ω′: rad/s5.01
2
rad/s21
2
' === ωω
The Electric Field 2: Continuous Charge Distributions
159
87 ••
Picture the Problem We can apply Gauss’s law to express E
r
as a function of r. We can
use the hint to think of the fields at points 1 and 2 as the sum of the fields due to a sphere
of radius a with a uniform charge distribution ρ and a sphere of radius b, centered at a/2
with uniform charge distribution −ρ.
(a) The electric field at a distance r
from the center of the sphere is
given by:
rE ˆE=r (1)
where rˆ is a unit vector pointing radially
outward.
Apply Gauss’s law to a spherical
surface of radius r centered at the
origin to obtain:
( )
0
enclosed2
S n
4 ∈==∫ QrEdAE π
Relate Qenclosed to the charge density
ρ:
3
3
4
enclosed
r
Q
πρ = ⇒
3
3
4
enclosed rQ ρπ=
Substitute for Qenclosed:
( )
0
3
3
4
24 ∈=
rrE ρππ
Solve for E to obtain:
03∈
= rE ρ
Substitute for E in equation (1) to
obtain:
rE ˆ
3 0
r∈=
ρr
(b) The electric field at point 1 is
the sum of the electric fields due to
the two charge distributions:
rrEEE ˆˆ1 ρρρρ −− +=+= EE
rrr
(2)
Apply Gauss’s law to relate the
magnitude of the field due to the
positive charge distribution to the
charge enclosed by the sphere:
( )
0
3
3
4
0
encl24 ∈=∈=
ρππρ aqaE
Solve for Eρ:
00 3
2
3 ∈=∈=
baE ρρρ
Proceed similarly for the spherical
hole to obtain: ( )
0
3
3
4
0
encl24 ∈−=∈=−
ρππρ bqbE
Solve for E−ρ:
03∈
−=− bE ρρ
Chapter 22
160
Substitute in equation (2) to obtain:
rrrE ˆ3
ˆ
3
ˆ
3
2
000
1 ∈=∈−∈=
bbb ρρρr
The electric field at point 2 is the
sum of the electric fields due to the
two charge distributions:
rrEEE ˆˆ2 ρρρρ −− +=+= EE
rrr
(3)
Because point 2 is at the center of
the larger sphere:
0=ρE
The magnitude of the field at point
2 due to the negative charge
distribution is:
03∈
=− bE ρρ
Substitute in equation (3) to obtain:
rrE ˆ
3
ˆ
3
0
00
2 ∈=∈+=
bb ρρr
88 •••
Picture the Problem The electric field in the cavity is the sum of the electric field due to
the uniform and positive charge distribution of the sphere whose radius is a and the
electric field due to any charge in the spherical cavity whose radius is b.
The electric field at any point inside
the cavity is the sum of the electric
fields due to the two charge
distributions:
rrEEE ˆˆ inside chargeinside charge EE +=+= ρρ
rrr
where rˆ is a unit vector pointing radially
outward.
Because there is no charge inside
the
cavity:
0inside charge =E
The magnitude of the field inside
the cavity due to the positive charge
distribution is:
03∈
= bE ρρ
Substitute in the expression for E
r
to obtain: rrE ˆ3
ˆ
3
0
00
bb ∈=∈+=
ρρr
89 ••
Picture the Problem We can use the hint given in Problem 87 to think of the fields at
points 1 and 2 as the sum of the fields due to a sphere of radius a with a uniform charge
distribution ρ and a sphere of radius b, centered at a/2 with charge Q spread uniformly
throughout its volume.
The electric field at point 1 is the
sum of the electric fields due to the
two charge distributions:
rrEEE ˆˆ1 QQ EE +=+= ρρ
rrr
(1)
where rˆ is a unit vector pointing radially
outward.
The Electric Field 2: Continuous Charge Distributions
161
Apply Gauss’s law to relate the
field due to the positive charge
distribution to the charge of the
sphere:
( )
0
3
3
4
0
encl24 ∈=∈=
ρππρ aqaE
Solve for Eρ:
00 3
2
3 ∈=∈=
baE ρρρ
Apply Gauss’s law to relate the
field due to the negative charge
distributed uniformly throughout
the volume of the cavity :
( )
00
encl24 ∈=∈=
QqbEQ π
where 334'' bVQ πρρ ==
Substitute for Q to obtain:
( )
0
3
3
4
2 '4 ∈=
bbEQ
πρπ
Solve for EQ:
03
'
∈=
bEQ
ρ
Substitute in equation (1) to obtain:
( ) rrrE ˆ
3
'2ˆ
3
'ˆ
3
2
000
1 ∈
+=∈+∈=
bbb ρρρρr
The electric field at point 2 is the
sum of the electric fields due to the
two charge distributions:
rrEEE ˆˆ2 QQ EE +=+= ρρ
rrr
(2)
Because point 2 is at the center of
the larger sphere:
0=ρE
The magnitude of the field at point
2 due to the uniformly distributed
charge Q was shown above to be:
03
'
∈=
bEQ
ρ
Substitute in equation (2) to obtain:
rrE ˆ
3
'ˆ
3
'0
00
2 b
b
∈=∈+=
ρρr
90 ••
Picture the Problem Let the length of the cylinder be L, its radius R, and charge Q. Let
P be a generic point of interest on the x axis. We can find the electric field at P by
expressing the field due to an elemental disk of radius R, thickness dx, and charge dq and
then integrating ( )2212 RxxkEx +−= σπ .
Chapter 22
162
Express the electric field on the x
axis due to the charge carried by the
disk of thickness dx:
dx
Rx
xkdEx ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 ρπ
Integrate dEx for P beyond the end of the cylinder:
⎥⎥⎦
⎤
+⎟⎠
⎞⎜⎝
⎛ −+⎢⎢⎣
⎡
+⎟⎠
⎞⎜⎝
⎛ +−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= ∫
+
−
2
2
2
2
2
2
22
22
2
12
RxLRxLLk
dx
Rx
xkE
Lx
Lx
x
ρπ
ρπ
Integrate dEx for P inside the cylinder:
⎥⎥⎦
⎤
+⎟⎠
⎞⎜⎝
⎛ −+⎢⎢⎣
⎡
+⎟⎠
⎞⎜⎝
⎛ +−=
⎥⎥⎦
⎤
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−−⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= ∫∫
−+
2
2
2
2
2
0
22
2
0
22
22
22
112
RxLRxLxk
dx
Rx
xdx
Rx
xkE
xLxL
x
ρπ
ρπ
The effective charge density of the disk
is given by:
2R
LQ
πρ =
Substitute numerical values and
evaluate ρ:
( ) ( ) 32 C/m53.5m2m2.1
C50 µπ
µρ ==
Evaluate 2πkρ :
( )( ) mN/C1012.3C/m53.5/CmN1099.822 53229 ⋅×=⋅×= µπρπk
The Electric Field 2: Continuous Charge Distributions
163
(a) Evaluate Ex(0.5 m):
( ) ( )
( ) ( ) ( )
kN/C118
m2.1m5.0
2
m2m2.1m5.0
2
m2m5.02
mN/C1012.3m5.0
2
2
2
2
5
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ +−×
⋅×=xE
(b) Evaluate Ex(2 m):
( ) ( )
( ) ( )
kN/C103
m2.1m2
2
m2m2.1m2
2
m2m2
mN/C1012.3m2
2
2
2
2
5
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ +−×
⋅×=xE
(c) Evaluate Ex(20 m):
( ) ( )
( ) ( )
kN/C12.1
m2.1m20
2
m2m2.1m20
2
m2m2
mN/C1012.3m20
2
2
2
2
5
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ +−×
⋅×=xE
Remarks: Note that, in (c), the distance of 20 m is much greater than the length of
the cylinder that we could have used Ex = kQ/x2.
91 ••
Picture the Problem We can use ( )[ ]LxxkQEx −= 00 to express the electric fields at
x0 = 2L and x0 = 3L and take the ratio of these expressions to find the field at x0 = 3L.
Express the electric field along the x
axis due to a uniform line charge on
the x axis:
( ) ( )Lxx
kQxEx −= 000
Evaluate Ex at x0 = 2L: ( ) ( ) 22222 L
kQ
LLL
kQLEx =−= (1)
Evaluate Ex at x0 = 3L: ( ) ( ) 26333 L
kQ
LLL
kQLEx =−= (2)
Chapter 22
164
Divide equation (2) by equation (1)
to obtain:
( )
( ) 3
1
2
6
2
3
2
2 ==
L
kQ
L
kQ
LE
LE
x
x
Solve for and evaluate ( )LEx 3 : ( ) ( ) ( )
N/C200
N/C600
3
12
3
13
=
== LELE xx
92 •••
Picture the Problem Let the coordinates of one corner of the cube be (x,y,z), and assume
that the sides of the cube are ∆x, ∆y, and ∆z and compute the flux through the faces of
the cube that are parallel to the yz plane. The net flux of the electric field out of the
gaussian surface is the difference between the flux out of the surface and the flux into the
surface.
The net flux out of the cube is given
by:
( ) ( )xxx φφφ −∆+=net
Use a Taylor series expansion to express the net flux through faces of the cube
that are parallel to the yz plane:
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ...'''...''' 221221net +∆+∆=−+∆+∆+= xxxxxxxxxx φφφφφφφ
Neglecting terms higher than first
order we have:
( )x'xφφ ∆=net
Because the electric field is in the x
direction, φ (x) is:
( ) zyEx ∆∆= xφ
and
( ) zy
x
Ex' ∆∆∂
∂= xφ
Substitute for φ ′(x) to obtain:
( )
( )
V
x
E
zyx
x
E
zy
x
Ex
x
x
x
∆∂
∂=
∆∆∆∂
∂=
∆∆∂
∂∆=netφ
The Electric Field 2: Continuous Charge Distributions
165
93 ••
Picture the Problem We can use the definition of electric flux in conjunction with the
result derived in Problem 92 to show that 0/∈=⋅∇ ρE
r
.
From Gauss’s law, the net flux
through any surface is: V
q
00
encl
net ∈=∈=
ρφ
Generalizing our result from
Problem 92 (see the remark
following Problem 92):
( )VV
z
E
y
E
x
E zyx E
r⋅∇=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂+∂
∂+∂
∂=netφ
Equate these two expressions to
obtain:
( ) VV
0∈
=⋅∇ ρEr or
0∈
=⋅∇ ρEr
*94 •••
Picture the Problem We can find the field due to the infinitely long line charge from
rkE λ2= and the force that acts on the dipole using drdEpF = .
Express the force acting on the
dipole: dr
dEpF =
The electric field at the location of
the dipole is given by: r
kE λ2=
Substitute to obtain:
2
22
r
pk
r
k
dr
dpF λλ −=⎥⎦
⎤⎢⎣
⎡=
where the minus sign indicates that the
dipole is attracted to the line charge.
95 ••
Picture the Problem We can find the distance from the center where the net force on
either charge is zero by setting the sum of the forces acting on either point charge equal
to zero. Each point charge experiences two forces; one a Coulomb force of repulsion due
to the other point charge, and the second due to that fraction of the sphere’s charge that is
between the point charge and the center of the sphere that creates an electric field at the
location of the point charge.
Apply 0=∑F to either of the
point charges:
0fieldCoulomb =− FF (1)
Express the Coulomb force on the
proton:
( ) 2
2
2
2
Coulomb 42 a
ke
a
keF ==
The force exerted by the field E is:
eEF =field
Chapter 22
166
Apply Gauss’s law to a spherical
surface
of radius a centered at the
origin:
( )
0
enclosed24 ∈=
QaE π
Relate the charge density of the
electron sphere to Qenclosed:
3
3
4
enclosed
3
3
4
2
a
Q
R
e
ππ = ⇒ 3
3
enclosed
2
R
eaQ =
Substitute for Qenclosed:
( ) 3
0
3
2 24
R
eaaE ∈=π
Solve for E to obtain:
3
02 R
eaE ∈= π ⇒ 30
2
field 2 R
aeF ∈= π
Substitute for FCoulomb and Ffield in
equation (1):
0
24 30
2
2
2
=∈− R
ae
a
ke
π
or
02
4 3
2
2
2
=−
R
ake
a
ke
Solve for a to obtain:
RRa 5.0
8
1
3 ==
96 •••
Picture the Problem We can use the result of Problem 96 to express the force acting on
both point charges when they are separated by 2a. We can then use this expression to
write the force function when the point charges are each displaced a small distance x
from their equilibrium positions and then expand this function binomially to show that
each point charge experiences a linear restoring force.
From Problem 95, the force function
at the equilibrium position is:
( ) 02
4 3
2
2
2
=−=
R
ake
a
keaF
When the charges are displaced a
distance x symmetrically from their
equilibrium positions, the force
function becomes:
( ) ( ) ( )xa
R
kexakexaF +−+=+ − 3
2
2
2 2
4
Expand this function binomially to obtain:
( ) ( )
x
R
kea
R
kex
a
ke
a
ke
x
R
kea
R
kexaakexaF
3
2
3
2
3
2
2
2
3
2
3
2
32
2
22
24
22...2
4
−−−≈
−−+−=+ −−
The Electric Field 2: Continuous Charge Distributions
167
Substitute for R using the result
obtained in Problem 96 and
simplify to obtain:
x
a
keF ⎟⎟⎠
⎞
⎜⎜⎝
⎛−= 3
2
restoring 4
3
Hence, we’ve shown that, for a small
displacement from equilibrium, the point
charges experience a linear restoring force.
Remarks: An alternative approach that you might find instructive is to expand the
force function using the Taylor series.
97 •••
Picture the Problem Because the restoring force found in Problem 96 is linear, we can
express the differential equation of the proton’s motion and then identify ω2 from this
equation.
Apply maFx =∑ to the displaced
proton to obtain:
2
2
3
2
4
3
dt
xdmx
r
ke =−
or
xx
mr
ke
dt
xd 2
3
2
2
2
4
3 ω−=−=
where 3
2
2
4
3
mr
ke=ω
Solve for ω :
3
2
4
3
mr
ke=ω
Substitute numerical values and evaluate ω:
( )( )( )( ) 114327
219229
s1049.4
nm08.0kg1067.14
C106.1C/mN1099.83 −
−
−
×=×
×⋅×=ω
Chapter 22
168
169
Chapter 23
Electrical Potential
Conceptual Problems
*1 •
Determine the Concept A positive charge will move in whatever direction reduces its
potential energy. The positive charge will reduce its potential energy if it moves toward a
region of lower electric potential.
2 ••
Picture the Problem A charged particle placed in an electric field experiences an
accelerating force that does work on the particle. From the work-kinetic energy theorem
we know that the work done on the particle by the net force changes its kinetic energy
and that the kinetic energy K acquired by such a particle whose charge is q that is
accelerated through a potential difference V is given by K = qV. Let the numeral 1 refer
to the alpha particle and the numeral 2 to the lithium nucleus and equate their kinetic
energies after being accelerated through potential differences V1 and V2.
Express the kinetic energy of the
alpha particle when it has been
accelerated through a potential
difference V1:
1111 2eVVqK ==
Express the kinetic energy of the
lithium nucleus when it has been
accelerated through a potential
difference V2:
2222 3eVVqK ==
Equate the kinetic energies to
obtain:
21 32 eVeV =
or
13
2
2 VV = and ( ) correct. is b
3 •
Determine the Concept If V is constant, its gradient is zero; consequently E
r
= 0.
4 •
Determine the Concept No. E can be determined from either ll d
dVE −= provided V is
known and differentiable or from ll ∆
∆−= VE provided V is known at two or more points.
Chapter 23
170
5 •
Determine the Concept Because the field lines are always perpendicular to equipotential
surfaces, you move always perpendicular to the field.
6 ••
Determine the Concept V along the axis of the ring does not depend on the charge
distribution. The electric field, however, does depend on the charge distribution, and the
result given in Chapter 21 is valid only for a uniform distribution.
*7 ••
Picture the Problem The electric field
lines, shown as solid lines, and the
equipotential surfaces (intersecting the
plane of the paper), shown as dashed lines,
are sketched in the adjacent figure. The
point charge +Q is the point at the right,
and the metal sphere with charge −Q is at
the left. Near the two charges the
equipotential surfaces are spheres, and the
field lines are normal to the metal sphere at
the sphere’s surface.
8 ••
Picture the Problem The electric field
lines, shown as solid lines, and the
equipotential surfaces (intersecting the
plane of the paper), shown as dashed lines,
are sketched in the adjacent figure. The
point charge +Q is the point at the right,
and the metal sphere with charge +Q is at
the left. Near the two charges the
equipotential surfaces are spheres, and the
field lines are normal to the metal sphere at
the sphere’s surface. Very far from both
charges, the equipotential surfaces and
field lines approach those of a point charge
2Q located at the midpoint.
Electric Potential
171
9 ••
Picture the Problem The equipotential
surfaces are shown with dashed lines, the
field lines are shown in solid lines. It is
assumed that the conductor carries a
positive charge. Near the conductor the
equipotential surfaces follow the
conductor’s contours; far from the
conductor, the equipotential surfaces are
spheres centered on the conductor. The
electric field lines are perpendicular to the
equipotential surfaces.
10 ••
Picture the Problem The equipotential
surfaces are shown with dashed lines, the
electric field lines are shown with solid
lines. Near each charge, the equipotential
surfaces are spheres centered on each
charge; far from the charges, the
equipotential is a sphere centered at the
midpoint between the charges. The electric
field lines are perpendicular to the
equipotential surfaces.
*11 •
Picture the Problem We can use Coulomb’s law and the superposition of fields to find E
at the origin and the definition of the electric potential due to a point charge to find V at
the origin.
Apply Coulomb’s law and the
superposition of fields to find the
electric field E at the origin:
0ˆˆ 22
atat
=−=
+= +−+
ii
EEE
a
kQ
a
kQ
aQaQ
rrr
Express the potential V at the origin:
a
kQ
a
kQ
a
kQ
VVV aQaQ
2
atat
=+=
+= +−+
and correct. is )(b
Chapter 23
172
12 •
Picture the Problem We can use iE ˆ
x
V
∂
∂−=r to find the electric field corresponding the
given potential and then compare its form to those produced by the four alternatives
listed.
Find the electric field corresponding to
this potential function:
[ ]
[ ]
i
ii
iiE
ˆ
0 if4
0if4
ˆ
0 if1
0if1
4ˆ4
ˆ4ˆ 0
⎥⎦
⎤⎢⎣
⎡
<
≥−=
⎥⎦
⎤⎢⎣
⎡
<−
≥−=∂
∂−=
+∂
∂−=∂
∂−=
x
x
x
x
x
x
Vx
xx
Vr
Of
the alternatives provided above, only a uniformly charged sheet in the yz plane would
produce a constant electric field whose direction changes at the origin. correct. is )(c
13 •
Picture the Problem We can use Coulomb’s law and the superposition of fields to find E
at the origin and the definition of the electric potential due to a point charge to find V at
the origin.
Apply Coulomb’s law and the
superposition of fields to find the
electric field E at the origin:
iii
EEE
ˆ2ˆˆ
222
atat
a
kQ
a
kQ
a
kQ
aQaQ
=+=
+= −−+
rrr
Express the potential V at the origin:
( ) 0
atat
=−+=
+= −−+
a
Qk
a
kQ
VVV aQaQ
and correct is )(c
14 ••
(a) False. As a counterexample, consider two equal charges at equal distances from the
origin on the x axis. The electric field due to such an array is zero at the origin but the
electric potential is not zero.
(b) True.
(c) False. As a counterexample, consider two equal-in-magnitude but opposite-in-sign
charges at equal distances from the origin on the x axis. The electric potential due to such
an array is zero at the origin but the electric field is not zero.
Electric Potential
173
(d) True.
(e) True.
(f) True.
(g) False. Dielectric breakdown occurs in air at an electric field strength of approximately
3×106 V/m.
15 ••
(a) No. The potential at the surface of a conductor also depends on the local radius of the
surface. Hence r and σ can vary in such a way that V is constant.
(b) Yes; yes.
*16 •
Determine the Concept When the two spheres are connected, their charges will
redistribute until the two-sphere system is in electrostatic equilibrium. Consequently, the
entire system must be an equipotential. corrent. is )(c
Estimation and Approximation Problems
17 •
Picture the Problem The field of a thundercloud must be of order 3×106 V/m just before
a lightning strike.
Express the potential difference
between the cloud and the earth as a
function of their separation d and
electric field E between them:
EdV =
Assuming that the thundercloud is at
a distance of about 1 km above the
surface of the earth, the potential
difference is approximately:
( )( )
V1000.3
m10V/m103
9
36
×=
×=V
Note that this is an upper bound, as there will be localized charge distributions on the
thundercloud which raise the local electric field above the average value.
*18 •
Picture the Problem The potential difference between the electrodes of the spark plug is
the product of the electric field in the gap and the separation of the electrodes. We’ll
assume that the separation of the electrodes is 1 mm.
Express the potential difference
between the electrodes of the spark
EdV =
Chapter 23
174
plug as a function of their separation
d and electric field E between them:
Substitute numerical values and
evaluate V:
( )( )
kV0.20
m10V/m102 37
=
×= −V
19 ••
Picture the Problem We can use conservation of energy to relate the initial kinetic
energy of the protons to their electrostatic potential energy when they have approached
each other to the given "radius".
(a) Apply conservation of energy to
relate the initial kinetic energy of the
protons to their electrostatic
potential when they are separated by
a distance r:
ffii UKUK +=+
or, because Ui = Kf = 0,
fi UK =
Because each proton has kinetic
energy K:
r
eK
0
2
4
2 ∈= π ⇒ r
eK
0
2
8 ∈= π
Substitute numerical values and evaluate K:
( )( )( )
MeV719.0
J106.1
eV1J1015.1
m10mN/C1085.88
C106.1
19
13
152212
219
=
×××=⋅×
×= −−−−
−
πK
(b) Express and evaluate the ratio of
the two energies:
%0767.0
MeV938
MeV719.0
rest
===
E
Kf
20 ••
Picture the Problem The magnitude of the electric field for which dielectric breakdown
occurs in air is about 3 MV/m. We can estimate the potential difference between you and
your friend from the product of the length of the spark and the dielectric constant of air.
Express the product of the length of
the spark and the dielectric constant of
air:
( )( ) V6000mm2MV/m3 ==V
Electric Potential
175
Potential Difference
21 •
Picture the Problem We can use the definition of finite potential difference to find the
potential difference V(4 m) − V(0) and conservation of energy to find the kinetic energy
of the charge when it is at x = 4 m. We can also find V(x) if V(x) is assigned various
values at various positions from the definition of finite potential difference.
(a) Apply the definition of finite
potential difference to obtain: ( ) ( )
( )( )
kV8.00
m4kN/C2
0m4
m4
0
−=
−=
−=⋅−=− ∫∫ llrr EddVV b
a
E
(b) By definition, ∆U is given by: ( )( )
mJ0.24
kV8C3
−=
−=∆=∆ µVqU
(c) Use conservation of energy to
relate ∆U and ∆K:
0=∆+∆ UK
or
00m4 =∆+− UKK
Because K0 = 0: mJ0.24m4 =∆−= UK
Use the definition of finite potential
difference to obtain:
( ) ( ) ( )
( )( )0
00
kV/m2 xx
xxExVxV x
−−=
−−=−
(d) For V(0) = 0: ( ) ( )( )0kV/m20 −−=− xxV
or
( ) ( )xxV kV/m2−=
(e) For V(0) = 4 kV: ( ) ( )( )0kV/m2kV4 −−=− xxV
or
( ) ( )xxV kV/m2kV4 −=
(f) For V(1m) = 0: ( ) ( )( )1kV/m20 −−=− xxV
or
( ) ( )xxV kV/m2kV2 −=
Chapter 23
176
22 •
Picture the Problem Because the electric field is uniform, we can find its magnitude
from E = ∆V/∆x. We can find the work done by the electric field on the electron from the
difference in potential between the plates and the charge of the electron and find the
change in potential energy of the electron from the work done on it by the electric field.
We can use conservation of energy to find the kinetic energy of the electron when it
reaches the positive plate.
(a) Express the magnitude of the
electric field between the plates in
terms of their separation and the
potential difference between them:
kV/m5.00
m0.1
V500 ==∆
∆=
x
VE
potential.higher
at the is plate positive theplate, negative the towardand plate
positive thefromaway is charge test aon force electric theBecause
(b) Relate the work done by the
electric field on the electron to the
difference in potential between the
plates and the charge of the electron:
( )( )
J1001.8
V005C106.1
17
19
−
−
×=
×=∆= VqW
Convert 8.01×10−17 J to eV: ( )
eV500
J101.6
eV1J108.01 19
17
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
××= −
−W
(c) Relate the change in potential
energy of the electron to the work
done on it as it moves from the
negative plate to the positive plate:
eV500−=−=∆ WU
Apply conservation of energy to
obtain:
eV500=∆−=∆ UK
23 •
Picture the Problem The Coulomb potential at a distance r from the origin relative to V
= 0 at infinity is given by V = kq/r where q is the charge at the origin. The work that must
be done by an outside agent to bring a charge from infinity to a position a distance r from
the origin is the product of the magnitude of the charge and the potential difference due to
the charge at the origin.
Electric Potential
177
(a) Express and evaluate the Coulomb
potential of the charge:
( )( )
kV50.4
m4
C2/CmN1099.8 229
=
⋅×=
=
µ
r
kqV
(b) Relate the work that must be
done to the magnitude of the charge
and the potential difference through
which the charge is moved:
( )( )
mJ5.13
kV50.4C3
=
=∆= µVqW
(c) Express the work that must be
done by the outside agent in terms
of the potential difference through
which the 2-µC is to be moved:
r
qkqVqW 3232 =∆=
Substitute numerical values and
evaluate W:
( )( )( )
mJ5.13
m4
C3C2/CmN1099.8 229
=
⋅×= µµW
24 ••
Picture the Problem In general, the work done by an external agent in separating the
two ions changes both their kinetic and potential energies. Here we’re assuming that they
are at rest initially and that they will be at rest when they are infinitely far apart. Because
their potential energy is also zero when they are infinitely far apart, the energy Wext
required to separate the ions to an infinite distance apart is the negative of their potential
energy when they are a distance r apart.
Express the energy required to separate
the ions in terms of the work required
by an external agent to bring about this
separation:
( )
r
ke
r
eek
r
qkq
UUKW
2
iext 0
=−−=−=
−=∆+∆=
+−
Substitute numerical values and evaluate Wext:
( )( ) J1024.8
m102.80
C106.1/CmN1099.8 19
10
219229
ext
−
−
−
×=×
×⋅×=W
Chapter 23
178
Convert Wext to eV: ( )
eV14.5
J101.6
eV1J1024.8 19
19
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
××= −
−W
25 ••
Picture the Problem We can find the final speeds of the protons from the potential
difference through which they are accelerated and use E = ∆V/∆x to find the accelerating
electric field.
(a) Apply the work-kinetic energy
theorem to the accelerated protons:
fKKW =∆=
or
2
2
1 mvVe =∆
Solve for v to obtain:
m
Vev ∆= 2
Substitute numerical values and
evaluate v:
( )( )
m/s1010.3
kg101.67
MV5C101.62
7
27
19
×=
×
×= −
−
v
(b) Assuming the same potential
change occurred uniformly over the
distance of 2.0 m, we can use the
relationship between E, ∆V, and ∆x
express and evaluate E:
MV/m2.50
m2
MV5 ==∆
∆=
x
VE
*26 ••
Picture the Problem The work done on the electrons by the electric field changes their
kinetic energy. Hence we can use the work-kinetic energy theorem to find the kinetic
energy and the speed of impact of the electrons.
Use the work-kinetic energy
theorem to relate the work done by
the electric field to the change in the
kinetic energy of the electrons:
fKKW =∆=
or
VeK ∆=f (1)
(a) Substitute numerical values and
evaluate Kf:
( )( ) eV103kV301 4f ×== eK
Electric Potential
179
(b) Convert this energy to eV: ( )
J1080.4
eV
J101.6eV103
15
19
4
f
−
−
×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ××=K
(c) From equation (1) we have:
Vemv ∆=2f21
Solve for vf to obtain:
m
Vev ∆= 2f
Substitute numerical values and
evaluate vf:
( )( )
m/s1003.1
kg1011.9
kV03C101.62
8
13
19
f
×=
×
×= −
−
v
Remarks: Note that this speed is about one-third that of light.
27 ••
Picture the Problem We know that energy is conserved in the interaction between the α
particle and the massive nucleus. Under the assumption that the recoil of the massive
nucleus is negligible, we know that the initial kinetic energy of the α particle will be
transformed into potential energy of the two-body system when the particles are at their
distance of closest approach.
(a) Apply conservation of energy to
the system consisting of the α particle
and the massive nucleus:
0=∆+∆ UK
or
0ifif =−+− UUKK
Because Kf = Ui = 0 and Ki = E: 0f =+− UE
Letting r be the separation of the
particles at closest approach, express
Uf:
( )( )
r
kZe
r
eZek
r
qkqU
2
nucleus
f
22 === α
Substitute to obtain:
02
2
=+−
r
kZeE
Solve for r to obtain:
E
kZer
22=
(b) For a 5.0-MeV α particle and a gold nucleus:
Chapter 23
180
( )( )( )
( )( ) fm45.4m1055.4J/eV106.1MeV5 C101.679/CmN108.992 1419
219229
5 =×=×
×⋅×= −−
−
r
For a 9.0-MeV α particle and a gold nucleus:
( )( )( )
( )( ) fm25.3m1053.2J/eV106.1MeV9 C101.679/CmN108.992 1419
219229
9 =×=×
×⋅×= −−
−
r
Potential Due to a System of Point Charges
28 •
Picture the Problem Let the numerals 1, 2, 3, and 4 denote the charges at the four
corners of square and r the distance from each charge to the center of the square. The
potential at the center of square is the algebraic sum of the potentials due to the four
charges.
Express the potential at the center of
the square:
( )
∑
=
=
+++=
+++=
4
1
4321
4321
i
iqr
k
qqqq
r
k
r
kq
r
kq
r
kq
r
kqV
(a) If the charges are positive: ( )( )
kV4.25
C24
m22
/CmN108.99 229
=
⋅×= µV
(b) If three of the charges are positive
and one is negative:
( )( )
kV7.12
C22
m22
/CmN108.99 229
=
⋅×= µV
(c) If two are positive and two are
negative:
0=V
29 •
Picture the Problem The potential at the point whose coordinates are (0, 3 m) is the
algebraic sum of the potentials due to the charges at the three locations given.
Electric Potential
181
Express the potential at the point
whose coordinates are (0, 3 m): ⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++== ∑
= 3
3
2
2
1
1
3
1 r
q
r
q
r
qk
r
qkV
i i
i
(a) For q1 = q2 = q3 = 2 µC:
( )( ) kV9.12
m53
1
m23
1
m3
1C2/CmN1099.8 229 =⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++⋅×= µV
(b) For q1 = q2 = 2 µC and q3 = −2 µC:
( )( ) kV55.7
m53
1
m23
1
m3
1C2/CmN1099.8 229 =⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+⋅×= µV
(c) For q1 = q3 = 2 µC and q2 = −2 µC:
( )( ) kV44.4
m53
1
m23
1
m3
1C2/CmN1099.8 229 =⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−⋅×= µV
30 •
Picture the Problem The potential at point C is the algebraic sum of the potentials due to
the charges at points A and B and the work required to bring a charge from infinity to
point C equals the change in potential energy of the system during this process.
(a) Express the potential at point C
as the sum of the potentials due to
the charges at points A and B:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
B
B
A
A
C r
q
r
qkV
Substitute numerical values and evaluate VC:
( )( ) kV0.12
m3
1
m3
1C2/CmN1099.811 229
BA
C =⎟⎟⎠
⎞
⎜⎜⎝
⎛ +⋅×=⎟⎟⎠
⎞
⎜⎜⎝
⎛ += µ
rr
kqV
(b) Express the required work in
terms of the change in the potential
energy of the system:
( )( ) mJ60.0kV12.0µC5
C5
==
=∆= VqUW
(c) Proceed as in (a) with qB = −2 µC:
Chapter 23
182
( ) 0
m3
C2
m3
C2/CmN1099.8 229
B
B
A
A
C =⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+⋅×=⎟⎟⎠
⎞
⎜⎜⎝
⎛ += µµ
r
q
r
qkV
and ( )( ) 00µC5C5 ===∆= VqUW
31 •
Picture the Problem The electric potential at the origin and at the north pole is the
algebraic sum of the potentials at those points due to the individual charges distributed
along the equator.
(a) Express the potential at the
origin as the sum of the potentials
due to the charges placed at 60°
intervals along the equator of the
sphere:
r
qk
r
qkV
i i
i 6
6
1
== ∑
=
Substitute numerical values and
evaluate V:
( )
kV270
m6.0
C3/CmN1099.86 229
=
⋅×= µV
(b) Using geometry, find the
distance from each charge to the
north pole:
m26.0'=r
Proceed as in (a) with m26.0'=r :
( )
kV191
m26.0
C3/CmN1099.86
'
6
229
6
1
'
=
⋅×=
== ∑
=
µ
r
qk
r
qkV
i i
i
*32 •
Picture the Problem We can use the fact that the electric potential at the point of interest
is the algebraic sum of the potentials at that point due to the charges q and q′ to find the
ratio q/q'.
Express the potential at the point of
interest as the sum of the potentials
due to
the two charges:
0
323
=+
a
kq'
a
kq
Electric Potential
183
Simplify to obtain: 0
2
=+ q'q
Solve for the ratio q/q':
2
1−=
q'
q
33 ••
Picture the Problem For the two charges, axr −= and ax + respectively and the
electric potential at x is the algebraic sum of the potentials at that point due to the charges
at x = +a and x = −a.
(a) Express V(x) as the sum of the
potentials due to the charges at
x = +a and x = −a:
⎟⎟⎠
⎞
⎜⎜⎝
⎛
++−= axaxkqV
11
(b) The following graph of V(x) versus x for kq = 1 and a = 1 was plotted using a
spreadsheet program:
0
2
4
6
8
10
-3 -2 -1 0 1 2 3
x (m)
V
(V
)
(c) At x = 0: 0=
dx
dV
and 0=−=
dx
dVEx
*34 ••
Picture the Problem For the two charges, axr −= and x respectively and the electric
potential at x is the algebraic sum of the potentials at that point due to the charges at x = a
and x = 0. We can use the graph and the function found in part (a) to identify the points at
which V(x) = 0. We can find the work needed to bring a third charge +e to the point
Chapter 23
184
ax 21= on the x axis from the change in the potential energy of this third charge.
Express the potential at x: ( ) ( ) ( )
ax
ek
x
ekxV −
−+= 23
The following graph of V(x) for ke = 1 and a =1 was plotted using a spreadsheet
program.
-15
-10
-5
0
5
10
15
20
25
-3 -2 -1 0 1 2 3
x (m)
V
(V
)
(b) From the graph we can see that
V(x) = 0 when:
∞±=x
Examining the function, we see that
V(x) is also zero provided:
023 =−− axx
For x > 0, V(x) = 0 when:
ax 3=
For 0 < x < a, V(x) = 0 when:
ax 6.0=
(c) Express the work that must be
done in terms of the change in
potential energy of the charge:
( )aqVUW 21=∆=
Evaluate the potential at ax 21= : ( ) ( ) ( )
a
ke
a
ke
a
ke
aa
ek
a
ekaV
246
23
2
1
2
12
1
=−=
−
−+=
Electric Potential
185
Substitute to obtain:
a
ke
a
keeW
222 =⎟⎠
⎞⎜⎝
⎛=
Computing the Electric Field from the Potential
35 •
Picture the Problem We can use the relationship Ex = − (dV/dx) to decide the sign of Vb
− Va and E = ∆V/∆x to find E.
(a) Because Ex = − (dV/dx), V is
greater for larger values of x. So:
positive. is ab VV −
(b) Express E in terms of Vb − Va and
the separation of points a and b:
x
VV
x
VE abx ∆
−=∆
∆=
Substitute numerical values and
evaluate Ex:
kV/m25.0
m4
V105 ==xE
*36 •
Picture the Problem Because Ex = −dV/dx, we can find the point(s) at which
Ex = 0 by identifying the values for x for which dV/dx = 0.
Examination of the graph indicates
that dV/dx = 0 at x = 4.5 m. Thus Ex =
0 at:
m5.4=x
37 •
Picture the Problem We can use V(x) = kq/x to find the potential V on the x axis at x =
3.00 m and at x = 3.01 m and E(x) = kq/r2 to find the electric field at
x = 3.00 m. In part (d) we can express the off-axis potential using V(x) = kq/r, where
22 yxr += .
(a) Express the potential on the x axis
as a function of x and q:
( )
x
kqxV =
Evaluate V at x = 3 m: ( ) ( )( )
kV99.8
m3
C3/CmN1099.8m3
229
=
⋅×= µV
Chapter 23
186
Evaluate V at x = 3.01 m: ( ) ( )( )
kV96.8
m01.3
C3/CmN1099.8m01.3
229
=
⋅×= µV
(b) The potential decreases as x
increases and:
kV/m00.3
m3.00m3.01
kV8.99kV8.96
=
−
−−=∆
∆−
x
V
(c) Express the Coulomb field as a
function of x:
( ) 2x
kqxE =
Evaluate this expression at
x = 3.00 m to obtain:
( ) ( )( )( )
kV/m00.3
m3
C3/CmN1099.8m3 2
229
=
⋅×= µE
in agreement with our result in (b).
(d) Express the potential at (x, y)
due to a point charge q at the origin:
( )
22
,
yx
kqyxV +=
Evaluate this expression at (3.00 m, 0.01 m):
( ) ( )( )( ) ( ) kV99.8m01.0m00.3
C3/CmN1099.8mm,0.01.003
22
229
=
+
⋅×= µV
For y << x, V is independent of y and the points (x, 0) and (x, y) are at the same potential,
i.e., on an equipotential surface.
38 •
Picture the Problem We can find the potential on the x axis at x = 3.00 m by expressing
it as the sum of the potentials due to the charges at the origin and at
x = 6 m. We can also express the Coulomb field on the x axis as the sum of the fields due
to the charges q1 and q2 located at the origin and at x = 6 m.
(a) Express the potential on the x
axis as the sum of the potentials due
to the charges q1 and q2 located at
the origin and at
x = 6 m:
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
2
2
1
1
r
q
r
qkxV
Electric Potential
187
Substitute numerical values and
evaluate V(3 m):
( ) ( )
0
m3
C3
m3
C3
/CmN1099.8 229
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+×
⋅×=
µµ
xV
(b) Express the Coulomb field on the
x axis as the sum of the fields due to
the charges q1 and q2 located at the
origin and at
x = 6 m:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+= 2
2
2
2
1
1
2
2
2
2
1
1
r
q
r
qk
r
kq
r
kqEx
Substitute numerical values and
evaluate E(3 m):
( )
( ) ( )
kV/m99.5
m3
C3
m3
C3
/CmN1099.8
22
229
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−×
⋅×=
µµ
xE
(c) Express the potential on the x
axis as the sum of the potentials due
to the charges q1 and q2 located at
the origin and at
x = 6 m:
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
2
2
1
1
r
q
r
qkxV
Substitute numerical values and evaluate
V(3.01 m):
( ) ( )
V9.59
m99.2
C3
m01.3
C3
/CmN1099.8m01.3 229
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+×
⋅×=
µµ
V
Compute −∆V/∆x:
( )m00.3
kV/m99.5
m3.00m3.01
0V59.9
xE
x
V
=
=
−
−−−=∆
∆−
39 •
Picture the Problem We can use the relationship Ey = − (dV/dy) to decide the sign of Vb
− Va and E = ∆V/∆y to find E.
(a) Because Ex = − (dV/dx), V is
smaller for larger values of y. So:
negative. is ab VV −
Chapter 23
188
(b) Express E in terms of Vb − Va and
the separation of points a and b:
y
VV
y
VE aby ∆
−=∆
∆=
Substitute numerical values and
evaluate Ey:
kV/m00.5
m4
V102 4 =×=yE
40 •
Picture the Problem Given V(x), we can find Ex from −dV/dx.
(a) Find Ex from −dV/dx: [ ]
kV/m00.3
30002000
−=
+−= x
dx
dEx
(b) Find Ex from −dV/dx: [ ]
kV/m00.3
30004000
−=
+−= x
dx
dEx
(c) Find Ex from −dV/dx: [ ]
kV/m00.3
30002000
=
−−= x
dx
dEx
(d) Find Ex from −dV/dx: [ ] 02000 =−−=
dx
dEx
41 ••
Picture the Problem We can express the potential at a general point on the x axis as the
sum of the potentials due to the charges at x = 0 and x = 1 m. Setting this expression
equal to zero will identify the points at which V(x) = 0. We can find the electric field at
any point on the x axis from Ex = −dV/dx.
(a) Express V(x) as the sum of the
potentials due to the point charges at x
= 0 and x = 1 m:
( ) ( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−=
−
−+=
1
3
1
3
x
q
x
qk
x
qk
x
kqxV
(b) Set V(x) = 0:
0
1
3 =⎟⎟⎠
⎞
⎜⎜⎝
⎛
−− x
q
x
qk
or
Electric Potential
189
0
1
31 =−− xx
For x < 0: ( ) m500.001
31 −=⇒=−−−− xxx
For 0 < x < 1: ( ) m250.001
31 =⇒=−−− xxx
Note also that:
( ) ±∞→→ xxV as0
(c) Evaluate V(x) for 0 < x < 1:
( ) ⎟⎠
⎞⎜⎝
⎛
−+=<< 1
310
x
q
x
qkxV
Apply Ex = −dV/dx to find Ex in this
region:
( )
( ) ⎥⎦
⎤⎢⎣
⎡
−+=
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
−+−=<<
22 1
31
1
310
xx
kq
x
q
x
qk
dx
dxEx
Evaluate this expression at
x = 0.25 m to obtain:
( ) ( ) ( )
( )kq
kqEx
2
22
m3.21
m75.0
3
m25.0
1m25.0
−=
⎥⎦
⎤⎢⎣
⎡ +=
Evaluate V(x) for x < 0:
( ) ⎟⎠
⎞⎜⎝
⎛
−+−=< xxkqxV 1
310
Apply Ex = −dV/dx to find Ex in this
region:
( )
( ) ⎥⎦
⎤⎢⎣
⎡
−+−=
⎥⎦
⎤⎢⎣
⎡
−+−=<
22 1
31
1
310
xx
kq
xxdx
dkqxEx
Evaluate this expression at x = −0.5 m to obtain:
( ) ( ) ( ) ( )kqkqEx 222 m67.2m5.1 3m5.0 1m5.0 −−=⎥⎦
⎤⎢⎣
⎡ +−−=−
(d) The following graph of V(x) for kq = 1 and a = 1 was plotted using a spreadsheet
Chapter 23
190
program:
-25
-20
-15
-10
-5
0
5
-2 -1 0 1 2 3
x (m)
V
(V
)
*42 ••
Picture the Problem Because V(x) and Ex are related through Ex = − dV/dx, we can find
V from E by integration.
Separate variables to obtain: ( )dxxdxEdV x kN/C0.2 3−=−=
Integrate V from V1 to V2 and x from
1 m to 2 m: ( )
( )[ ] m2m1441
3
kN/C0.2
kN/C0.2
2
1
2
1
x
dxxdV
x
x
V
V
−=
−= ∫∫
Simplify to obtain: kV50.712 −=−VV
43 ••
Picture the Problem Let r1 be the distance from (0, a) to (x, 0), r2 the distance from (0,
−a), and r3 the distance from (a, 0) to (x, 0). We can express V(x) as the sum of the
potentials due to the charges at (0, a), (0, −a), and (a, 0) and then find Ex from −dV/dx.
(a) Express V(x) as the sum of the
potentials due to the charges at (0, a),
(0, −a), and (a, 0):
( )
3
3
2
2
1
1
r
kq
r
kq
r
kqxV ++=
where q1 = q2 = q3 = q
At x = 0, the fields due to q1 and q2 cancel, so Ex(0) = −kq/a2; this is also obtained from
(b) if x = 0.
Electric Potential
191
As x→∞, i.e., for x >> a, the three charges appear as a point charge 3q, so
Ex = 3kq/x2; this is also the result one obtains from (b) for x >> a.
Substitute for the ri to obtain:
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−++=⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
−++++= axaxkqaxaxaxkqxV
12111
222222
(b) For x > a, x − a > 0 and: axax −=−
Use Ex = −dV/dx to find Ex:
( ) ( ) ( )2232222 212 ax kqax kqxaxaxkqdxdaxEx −++=⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−++−=>
For x < a, x − a < 0 and: ( ) xaaxax −=−−=−
Use Ex = −dV/dx to find Ex:
( ) ( ) ( )2232222 212 xa kqax kqxxaaxkqdxdaxEx −−+=⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−++−=<
Calculations of V for Continuous Charge Distributions
44 •
Picture the Problem We can construct Gaussian surfaces just inside and just outside the
spherical shell and apply Gauss’s law to find the electric field at these locations. We can
use the expressions for the electric potential inside and outside a spherical shell to find
the potential at these locations.
(a) Apply Gauss’s law to a spherical
Gaussian surface of radius r < 12 cm:
0
0
enclosed
S
==⋅∫ ∈QdAE
rr
because the charge resides on the outer
surface of the spherical surface. Hence
( ) 0cm12 =<rEr
Apply Gauss’s law to a spherical
Gaussian surface of radius
( )
0
24 ∈π
qrE =
Chapter 23
192
r > 12 cm: and
( ) 2
0
24
cm12
r
kq
r
qrE ==> ∈π
Substitute numerical values and evaluate ( )cm12>rE :
( ) ( )( )( ) kV/m24.6m0.12 C10/CmN108.99cm12 2
8229
=⋅×=>
−
rE
(b) Express and evaluate the potential just inside the spherical shell:
( ) ( )( ) V749
m0.12
C10/CmN108.99 8229 =⋅×==≤
−
R
kqRrV
Express and evaluate the potential just outside the spherical shell:
( ) ( )( ) V749
m0.12
C10/CmN108.99 8229 =⋅×==≥
−
r
kqRrV
(c) The electric potential inside a uniformly charged spherical shell is constant and
given by:
( ) ( )( ) V749
m0.12
C10/CmN108.99 8229 =⋅×==≤
−
R
kqRrV
In part (a) we showed that: ( ) 0cm12 =<rEr
45 •
Picture the Problem We can use the expression for the potential due to a line
charge
a
rkV ln2 λ−= , where V = 0 at some distance r = a, to find the potential at these
distances from the line.
Express the potential due to a line
charge as a function of the distance
from the line:
a
rkV ln2 λ−=
Because V = 0 at r = 2.5 m:
a
k m5.2ln20 λ−= ,
Electric Potential
193
a
m5.2ln0 = ,
and
10lnm5.2 1 == −
a
Thus we have a = 2.5 m and:
( )( ) ( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×−=⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×−=
m5.2
lnm/CN1070.2
m5.2
lnC/m5.1/CmN1099.82 4229 rrV µ
(a) Evaluate V at r = 2.0 m: ( )
kV02.6
m5.2
m2lnm/CN1070.2 4
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×−=V
(b) Evaluate V at r = 4.0 m: ( )
kV7.12
m5.2
m4lnm/CN1070.2 4
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×−=V
(c) Evaluate V at r = 12.0 m: ( )
kV3.42
m5.2
m12lnm/CN1070.2 4
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×−=V
46 ••
Picture the Problem The electric field on the x axis of a disk charge of radius R is given
by ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−= 2212 Rx
xkEx σπ . We’ll choose V(∞) = 0 and integrate from x′ = ∞ to x′ =
x to obtain Equation 23-21.
Relate the electric potential on the
axis of a disk charge to the electric
field of the disk:
dxEdV x−=
Express the electric field on the x
axis of a disk charge: ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−=
22
12
Rx
xkEx σπ
Chapter 23
194
Substitute to obtain:
dx
Rx
xkdV ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−−= 2212 σπ
Let V(∞) = 0 and integrate from x′ =
∞ to x′ = x:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+=
−+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−−= ∫∞
112
2
'
'12
2
2
22
22
x
Rxk
xRxk
dx'
Rx
xkV
x
σπ
σπ
σπ
which is Equation 23-21.
*47 ••
Picture the Problem Let the charge per
unit length be λ = Q/L and dy be a line
element with charge λdy. We can express
the potential dV at any point on the x axis
due to λdy and integrate of find V(x, 0).
(a) Express the element of potential
dV due to the line element dy:
dy
r
kdV λ=
where 22 yxr +=
Integrate dV from y = −L/2 to
y = L/2:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−+
++=
+
= ∫
−
24
24
ln
0,
22
22
2
2
22
LLx
LLx
L
kQ
yx
dy
L
kQxV
L
L
(b) Factor x from the numerator and
denominator within the parentheses to
obtain:
( )
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
−+
++
=
x
L
x
L
x
L
x
L
L
kQxV
24
1
24
1
ln0,
2
2
2
2
Use ba
b
a lnlnln −= to obtain:
( )
⎪⎭
⎪⎬
⎫
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+−
⎪⎩
⎪⎨
⎧
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
x
L
x
L
x
L
x
L
L
kQxV
24
1ln
24
1ln0, 2
2
2
2
Electric Potential
195
Let 2
2
4x
L=ε and use ( ) ...11 2812121 +−+=+ εεε to expand 2
2
4
1
x
L+ :
1...
48
1
42
11
4
1
2
2
2
2
221
2
2
≈+⎟⎟⎠
⎞
⎜⎜⎝
⎛−+=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
x
L
x
L
x
L
for x >> L.
Substitute to obtain:
( ) ⎭⎬
⎫⎟⎠
⎞⎜⎝
⎛ −−⎩⎨
⎧ ⎟⎠
⎞⎜⎝
⎛ +=
x
L
x
L
L
kQxV
2
1ln
2
1ln0,
Let
x
L
2
=δ and use ( ) ...1ln 221 +−=+ δδδ to expand ⎟⎠
⎞⎜⎝
⎛ ±
x
L
2
1ln :
2
2
422
1ln
x
L
x
L
x
L −≈⎟⎠
⎞⎜⎝
⎛ + and 2
2
422
1ln
x
L
x
L
x
L −−≈⎟⎠
⎞⎜⎝
⎛ − for x >> L.
Substitute and simplify to obtain:
( )
x
kQ
x
L
x
L
x
L
x
L
L
kQxV =
⎭⎬
⎫
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−−
⎩⎨
⎧ −= 2
2
2
2
4242
0,
48 ••
Picture the Problem We can find Q by
integrating the charge on a ring of radius r
and thickness dr from r = 0 to
r = R and the potential on the axis of the
disk by integrating the expression for the
potential
on the axis of a ring of charge
between the same limits.
(a) Express the charge dq on a ring
of radius r and thickness dr:
Rdr
dr
r
Rrdrrdq
0
0
2
22
πσ
σπσπ
=
⎟⎠
⎞⎜⎝
⎛==
Integrate from r = 0 to r = R to obtain: 2
0
0
0 22 RdrRQ
R
πσπσ == ∫
Chapter 23
196
(b) Express the potential on the axis
of the disk due to a circular element
of charge drrdq σπ2= :
22
02
' rx
Rdrk
r
kdqdV
+
== σπ
Integrate from r = 0 to r = R to obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
+
= ∫
x
RxRRk
rx
drRkV
R
22
0
0
220
ln2
2
σπ
σπ
49 ••
Picture the Problem We can find Q by
integrating the charge on a ring of radius r
and thickness dr from r = 0 to
r = R and the potential on the axis of the
disk by integrating the expression for the
potential on the axis of a ring of charge
between the same limits.
(a) Express the charge dq on a ring
of radius r and thickness dr:
drr
R
dr
R
rrdrrdq
3
2
0
2
2
0
2
22
πσ
σπσπ
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛==
Integrate from r = 0 to r = R to obtain: 2
02
1
0
3
2
02 Rdrr
R
Q
R
πσπσ == ∫
(b)Express the potential on the axis of
the disk due to a circular element of
charge drr
R
dq 32
02πσ= :
dr
rx
r
R
k
r
kdqdV
22
3
2
02
' +==
σπ
Integrate from r = 0 to r = R to obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++−=+= ∫ 3
2
3
222 32222
2
0
0
22
3
2
0 xRxxR
R
k
rx
drr
R
kV
R σπσπ
Electric Potential
197
50 ••
Picture the Problem Let the charge per
unit length be λ = Q/L and dy be a line
element with charge λdy. We can express
the potential dV at any point on the x axis
due to λdy and integrate to find V(x, 0).
Express the element of potential dV
due to the line element dy:
dy
r
kdV λ=
where 22 yxr +=
Integrate dV from y = −L/2 to
y = L/2:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−+
++=
+
= ∫
−
24
24
ln
0,
22
22
2
2
22
LLx
LLx
L
kQ
yx
dy
L
kQxV
L
L
*51 ••
Picture the Problem The potential at any
location on the axis of the disk is the sum
of the potentials due to the positive and
negative charge distributions on the disk.
Knowing that the total charge on the disk is
zero and the charge densities are equal in
magnitude will allow us to find the radius
of the region that is positively charged. We
can then use the expression derived in the
text to find the potential due to this charge
closest to the axis and integrate dV from
2Rr = to r = R to find the potential at x
due to the negative charge distribution.
(a) Express the potential at a
distance x along the axis of the disk
as the sum of the potentials due to
the positively and negatively
charged regions of the disk:
( ) ( ) ( )xVxVxV -+= +
We know that the charge densities
are equal in magnitude and that the
arar QQ >< =
or
Chapter 23
198
total charge carried by the disk is
zero. Express this condition in terms
of the charge in each of two regions
of the disk:
2
0
2
0
2
0 aRa πσπσπσ −=
Solve for a to obtain:
2
Ra =
Use this result and the general
expression for the potential on the
axis of a charged disk to express
V+(x):
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+=+ xRxkxV 22
2
2
0σπ
Express the potential on the axis of
the disk due to a ring of charge a
distance r > a from the axis of the
ring:
( ) dr
r
rkxdV
'
2 0σπ−=−
where 22' rxr += .
Integrate this expression from
2Rr = to r = R to obtain:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−+−=
+−= ∫−
2
2
2
2
222
0
2
220
RxRxk
dr
rx
rkxV
R
R
σπ
σπ
Substitute and simplify to obtain:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+−+=
⎟⎟⎠
⎞+++−⎜⎜⎝
⎛ −+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−+−⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+=
xRxRxk
RxRxxRxk
RxRxkxRxkxV
22
2
2
0
2
222
2
2
0
2
222
0
2
2
0
2
22
22
2
2
2
2
2
σπ
σπ
σπσπ
(b) To determine V for
x >> R, factor x from the square roots
and expand using the binomial
expansion:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+
4
4
2
2
21
2
22
2
324
1
2
1
2
x
R
x
Rx
x
RxRx
and
Electric Potential
199
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+
4
4
2
2
21
2
2
22
82
1
1
x
R
x
Rx
x
RxRx
Substitute to obtain:
( ) 3
4
0
4
4
2
2
4
4
2
2
0 882
1
324
122
x
Rkx
x
R
x
Rx
x
R
x
RxkxV σπσπ =⎟⎟⎠
⎞−⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+−⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+≈
52 ••
Picture the Problem Given the potential function
( ) ( )xRxRxkxV −+−+= 22220 222 σπ found in Problem 51(a), we can find Ex
from −dV/dx. In the second part of the problem, we can find the electric field on the axis
of the disk by integrating Coulomb’s law for the oppositely charged regions of the disk
and expressing the sum of the two fields.
Relate Ex to dV/dx:
dx
dVEx −=
From Problem 51(a) we have:
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+−+= xRxRxkxV 22
2
2
0 2
22 σπ
Evaluate the negative of the derivative of V(x) to obtain:
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
−+−+
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+−+−=
1
2
22
2
22
222
2
0
22
2
2
0
Rx
x
Rx
xk
xRxRx
dx
dkEx
σπ
σπ
Express the field on the axis of the
disk as the sum of the field due to
the positive charge on the disk and
the field due to the negative charge
+− += xxx EEE
Chapter 23
200
on the disk:
The field due to the positive charge
(closest to the axis) is:
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
+
−=+
2
12
2
2
0
Rx
xkEx σπ
To determine Ex− we integrate the
field due to a ring charge:
( )
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
+−+
−=
+−= ∫−
222
2
0
2
23220
2
2
2
Rx
x
Rx
xk
rx
rdrkE
R
R
x
σπ
σπ
Substitute and simplify to obtain:
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
−+−+
−=
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
+
−+
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
+−+
−=
1
2
22
2
12
2
2
222
2
0
2
2
0222
2
0
Rx
x
Rx
xk
Rx
xk
Rx
x
Rx
xkEx
σπ
σπσπ
53 ••
Picture the Problem We can express the electric potential dV at x due to an elemental
charge dq on the rod and then integrate over the length of the rod to find V(x). In the
second part of the problem we use a binomial expansion to show that, for x >> L/2, our
result reduces to that due to a point charge Q.
Electric Potential
201
(a) Express the potential at x due to
the element of charge dq located at
u:
ux
duk
r
kdqdV −==
λ
or, because λ = Q/L,
ux
du
L
kQdV −=
Integrate V from u = −L/2 to L/2 to
obtain: ( )
( )
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
+
=
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ ++⎟⎠
⎞⎜⎝
⎛ −−=
−=
−=
−
−
∫
2
2ln
2
ln
2
ln
ln 2
2
2
2
Lx
Lx
L
kQ
LxLx
ux
L
kQ
ux
du
L
kQxV
L
L
L
L
(b) Divide the numerator and
denominator of the argument of the
logarithm by x to obtain:
⎟⎠
⎞⎜⎝
⎛
−
+=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
+
=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
+
a
a
x
L
x
L
Lx
Lx
1
1ln
2
1
2
1
ln
2
2ln
where a = L/2x.
Divide 1 + a by 1 − a to obtain:
⎟⎠
⎞⎜⎝
⎛ +≈
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
−
++=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−++=⎟⎠
⎞⎜⎝
⎛
−
+
x
L
x
L
x
L
x
L
a
aa
a
a
1ln
2
1ln
1
221ln
1
1ln
2
2
2
provided x >> L/2.
Expand ln(1 + L/x) binomially to
obtain: x
L
x
L ≈⎟⎠
⎞⎜⎝
⎛ +1ln
provided x >> L/2.
Substitute to express V(x) for
x >> L/2:
( )
x
kQ
x
L
L
kQxV == , the field due to a
Chapter 23
202
point charge Q.
54 ••
Picture the Problem The diagram is a
cross-sectional view showing the charges
on the sphere and the spherical conducting
shell. A portion of the Gaussian surface
over which we’ll integrate E in order to
find V in the region r > b is also shown. For
a < r < b, the sphere acts like point charge
Q and the potential of the metal sphere is
the sum of the potential due to a point
charge at its center and the potential at its
surface due to the charge on the inner
surface of the spherical shell.
(a) Express Vr > b: ∫ >> −= drEV brbr
Apply Gauss’s law for r > b: 0ˆ
0
enclosed
S
==⋅∫ εQdAr nE
r
and Er>b = 0 because Qenclosed = 0 for
r > b.
Substitute to obtain:
( ) 00 =−= ∫> drV br
(b) Express the potential of the metal
sphere:
surfacecenter itsat VVV Qa +=
Express the potential at the surface
of the metal sphere:
( )
b
kQ
b
QkV −=−=surface
Substitute and simplify to obtain: ⎟⎠
⎞⎜⎝
⎛ −=−=
ba
kQ
b
kQ
a
kQVa
11
Electric Potential
203
55 ••
Picture the Problem The diagram is a
cross-sectional view showing the charges
on the inner and outer conducting shells. A
portion of the Gaussian surface over which
we’ll integrate E in order to find V in the
region a < r < b is also shown. Once
we’ve determined how E varies with r, we
can find Vb – Va from ∫−=− drEVV rab .
Express the potential difference
Vb – Va:
∫−=− drEVV rab
Apply Gauss’s law to cylindrical
Gaussian surface of radius r and
length L:
( )
0
S
2ˆ επ
qrLEdA r ==⋅∫ nEr
Solve for Er:
rL
qEr
02πε=
Substitute for Er and integrate from r
= a to b:
⎟⎠
⎞⎜⎝
⎛−=
−=− ∫
a
b
L
kq
r
dr
L
qVV
b
a
ab
ln2
2 0πε
56 ••
Picture the Problem Let R be the radius of the sphere and Q its charge. We can express
the potential at the two locations given and solve the resulting equations simultaneously
for R and Q.
Relate the potential of the sphere at
its surface to its radius:
V450=
R
kQ
(1)
Express the potential at a distance of
20 cm from its surface:
V150
m2.0
=+R
kQ
(2)
Chapter 23
204
Divide equation (1) by equation (2)
to obtain:
V150
V450
m2.0
=
+R
kQ
R
kQ
or
3m2.0 =+
R
R
Solve for R to obtain:
m100.0=R
Solve equation (1) for Q: ( )
k
RQ V450=
Substitute numerical values and evaluate
Q:
( ) ( )( )
nC01.5
/CmN108.99
m0.1V450 229
=
⋅×=Q
57 ••
Picture the Problem Let the charge
density on the infinite plane at x = a be σ1
and that on the infinite plane at x = 0 be σ2.
Call that region in space for which x < 0,
region I, the region for which 0 < x < a
region II, and the region for which a < x
region III. We can integrate E due to the
planes of charge to find the electric
potential in each of these regions.
(a) Express the potential in region I
in terms of the electric field in that
region:
∫ ⋅−= x dV
0
II xE
rr
Express the electric field in region I
as the sum of the fields due to the
charge densities σ1 and σ2:
i
iiiiE
ˆ
ˆ
2
ˆ
2
ˆ
2
ˆ
2
0
000
2
0
1
I
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
−=
−−=−−=r
Electric Potential
205
Substitute and evaluate VI: ( )
xx
VxdxV
x
00
00 0
I
0
0
∈
σ
∈
σ
∈
σ
∈
σ
=+=
+=⎟⎟⎠
⎞
⎜⎜⎝
⎛−−= ∫
Express the potential in region II in
terms of the electric field in that
region:
( )0IIII VdV +⋅−= ∫ xE rr
Express the electric field in region II
as the sum of the fields due to the
charge densities σ1 and σ2:
0
ˆ
2
ˆ
2
ˆ
2
ˆ
2 000
2
0
1
II
=
+−=+−= iiiiE ∈
σ
∈
σ
∈
σ
∈
σr
Substitute and evaluate VII: ( ) ( ) 0000
0
II =+=−= ∫ VdxV x
Express the potential in region III in
terms of the electric field in that
region:
∫ ⋅−= x
a
dV xE r
r
IIIIII
Express the electric field in region
III as the sum of the fields due to the
charge densities σ1 and σ2:
i
iiiiE
ˆ
ˆ
2
ˆ
2
ˆ
2
ˆ
2
0
000
2
0
1
III
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
=
+=+=r
Substitute and evaluate VIII:
( )xa
axdxV
x
a
−=
+−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−= ∫
0
000
III
∈
σ
∈
σ
∈
σ
∈
σ
(b) Proceed as in (a) with σ1 = −σ and
σ2 = σ to obtain:
0I =V ,
xV
0
II ∈
σ−= and aV
0
III ∈
σ−=
*58 ••
Picture the Problem The potential on the axis of a disk charge of radius R and charge
density σ is given by ( )[ ]xRxkV −+= 21222 σπ .
Chapter 23
206
Express the potential on the axis of the
disk charge:
( )[ ]xRxkV −+= 21222 σπ
Factor x from the radical and use the binomial expansion to obtain:
( )
⎥⎦
⎤⎢⎣
⎡ −+≈
⎥⎦
⎤+⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛+⎢⎣
⎡ +=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+
4
4
2
2
4
4
2
221
2
2
2122
82
1
...
2
1
2
1
2
1
2
11
x
R
x
Rx
x
R
x
Rx
x
RxRx
Substitute for the radical term to
obtain:
x
kQ
x
Rk
x
R
x
Rk
x
x
R
x
RxkV
=⎟⎟⎠
⎞
⎜⎜⎝
⎛≈
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
⎭⎬
⎫
⎩⎨
⎧ −⎥⎦
⎤⎢⎣
⎡ −+=
2
2
82
2
82
12
2
3
42
4
4
2
2
σπ
σπ
σπ
provided x >> R.
59 ••
Picture the Problem The diagram shows a
sphere of radius R containing a charge Q
uniformly distributed. We can use the
definition of density to find the charge q′
inside a sphere of radius r and the potential
V1 at r due to this part of the charge. We
can express the potential dV2 at r due to the
charge in a shell of radius r′ and thickness
dr′ at r′ > r using rkdq'dV =2 and then
integrate this expression from r′ = r to r′ =
R to find V2.
(a) Express the potential V1 at r due to
q′:
r
kq'V =1
Use the definition of density and the
fact that the charge density is uniform
to relate q′ to Q:
3
3
43
3
4 R
Q
r
q'
ππρ ==
Electric Potential
207
Solve for q′: Q
R
rq' 3
3
=
Substitute to express V1: 2
33
3
1 rR
kQQ
R
r
r
kV =⎟⎟⎠
⎞
⎜⎜⎝
⎛=
(b) Express the potential dV2 at r due
to the charge in a shell of radius r′
and thickness dr′ at
r′ > r:
r
kdq'dV =2
Express the charge dq′ in a shell of
radius r′ and thickness dr′ at
r′ > r:
dr'r'
R
Q
dr'
R
Qr'drr'dq'
2
3
3
22
3
4
34'4
=
⎟⎠
⎞⎜⎝
⎛== ππρπ
Substitute to obtain: r'dr'
R
kQdV 32
3=
(c) Integrate dV2 from r′ = r to
r′ = R to find V2:
( )22332 233 rRRkQr'dr'RkQV
R
r
−== ∫
(d) Express the potential V at r as the
sum of V1 and V2: ( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
−+=
+=
2
2
22
3
2
3
21
3
2
2
3
R
r
R
kQ
rR
R
kQr
R
kQ
VVV
60 •
Picture the Problem We can equate the expression for the electric field due to an infinite
plane of charge and −∆V/∆x and solve the resulting equation for the separation of the
equipotential surfaces.
Express
the electric field due to the
infinite plane of charge:
02∈
σ=E
Relate the electric field to the
potential:
x
VE ∆
∆−=
Chapter 23
208
Equate these expressions and solve
for ∆x to obtain:
σ
∈ Vx ∆=∆ 02
Substitute numerical values and
evaluate x∆ :
( )( )
mm0.506
µC/m3.5
V100m/NC108.852
2
2212
=
⋅×=∆
−
x
61 •
Picture the Problem The equipotentials are spheres centered at the origin with radii ri =
kq/Vi.
Evaluate r for V = 20 V: ( )( )
m499.0
V20
C10/CmN108.99 891
229
V20
=
×⋅×=
−
r
Evaluate r for V = 40 V: ( )( )
m250.0
V40
C10/CmN108.99 891
229
V40
=
×⋅×=
−
r
Evaluate r for V = 60 V: ( )( )
m166.0
V60
C10/CmN108.99 891
229
V60
=
×⋅×=
−
r
Evaluate r for V = 80 V: ( )( )
m125.0
V80
C10/CmN108.99 891
229
V80
=
×⋅×=
−
r
Evaluate r for V = 100 V: ( )( )
m0999.0
V100
C10/CmN108.99 891
229
V100
=
×⋅×=
−
r
Electric Potential
209
The equipotential surfaces are
shown in cross-section to the right:
spaced.equally not are surfaces ialequipotent The
62 •
Picture the Problem We can relate the dielectric strength of air (about 3 MV/m) to the
maximum net charge that can be placed on a spherical conductor using the expression for
the electric field at its surface. We can find the potential of the sphere when it carries its
maximum charge using RkQV max= .
(a) Express the dielectric strength of
a spherical conductor in terms of the
charge on the sphere:
2
max
breakdown R
kQE =
Solve for Qmax:
k
REQ
2
breakdown
max =
Substitute numerical values and
evaluate Qmax:
( )( )
C54.8
/CmN108.99
m0.16MV/m3
229
2
max
µ=
⋅×=Q
(b) Because the charge carried by the
sphere could be either positive or
negative:
( )( )
kV480
m16.0
C54.8/CmN1099.8 229
max
max
±=
⋅×±=
±=
µ
R
kQV
*63 •
Picture the Problem We can solve the equation giving the electric field at the surface of
a conductor for the greatest surface charge density that can exist before dielectric
breakdown of the air occurs.
Relate the electric field at the surface
of a conductor to the surface charge
density:
0∈
σ=E
Chapter 23
210
Solve for σ under dielectric
breakdown of the air conditions:
breaddown0max E∈σ =
Substitute numerical values and
evaluate σmax:
( )( )
2
2212
max
C/m6.26
MV/m3m/NC108.85
µ
σ
=
⋅×= −
64 ••
Picture the Problem Let L and S refer to the larger and smaller spheres, respectively.
We can use the fact that both spheres are at the same potential to find the electric fields
near their surfaces. Knowing the electric fields, we can use E0=∈σ to find the surface
charge density of each sphere.
Express the electric fields at the
surfaces of the two spheres:
2
S
S
S R
kQE = and 2
L
L
L R
kQE =
Divide the first of these equations by
the second to obtain:
2SL
2
LS
2
L
L
2
S
S
L
S
RQ
RQ
R
kQ
R
kQ
E
E ==
Because the potentials are equal at the
surfaces of the spheres:
S
S
L
L
R
kQ
R
kQ = and
L
S
L
S
R
R
Q
Q =
Substitute to obtain:
S
L
2
SL
2
LS
L
S
R
R
RR
RR
E
E ==
Solve for ES: ( )
kV/m480
kV/m200
cm5
cm12
L
S
L
S
=
== E
R
RE
Use E0∈σ = to find the surface charge density of each sphere:
( )( ) 22212cm120cm12 C/m77.1kV/m200m/NC108.85 µ∈σ =⋅×== −E
and ( )( ) 22212cm50cm5 C/m25.4kV/m804m/NC108.85 µ∈σ =⋅×== −E
Electric Potential
211
65 ••
Picture the Problem The diagram is a
cross-sectional view showing the charges
on the concentric spherical shells. The
Gaussian surface over which we’ll
integrate E in order to find V in the region r
≥ b is also shown. We’ll also find E in the
region for which a < r < b. We can then
use the relationship ∫−= EdrV to find Va
and Vb and their difference.
Express Vb: ∫
∞
≥−=
b
arb drEV
Apply Gauss’s law for r ≥ b: 0ˆ
0
enclosed
S
==⋅∫ ∈QdAr nE
r
and Er≥b = 0 because Qenclosed = 0 for
r ≥ b.
Substitute to obtain:
( ) 00 =−= ∫∞
b
b drV
Express Va: ∫ ≥−= a
b
ara drEV
Apply Gauss’s law for r ≥ a: ( )
0
24 ∈π
qrE ar =≥
and
22
04 r
kq
r
qE ar ==≥ ∈π
Substitute to obtain:
b
kq
a
kq
r
drkqV
a
b
a −=−= ∫ 2
The potential difference between the
shells is given by: ⎟⎠
⎞⎜⎝
⎛ −==−
ba
kqVVV aba
11
*66 •••
Picture the Problem We can find the potential relative to infinity at the center of the
sphere by integrating the electric field for 0 to ∞. We can apply Gauss’s law to find the
Chapter 23
212
electric field both inside and outside the spherical shell.
The potential relative to infinity the
center of the spherical shell is:
drEdrEV
R
Rr
R
Rr ∫∫ ∞ >< +=
0
(1)
Apply Gauss’s law to a spherical
surface of radius r < R to obtain:
( )
0
inside2
S n
4 ∈== <∫ QrEdAE Rr π
Using the fact that the sphere is
uniformly charged, express Qinside in
terms of Q:
3
3
43
3
4
inside
R
Q
r
Q
ππ = ⇒ QR
rQ 3
3
inside =
Substitute for Qinside to obtain:
( ) Q
R
rrE Rr 3
0
3
24 ∈=< π
Solve for Er < R:
r
R
kQQ
R
rE Rr 33
04
=∈=< π
Apply Gauss’s law to a spherical
surface of radius r > R to obtain:
( )
00
inside2
S n
4 ∈=∈== >∫ QQrEdAE Rr π
Solve for Er>R to obtain:
2204 r
kQ
r
QE Rr =∈=> π
Substitute for Er<R and Er>R in
equation (1) and evaluate the
resulting integral:
R
kQ
r
kQr
R
kQ
r
drkQdrr
R
kQV
R
R
R
R
2
31
2 0
2
3
2
0
3
=⎥⎦
⎤⎢⎣
⎡−+⎥⎦
⎤⎢⎣
⎡=
+=
∞
∞∫∫
67 ••
Picture the Problem
(a) The field lines are shown on the figure.
The charged spheres induce charges of
opposite sign on the spheres near them so
that sphere 1 is negatively charged, and
sphere 2 is positively charged. The total
charge of the system is zero.
(b)
. that followsit lines field electric the
ofdirection theFrom connected. are spheres thebecause
13
21
VV
VV
>
=
Electric Potential
213
(c)
zero. is sphereeach on
charge ly theConsequent zero. are potentials all if satisfied beonly
can )(part of conditions theand connected, are 4 and 3 If 43 bVV =
General Problems
68 •
Picture the Problem Because the charges at either end of the electric dipole are point
charges, we can use the expression for the Coulomb potential to find the field at any
distance from the dipole charges.
Using the expression for the potential
due to a system of point charges,
express the potential at the point
9.2×10−10 m from each of the two
charges:
( )−+
−+
+=
+=
qq
d
k
d
kq
d
kqV
Because q+ = −q−: 0=+ −+ qq , 0=V and correct. is )(b
69 •
Picture the Problem The potential V at
any point on the x axis is the sum of the
Coulomb potentials due to the two point
charges. Once we have found V, we can
use Vgrad−=Er to find the electric field
at any point on the x axis.
(a) Express the potential due to a system of
point charges: ∑= i i ir
kqV
Substitute to obtain: ( )
22
2222
-at chargeat charge
2
ax
kq
ax
kq
ax
kq
VVxV aa
+=
+++=
+= +
Chapter 23
214
(b) The electric field at any point on
the x axis
is given by:
( )
( ) i
iE
ˆ2
ˆ2grad
2322
22
ax
kqx
ax
kq
dx
dVx
+=
⎥⎦
⎤⎢⎣
⎡
+−=−=
r
70 •
Picture the Problem The radius of the sphere is related to the electric field and the
potential at its surface. The dielectric strength of air is about 3 MV/m.
Relate the electric field at the surface
of a conducting sphere to the potential
at the surface of the sphere:
( )
r
rVEr =
Solve for r: ( )
rE
rVr =
When E is a maximum, r is a
minimum:
( )
max
min E
rVr =
Substitute numerical values and
evaluate rmin:
mm3.33
MV/m3
V104
min ==r
*71 ••
Picture the Problem The geometry of the
wires is shown to the right. The potential at
the point whose coordinates are (x, y) is the
sum of the potentials due to the charge
distributions on the wires.
(a) Express the potential at the point
whose coordinates are
(x, y):
( )
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+⎟⎟⎠
⎞
⎜⎜⎝
⎛=
+= −
1
2
0
2
ref
1
ref
2
ref
1
ref
at wireat wire
ln
2
lnln2
ln2ln2
,
r
r
r
r
r
rk
r
rk
r
rk
VVyxV aa
π
λ
λ
λλ
where V(0) = 0.
Electric Potential
215
Because ( ) 221 yaxr ++= and
( ) :222 yaxr +−=
( ) ( )( ) ⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
++
+−
∈= 22
22
0
ln
2
,
yax
yax
yxV π
λ
On the y-axis, x = 0 and:
( )
( ) 01ln
2
ln
2
,0
0
22
22
0
=∈=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
+
∈=
π
λ
π
λ
ya
ya
yV
(b) Evaluate the potential at ( ) ( ) :0cm,25.10,41 =a ( ) ( )( )
⎟⎠
⎞⎜⎝
⎛
∈=
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
+
−
∈=
5
3ln
2
ln
2
0,
0
2
4
1
2
4
1
0
4
1
π
λ
π
λ
aa
aa
aV
Equate V(x,y) and ( )0,41 aV :
( )
( ) 22
22
5
5
5
3
yx
yx
++
+−=
Solve for y to obtain: 2525.21 2 −−±= xxy
A spreadsheet program to plot 2525.21 2 −−±= xxy is shown below. The formulas
used to calculate the quantities in the columns are as follows:
Cell Content/Formula Algebraic Form
A2 1.25 a41
A3 A2 + 0.05 x + ∆x
B2 SQRT(21.25*A2 − A2^2 − 25) 2525.21 2 −−= xxy
B4 −SQRT(21.25*A2 − A2^2 − 25) 2525.21 2 −−−= xxy
A B C
1 x y_pos y_neg
2 1.25 0.00 0.00
3 1.30 0.97 −0.97
4 1.35 1.37 −1.37
5 1.40 1.67 −1.67
6 1.45 1.93 −1.93
7 1.50 2.15 −2.15
370 19.65 2.54 −2.54
371 19.70 2.35 −2.35
372 19.75 2.15 −2.15
Chapter 23
216
373 19.80 1.93 −1.93
374 19.85 1.67 −1.67
375 19.90 1.37 −1.37
376 19.95 0.97 −0.97
The following graph shows the equipotential curve in the xy plane for
( ) ⎟⎠
⎞⎜⎝
⎛
∈= 5
3ln
2
0,
0
4
1
π
λaV .
-10
-8
-6
-4
-2
0
2
4
6
8
10
0 5 10 15 20
x (cm)
y
(c
m
)
72 ••
Picture the Problem We can use the expression for the potential at any point in the xy
plane to show that the equipotential curve is a circle.
(a) Equipotential surfaces must satisfy
the condition: ⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈= 1
2
0
ln
2 r
rV π
λ
Solve for r2/r1:
Ce
r
r V ==
∈
λ
π 02
1
2 or 12 Crr =
where C is a constant.
Substitute for r1 and r2 to obtain: ( ) ( )[ ]22222 yaxCyax ++=+−
Expand this expression, combine like
terms, and simplify to obtain:
22
2
2
2
1
12 ayx
C
Cax −=+−
++
Complete the square by adding ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
+ 2
2
2
2
1
1
C
Ca to both sides of the equation:
Electric Potential
217
( )22
22
2
2
2
2
22
2
2
2
2
2
2
2
1
4
1
1
1
1
1
12 −=−⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
+=+⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
++−
++
C
Caa
C
Cay
C
Cax
C
Cax
Let
1
12 2
2
−
+=
C
Caα and
1
2 2 −= C
Caβ
to obtain:
( ) ,222 βα =++ yx the equation of
circle in the xy plane with its center at
(−α,0).
(b) wires. the toparallel cylinders are surfaces ldimensiona- threeThe
73 ••
Picture the Problem Expressing the charge dq in a spherical shell of volume 4πr2dr
within a distance r of the proton and setting the integral of this expression equal to e will
allow us to solve for the value of ρ0 needed for charge neutrality. In part (b), we can use
the given charge density to express the potential function due to this charge and then
integrate this function to find V as a function of r.
Express the charge dq in a spherical
shell of volume 4πr2dr within a
distance r of the proton:
( )( )
drer
drredVdq
ar
ar
22
0
22
0
4
4
−
−
=
==
πρ
πρρ
Express the condition for charge
neutrality:
drere ar2
0
2
04
−
∞∫= πρ
Integrate by parts twice to obtain:
3
0
3
0 4
4 aae πρπρ ==
Solve for ρ0:
30 a
e
πρ =
74 •
Picture the Problem Let Q be the sphere’s charge, R its radius, and n the number of
electrons that have been removed. Then neQ = , where e is the electronic charge. We can
use the expression for the Coulomb potential of the sphere to express Q and then
neQ = to find n.
Letting n be the number of electrons
that have been removed, express the
sphere’s charge Q in terms of the
electronic charge e:
neQ =
Solve for n:
e
Qn = (1)
Chapter 23
218
Relate the potential of the sphere to
its charge and radius:
R
kQV =
Solve for the sphere’s charge:
k
VRQ =
Substitute in equation (1) to obtain:
ke
VRn =
Substitute numerical values and evaluate n:
( )( )( )( ) 1019229 1039.1C101.6/CmN108.99 m0.05V400 ×=×⋅×= −n
75 •
Picture the Problem We can use conservation of energy to relate the change in the
kinetic energy of the particle to the change in potential energy of the charge-and-particle
system as the particle moves from x = 1.5 m to x = 1 m. The change in potential energy
is, in turn, related to the change in electric potential.
Apply conservation of energy to the
point charge Q and particle system:
0=∆+∆ UK
or, because Ki = 0,
0iff =∆+ UK
Solve for Kf: iff UK ∆−=
Relate the difference in potential
between points i and f to the change in
potential energy of the system as the
body whose charge is q moves from i
to f:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
−−=∆−=∆
ifif
ififif
11
xx
kqQ
x
kQ
x
kQq
VVqVqU
Substitute to obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
if
f
11
xx
kqQK
Electric Potential
219
Solve for Q:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
−=
if
f
11
xx
kq
KQ
Substitute numerical values and evaluate Q:
( )( ) C0.20
m1.5
1
m1
1µC4/CmN108.99
J0.24
229
µ−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⋅×
−=Q
*76 ••
Picture the Problem We can use the definition of power and the expression for the work
done in moving a charge through a potential difference to find the minimum power
needed to drive the moving belt.
Relate the power need to drive the
moving belt to the rate at which the
generator is doing work:
dt
dWP =
Express the work done in moving a
charge q through a potential
difference ∆V:
VqW ∆=
Substitute to obtain: [ ]
dt
dqVVq
dt
dP ∆=∆=
Substitute numerical values and
evaluate P:
( )( ) W250C/s200MV25.1 == µP
77 ••
Picture the Problem We can use fiq VqW →→ ∆=position final to find the work required to
move these charges between the given points.
(a) Express the required work in
terms of the charge being moved and
the potential due
to the charge at x =
+a:
( ) ( )[ ]
( )
a
kQ
a
kQQaQV
VaVQ
VQW aaQ
22
2
=⎟⎠
⎞⎜⎝
⎛==
∞−=
∆= +→∞+→+
Chapter 23
220
(b) Express the required work in
terms of the charge being moved and
the potentials due to the charges at
x = +a and x = −a:
( ) ( )[ ]
( )[ ]
a
kQ
a
kQ
a
kQQ
VVQ
QV
VVQ
VQW
aa
Q
2
at charge-at charge
00
2
0
0
−=⎟⎠
⎞⎜⎝
⎛ +−=
+−=
−=
∞−−=
∆−=
+
→∞→−
(c) Express the required work in
terms of the charge being moved and
the potentials due to the charges at
x = +a and x = −a:
( ) ( )[ ]
( )[ ]
a
kQ
a
kQ
a
kQ
a
kQQ
VVVQ
VaVQ
VQW
aa
aaQ
3
2
2
3
0
02
2
at charge-at charge
202
=
⎟⎠
⎞⎜⎝
⎛ −+−=
−+−=
−−=
∆−=
+
→→−
78 ••
Picture the Problem Let q represent the charge being moved from x = 50 cm to the
origin, Q the ring charge, and a the radius of the ring. We can use
fiq VqW →→ ∆=position final , where V is the expression for the axial field due to a ring
charge, to find the work required to move q from x = 50 cm to the origin.
Express the required work in terms of
the charge being moved and the
potential due to the ring charge at
x = 50 cm and x = 0:
( ) ( )[ ]m5.00 VVq
VqW
−=
∆=
The potential on the axis of a
uniformly charged ring is:
( )
22 ax
kQxV +=
Evaluate V(0): ( )
( )( )
V180
m0.1
nC2/CmN1099.8
0
229
2
=
⋅×=
=
a
kQV
Evaluate V(0.5 m): ( ) ( )( )( ) ( )
V3.35
m0.1m5.0
nC2/CmN1099.80
22
229
=
+
⋅×=V
Electric Potential
221
Substitute in the expression for W to
obtain:
( )( )
eV1006.9
J101.6
eV1J1045.1
J1045.1
V35.3V180nC1
11
19
7
7
×=
×××=
×=
−=
−
−
−
W
79 ••
Picture the Problem We can find the speed of the proton as it strikes the negatively
charged sphere from its kinetic energy and, in turn, its kinetic energy from the potential
difference through which it is accelerated.
Use the definition of kinetic energy
to express the speed of the proton
when it strikes the negatively
charged sphere:
p
p2
m
K
v = (1)
Use the work-kinetic energy
theorem to relate the kinetic energy
of the proton to the potential
difference through which it is
accelerated:
if KKKW −=∆=
or, because Ki = 0 and Kf = Kp,
pKKW =∆=
Express the work done on the proton
in terms of its charge e and the
potential difference ∆V between the
spheres:
VeW ∆=
Substitute to obtain: VeK ∆=p
Substitute in equation (1) to obtain:
p
2
m
Vev ∆=
Substitute numerical values and
evaluate v:
( )( )
m/s1038.1
kg101.67
V100C101.62
5
27
19
×=
×
×= −
−
v
Chapter 23
222
80 ••
Picture the Problem Equation 23-20 is 22 xakQV += .
(a) A spreadsheet solution is shown below for kQ = a = 1. The formulas used to calculate
the quantities in the columns are as follows:
Cell Content/Formula Algebraic Form
A4 A3 + 0.1 x + ∆x
B3 1/(1+A3^2)^(1/2)
22 xa
kQ
+
A B
1
2 x V(x)
3 −5.0 0.196
4 −4.8 0.204
5 −4.6 0.212
6 −4.4 0.222
7 −4.2 0.232
8 −4.0 0.243
9 −3.8 0.254
49 4.2 0.232
50 4.4 0.222
51 4.6 0.212
52 4.8 0.204
53 5.0 0.196
The following graph shows V as a function of x:
0.2
0.4
0.6
0.8
1.0
-5 -4 -3 -2 -1 0 1 2 3 4 5
x (arbit rary unit s)
V
(V
)
Electric Potential
223
(b) Examining the graph we see that
the maximum value of V occurs
where:
0=x
Because E = −dV/dx, examination of
the graph tells us that:
( ) 00 =E
81 ••
Picture the Problem Let R2 be the radius of the second sphere and Q1 and Q2 the charges
on the spheres when they have been connected by the wire. When the spheres are
connected, the charge initially on the sphere of radius R1 will redistribute until the
spheres are at the same potential.
Express the common potential of the
spheres when they are connected: 1
1kV12
R
kQ= (1)
and
2
2kV12
R
kQ= (2)
Express the potential of the first
sphere before it is connected to the
second sphere:
( )
1
21kV20
R
QQk += (3)
Solve equation (1) for Q1:
( )
k
RQ 11
kV12=
Solve equation (2) for Q2:
( )
k
RQ 22
kV12=
Substitute in equation (3) to obtain: ( ) ( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛+=
⎟⎠
⎞⎜⎝
⎛ +
=
1
2
1
21
kV12kV12
kV12kV12
kV20
R
R
R
k
R
k
Rk
or
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
1
2128
R
R
Solve for R2:
12 3
2 RR =
Chapter 23
224
*82 ••
Picture the Problem We can use the definition of surface charge density to relate the
radius R of the sphere to its charge Q and the potential function ( ) rkQrV = to relate Q
to the potential at r = 2 m.
Use its definition, relate the surface
charge density σ to the charge Q on
the sphere and the radius R of the
sphere:
24 R
Q
πσ =
Solve for R to obtain:
πσ4
QR =
Relate the potential at r = 2.0 m to the
charge on the sphere:
( )
r
kQrV =
Solve for Q to obtain:
( )
k
rrVQ =
Substitute to obtain: ( ) ( )
( )
σ
∈
πσ
∈π
σπ
rrV
rrV
k
rrVR
0
0
4
4
4
=
==
Substitute numerical values and evaluate R:
( )( )( ) m600.0
nC/m6.24
V500m2m/NC1085.8
2
2212
=⋅×=
−
R
83 ••
Picture the Problem We can use the definition of surface charge density to relate the
radius R of the sphere to its charge Q and the potential function ( ) rkQrV = to relate Q
to the potential at r = 2 m.
Use its definition, relate the surface
charge density σ to the charge Q on
the disk and the radius R of the disk:
2R
Q
πσ =
Solve for Q to obtain: 2RQ πσ= (1)
Electric Potential
225
Relate the potential at r to the charge
on the disk:
( ) ( )xRxkrV −+= 222 σπ
Substitute V(0.6 m) = 80 V: ( ) ⎟⎠⎞⎜⎝⎛ −+= m6.0m6.02V80 22 Rkσπ
Substitute V(1.5 m) = 40 V:
( ) ⎟⎠⎞⎜⎝⎛ −+= m5.1m5.12V40 22 Rkσπ
Divide the first of these equations
by the second to obtain:
( )
( ) m5.1m5.1
m6.0m6.0
2
22
22
−+
−+=
R
R
Solve for R to obtain:
m800.0=R
Express the electric field on the axis of
a disk charge: ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−=
22
12
Rx
xkEx σπ
Solve for σ to obtain:
22
0
22
1
2
12
Rx
x
E
Rx
xk
E
x
x
+−
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−
=
∈
π
σ
Evaluate σ using R = 0.8 m and
E(1.5 m) = 23.5 V/m:
( )( )
( ) ( )
2
22
2212
nC/m54.3
m0.8m1.5
m1.51
V/m23.5m/NC108.852
=
+
−
⋅×=
−
σ
Substitute in equation (1) and
evaluate Q:
( )( )
nC12.7
m0.8nC/m3.54 22
=
= πQ
84 ••
Picture the Problem We can use U = kq1q2/R to relate the electrostatic potential energy
of the particles to their separation.
Express the electrostatic potential
energy of the two particles in terms of
their charge and separation:
R
qkqU 21=
Chapter 23
226
Solve for R:
U
qkqR 21=
Substitute numerical values and evaluate R: ( )( )( )( ) fm6.44
eV
C101.6MeV30.5
C101.6822/CmN108.99
19
219229
=××
×⋅×= −
−
R
85 ••
Picture the Problem We can use l∆=∆ EV and the expression for the electric field due
to a plane of charge to find the potential difference between the two planes.
The
conducting slab introduced between the planes in part (b) will have a negative charge
induced on its surface closest to the plane with the positive charge density and a positive
charge induced on its other surface. We can proceed as in part (a) to find the potential
difference between the planes with the conducting slab in place.
(a) Express the potential difference
between the two planes:
EdEV =∆=∆ l
The electric field due to each plane
is:
02 ∈
σ=E
Because the charge densities are of
opposite sign, the fields are additive
and the resultant electric field
between the planes is:
000
2plane1plane
22 ∈
σ
∈
σ
∈
σ =+=
+= EEE
Substitute to obtain:
0∈
σ dV =∆
(b) The diagram shows the
conducting slab between the two
planes and the electric field lines in
the region between the original two
planes.
Electric Potential
227
Express the new potential difference
∆V′ between the planes in terms of the
potential differences ∆V1, ∆V2 and
∆V3:
23211
321'
ll EaEE
VVVV
++=
∆+∆+∆=∆
Express the electric fields in regions 1,
2 and 3:
0
31 ∈
σ== EE and 02 =E
Substitute to obtain:
( )21
0
2
0
1
0
'
ll
ll
+=
+=∆
∈
σ
∈
σ
∈
σV
Express 21 ll + in terms of a and d: ad −=+ 21 ll
Substitute to obtain: ( )adV −=∆
0
' ∈
σ
86 •••
Picture the Problem We need to consider three regions, as in Example 23-5. Region I, x
> a; region II, 0 < x < a; and region III, x < 0. We can find V in each of these regions and
then find E from lddVE −= .
(a) Relate EI to VI:
dx
dVE II −=
In region I we have:
ax
kq
x
kqV −+=
21
I
Substitute and evaluate EI:
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−+−= ax
kq
x
kq
dx
dE 21I
Because x > 0: xx =
For x > a: axax −=−
Substitute to obtain:
( )2221
21
I
ax
kq
x
kq
ax
kq
x
kq
dx
dE
−+=
⎥⎦
⎤⎢⎣
⎡
−+−=
Chapter 23
228
Proceed as above for regions II and III
to obtain: ( )2221II ax
kq
x
kqE −−=
and
( )2221III ax
kq
x
kqE −−−=
(b) The distance between q1 and a
point on y axis is y and the distance
between a point on the y axis and q2
is 22 ay + . Using these distances,
express the potential at a point on
the y axis:
( )
22
21
ay
kq
y
kqyV ++=
(c) To obtain the y component of
E
r
at a point on the y axis we take
the derivative of V(y). For y > 0:
( ) 2322 221
22
21
ay
ykq
y
kq
ay
kq
y
kq
dy
dEy
++=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++−=
For y < 0:
( ) 2322 221
22
21
ay
ykq
y
kq
ay
kq
y
kq
dy
dEy
++−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++−−=
law. sCoulomb' using
obtains one that and todue fields theof components theare These 21 qq
*87 •••
Picture the Problem We can consider the relationship between the potential and the
electric field to show that this arrangement is equivalent to replacing the plane by a point
charge of magnitude −q located a distance d beneath the plane. In (b) we can first find the
field at the plane surface and then use E0=∈σ to find the surface charge density. In (c)
the work needed to move the charge to a point 2d away from the plane is the product of
the potential difference between the points at distances 2d and 3d from −q multiplied by
the separation ∆x of these points.
Electric Potential
229
(a)
zero. isinfinity at potential thezero, is chargenet the
because Also, plane. in the everywhere potential same thegivemust
they so ts,arrangemenboth in plane thelar toperpendicu is field electric
theandt arrangemeneither in 0 is plane on the anywhere potential The
xy
(b) The surface charge density is
given by:
E0=∈σ (1)
At any point on the plane, the
electric field points in the negative x
direction and has magnitude:
θcos22 rd
kqE +=
where θ is the angle between the horizontal
and a vector pointing from the positive
charge to the point of interest on the xz
plane and r is the distance along the plane
from the origin (i.e., directly to the left of
the charge).
Because :cos
22 rd
d
+=θ
( )
( ) 23220
2322
2222
4 rd
qd
rd
kqd
rd
d
rd
kqE
+∈
=
+
=
++
=
π
Substitute for E in equation (1) to
obtain: ( ) 2/3224 rd qd+= πσ
88 •••
Picture the Problem We can express the potential due to the ring charges as the sum of
the potentials due to each of the ring charges. To show that V(x) is a minimum at x = 0,
we must show that the first derivative of V(x) = 0 at x = 0 and that the second derivative
is positive. In part (c) we can use a Taylor expansion to show that, for x << L, the
potential is of the form V(x) = V(0) + α x2. In part (d) we can obtain the potential energy
function from the potential function and, noting that it is quadratic in x, find the ″spring″
constant and the angular frequency of oscillation of the particle provided its displacement
from its equilibrium position is small.
(a) Express the potential due to the
ring charges as the sum of the
( ) rightthetoringleftthetoring VVxV +=
Chapter 23
230
potentials due to each of their
charges:
The potential for a ring of charge is:
( )
22 Rx
kQxV +=
where R is the radius of the ring and Q is
the charge of the ring.
For the ring to the left we have:
( ) 22leftthetoring LLx
kQV
++
=
For the ring to the right we have:
( ) 22rightthetoring LLx
kQV
+−
=
Substitute to obtain:
( ) ( ) ( ) 2222 LLx
kQ
LLx
kQxV
+−
+
++
=
(b) Evaluate dV/dx to obtain:
( )[ ] ( )[ ] extremafor 023222322 =⎪⎭⎪⎬
⎫
++
+−⎪⎩
⎪⎨
⎧
+−
−=
LxL
xL
LxL
xLkQ
dx
dV
Solve for x to obtain: x = 0
Evaluate d2V/dx2 to obtain:
( )
( )[ ] ( )[ ] ( )( )[ ]
( )[ ] ⎪⎭⎪⎬
⎫
+−
−
++
++
+−
−⎪⎩
⎪⎨
⎧
+−
−=
2322
2522
2
23222522
2
2
2
1
313
LxL
LxL
xL
LxLLxL
xLkQ
dx
Vd
Evaluate this expression for
x = 0 to obtain:
( ) 0
22
0
32
2
>=
L
kQ
dx
Vd
0. at maximum a is )( Hence =xxV
Electric Potential
231
(c) The Taylor expansion of V(x) is: ( ) ( ) ( ) ( )
sorder termhigher
0''0'0 221
+
++= xVxVVxV
For x << L:
( ) ( ) ( ) ( ) 221 0''0'0 xVxVVxV ++≈
Substitute our results from part (b)
to obtain:
( ) ( )
2
3
2
32
1
24
2
22
02
x
L
kQ
L
kQ
x
L
kQx
L
kQxV
+=
⎟⎠
⎞⎜⎝
⎛++=
or
( ) ( ) 20 xVxV α+=
where
( )
L
kQV 20 = and
324 L
kQ=α
(d) Express the angular frequency of
oscillation of a simple harmonic
oscillator:
m
k '=ω
where k′ is the restoring constant.
From our result for part (c) and the
definition of electric potential:
( ) ( )
( ) 221
2
3
'0
222
10
xkqV
x
L
kqQqVxU
+=
⎟⎠
⎞⎜⎝
⎛+=
where
322
'
L
kqQk =
Substitute for k′ in the expression
for ω: 322 Lm
kqQ=ω
89 •••
Picture the Problem The diagram shows
part of the shells in a cross-sectional view
under the conditions of part (a) of the
problem. We can use Gauss’s law to find
the electric field in the regions defined by
the three surfaces and then find the electric
potentials from the electric fields. In part
(b) we can use the redistributed charges to
find the charge on and potentials of the
three surfaces.
Chapter 23
232
(a) Apply Gauss’s law to a spherical
Gaussian surface of radius r ≥
c to
obtain:
( ) 04
0
enclosed2 == ∈π
QrEr
and Er = 0 because the net charge enclosed
by the Gaussian surface is zero.
Because Er(c) = 0:
( ) 0=cV
Apply Gauss’s law to a spherical
Gaussian surface of radius b < r < c
to obtain:
( )
0
24 ∈π
QrEr =
and
( ) 2r
kQcrbEr =<<
Use ( )crbEr << to find the
potential difference between c and b:
( ) ( )
⎟⎠
⎞⎜⎝
⎛ −=
−=− ∫
cb
kQ
r
drkQcVbV
b
c
11
2
Because V(c) = 0: ( ) ⎟⎠
⎞⎜⎝
⎛ −=
cb
kQbV 11
The inner shell carries no charge, so
the field between r = a and
r = b is zero and:
( ) ( ) ⎟⎠
⎞⎜⎝
⎛ −==
cb
kQbVaV 11
(b) When the inner and outer shells
are connected their potentials become
equal as a consequence of the
redistribution of charge.
The charges on surfaces a and c are
related according to:
QQQ ca −=+ (1)
Electric Potential
233
Qb does not change with the
connection of the inner and outer
shells:
QQ =b
Express the potentials of shells a
and c:
( ) ( ) 0== cVaV
In the region between the r = a and
r = b, the field is kQa/r2 and the
potential at r = b is then:
( ) ⎟⎠
⎞⎜⎝
⎛ −=
ab
kQbV a
11
(2)
The enclosed charge for b < r < c is
Qa + Q, and by Gauss’s law the field
in this region is:
( )
2r
QQkE acrb
+=<<
Express the potential difference
between b and c:
( ) ( ) ( )
( )bV
bc
QQkbVcV a
−=
⎟⎠
⎞⎜⎝
⎛ −+=− 11
because V(c) = 0.
Solve for V(b) to obtain:
( ) ( ) ⎟⎠
⎞⎜⎝
⎛ −+=
cb
QQkbV a
11
(3)
Equate equations (2) and (3) and
solve for Qa to obtain:
( )
( )acb
bcaQQa −
−−= (4)
Substitute equation (4) in equation (1)
and solve for Qc to obtain:
( )
( )acb
abcQQc −
−−= (5)
Substitute (4) and (5) in (3) to obtain: ( ) ( )( )( )acb
abbckQbV −
−−= 2
Chapter 23
234
*90 •••
Picture the Problem The diagram shows a
cross-sectional view of a portion of the
concentric spherical shells. Let the charge
on the inner shell be q. The dashed line
represents a spherical Gaussian surface
over which we can integrate dAnE ˆ⋅r in
order to find Er for r ≥ b. We can find V(b)
from the integral of Er between r = ∞ and
r = b. We can obtain a second expression
for V(b) by considering the potential
difference between a and b and solving the
two equations simultaneously for the
charge q on the inner shell.
Apply Gauss’s law to a spherical surface of
radius r ≥ b:
( )
0
24 επ
qQrEr
+=
Solve for Er to obtain: ( )
2r
qQkEr
+=
Use Er to find V(b): ( ) ( )
( )
b
qQk
r
drqQkbV
b
+=
+−= ∫
∞
2
We can also determine V(b) by
considering the potential difference
between a, i.e., 0 and b:
( ) ⎟⎠
⎞⎜⎝
⎛ −=
ab
kqbV 11
Equate these expressions for V(b) to
obtain:
( ) ⎟⎠
⎞⎜⎝
⎛ −=+
ab
ka
b
qQk 11
Solve for q to obtain: Q
b
aq −=
91 •••
Picture the Problem We can use the hint to derive an expression for the electrostatic
potential energy dU required to bring in a layer of charge of thickness dr and then
integrate this expression from r = 0 to R to obtain an expression for the required work.
Electric Potential
235
If we build up the sphere in layers,
then at a given radius r the net
charge on the sphere will be given
by:
Q(r) = Q r
R
⎛
⎝ ⎜
⎞
⎠ ⎟
3
When the radius of the sphere is r,
the potential relative to infinity is:
( ) ( ) 3
2
00 44 R
rQ
r
rQrV ∈=∈= ππ
Express the work dW required to
bring in charge dQ from infinity to
the surface of a uniformly charged
sphere of radius r:
( )
drr
R
Q
dr
R
Qr
R
rQ
dQrVdUdW
4
6
0
2
3
2
3
2
0
4
3
4
34
4
∈=
⎟⎠
⎞⎜⎝
⎛
∈=
==
π
πππ
Integrate dW from 0 to R to obtain:
R
Qr
R
Q
drr
R
QUW
R
R
0
2
0
5
6
0
2
0
4
6
0
2
20
3
54
3
4
3
∈=⎥⎦
⎤⎢⎣
⎡
∈=
∈== ∫
ππ
π
92 ••
Picture the Problem We can equate the rest energy of an electron and the result of
Problem 91 in order to obtain an expression that we can solve for the classical electron
radius.
From Problem 91 we have:
R
eU
0
2
20
3
∈= π
The rest mass of the electron is
given by:
2
00 cmE =
Equate these energies to obtain:
20
0
2
20
3 cm
R
e =∈π
Solve for R:
2
00
2
20
3
cm
eR ∈= π
Substitute numerical values and evaluate R:
( )( )( )( )
m1069.1
J/eV106.1eV1011.5mN/C1085.820
C106.13
15
1952212
219
−
−−
−
×=
××⋅×
×= πR
Chapter 23
236
repulsion. mutual
own itsagainst together holdselectron thehowexplain not does model This
93 ••
Picture the Problem Because the post-fission volumes of the fission products are equal,
we can express the post-fission radii in terms of the radius of the pre-fission sphere.
(a) Relate the initial volume V of the
uniformly charged sphere to the
volumes V′ of the fission products:
'2VV =
Substitute for V and V ′: ( )334334 '2 RR ππ =
Solve for and evaluate R′:
RRR 794.0
2
1'
3
==
(b) Express the difference ∆E in the
total electrostatic energy as a result
of fissioning:
'EEE −=∆
From Problem 91 we have:
R
QE
0
2
20
3
∈= π
After fissioning:
( )
E
R
Q
R
Q
R
QE
630.0
20
3
2
2
2
120
32
'20
'32'
0
23
30
2
2
1
0
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈=
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
∈
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈=
π
ππ
Substitute for E and E′ to obtain: EEEE 370.0630.0 =−=∆
*94 •••
Picture the Problem We can use the definition of density to express the radius R of a
nucleus as a function of its atomic mass N. We can then use the result derived in
Problem 91 to express the electrostatic energies of the 235U nucleus and the nuclei of the
fission fragments 140Xe and 94Sr.
The energy released by this fission
process is:
( )
SrXeU 94140235
UUUE +−=∆ (1)
Express the mass of a nucleus in
terms of its density and volume:
3
3
4 RNm ρπ=
where N is the nuclear number.
Electric Potential
237
Solve for R to obtain:
3
4
3
πρ
NmR =
Substitute numerical values and
evaluate R as a function of N:
( )( )
( ) 3116
313
317
27
m1097.9
kg/m1044
kg10660.13
N
NR
−
−
×=
×
×= π
The 'radius' of the 235U nucleus is
therefore:
( )( )
m1015.6
235m1097.9
15
3116
−
−
×=
×=UR
From Problem 91 we have:
R
QU
0
2
20
3
∈= π
Substitute numerical values and evaluate the electrostatic energy of the 235U
nucleus:
( )( )( )
MeV1189
J/eV106.1
eV1J1091.1
m1015.6mN/C1085.820
C106.1923
19
10
152212
219
U235
=×××=
×⋅×
××=
−
−
−−
−
πU
Proceed as above to find the electrostatic energy of the fission fragments 140Xe and 94Sr:
( )( )( )
MeV410
J/eV106.1
eV1J1057.6
m1015.6mN/C1085.820
C106.1543
19
11
152212
219
Xe140
=×××=
×⋅×
××=
−
−
−−
−
πU
and ( )( )( )
MeV203
J/eV10602.1
eV1J1025.3
m1015.6mN/C1085.820
C106.1383
19
11
152212
219
Sr94
=×××=
×⋅×
××=
−
−
−−
−
πU
Substitute for
U235
U ,
Xe140
U , and
Sr94
U in equation (1) and evaluate
∆E:
( )
MeV576
MeV203MeV410MeV1189
=
+−=∆E
95 •••
Picture
the Problem The geometry of the point charge and the sphere is shown below.
The charge is a distance R away from the center of a spherical shell of radius a.
Chapter 23
238
(a) The average potential over the
surface of the sphere is given by:
∫∫ == spheresphereav rdAkrkdqV σ
Substitute for k, σ, and dA to obtain: ( )( )∫∈=
π
π
θθπ
π 0 20av 4
sin2
4
1
ra
adaqV
Apply the law of cosines to the
triangle to obtain: θcos222 aRaRr −+=
Substitute for r and simplify to
obtain:
( )∫ −+∈=
π
θ
θθ
π 0 21220av cos2
sin
8 aRaR
dqV
Change variables by letting
u = cosθ. Then:
θθddu sin−=
and
( )∫
−
−+∈
−=
1
1
2122
0
av
28 aRuaR
duqV π (1)
To simplify the integrand, let: 22 aR +=α , aR2=β , and uv βα −=
Then dudv β−= and:
( )
[ ]βαβα
βαββ
−−+−=
−−=−=−=−+
−− ∫∫
aR
u
aR
v
v
dv
aRuaR
du
1
121
2
1
1
1
1
2122
2
1
2
1
l
l
l
l
Substitute for α and β to obtain:
Electric Potential
239
( ) [ ]
( ) ( )
( ) ( )[ ]
R
aRaR
aR
aRaR
aR
aRaRaRaR
aRaRuaR
du
21
1
221
2
22
2222
1
1
2122
−=−−+−=
⎥⎦
⎤⎢⎣
⎡ −−+−=
−+−++−=−+∫
−
Substitute in equation (1) to obtain:
R
q
R
qV
00
av 4
2
8 ∈=⎟⎠
⎞⎜⎝
⎛−∈
−= ππ
charge.point
the todue sphere theofcenter at the potential theisresult that thisNote
(b)
it. of outside charges ofion configuratany and sphereany for holdmust
result thissphere, theof propertiesany oft independen isresult this
Because space.in onsdistributi chargeany todue potentials theof sum
theispoint any at potential that theus tellsprincipleion superposit The
Chapter 23
240
241
Chapter 24
Electrostatic Energy and Capacitance
Conceptual Problems
*1 •
Determine the Concept The capacitance of a parallel-plate capacitor is a function of the
surface area of its plates, the separation of these plates, and the electrical properties of the
matter between them. The capacitance is, therefore, independent of the voltage across the
capacitor. correct. is )(c
2 •
Determine the Concept The capacitance of a parallel-plate capacitor is a function of the
surface area of its plates, the separation of these plates, and the electrical properties of the
matter between them. The capacitance is, therefore, independent of the charge of the
capacitor. correct. is )(c
3 •
Determine the Concept True. The energy density of an electrostatic field is given by
2
02
1
e Eu ∈= .
4 •
Picture the Problem The energy stored in the electric field of a parallel-plate capacitor is
related to the potential difference across the capacitor by .21 QVU =
Relate the potential energy stored in
the electric field of the capacitor to the
potential difference across the
capacitor:
QVU 21=
. doubles doubling Hence, . toalproportiondirectly is constant, With UVVUQ
*5 ••
Picture the Problem The energy stored in a capacitor is given by QVU 21= and the
capacitance of a parallel-plate capacitor by .0 dAC ∈= We can combine these
relationships, using the definition of capacitance and the condition that the potential
difference across the capacitor is constant, to express U as a function of d.
Express the energy stored in the
capacitor:
QVU 21=
Chapter 24
242
Use the definition of capacitance to
express the charge of the capacitor:
CVQ =
Substitute to obtain: 221 CVU =
Express the capacitance of a
parallel-plate capacitor in terms of
the separation d of its plates:
d
AC 0∈=
where A is the area of one plate.
Substitute to obtain:
d
AVU
2
2
0∈=
Because
d
U 1∝ , doubling the
separation of the plates will reduce
the energy stored in the capacitor to
1/2 its previous value:
correct. is )(d
6 ••
Picture the Problem Let V represent the initial potential difference between the plates, U
the energy stored in the capacitor initially, d the initial separation of the plates, and V ′, U
′, and d ′ these physical quantities when the plate separation has been doubled. We can
use QVU 21= to relate the energy stored in the capacitor to the potential difference
across it and V = Ed to relate the potential difference to the separation of the plates.
Express the energy stored in the
capacitor before the doubling of the
separation of the plates:
QVU 21=
Express the energy stored in the
capacitor after the doubling of the
separation of the plates:
QV'U' 21=
because the charge on the plates does not
change.
Express the ratio of U′ to U:
V
V'
U
U' =
Express the potential differences
across the capacitor plates before
and after the plate separation in
terms of the electric field E between
the plates:
EdV =
and
Ed'V' =
because E depends solely on the charge on
the plates and, as observed above, the
Electrostatic Energy and Capacitance
243
charge does not change during the
separation process.
Substitute to obtain:
d
d'
Ed
Ed'
U
U' ==
For d ′ = 2d: 22' ==
d
d
U
U
and correct is )(b
7 •
Determine the Concept Both statements are true. The total charge stored by two
capacitors in parallel is the sum of the charges on the capacitors and the equivalent
capacitance is the sum of the individual capacitances. Two capacitors in series have the
same charge and their equivalent capacitance is found by taking the reciprocal of the sum
of the reciprocals of the individual capacitances.
8 ••
(a) False. Capacitors connected in series carry the same charge.
(b) False. The voltage across the capacitor whose capacitance is C0 is Q/C0 and that
across the second capacitor is Q/2C0.
(c) False. The energy stored by the capacitor whose capacitance is C0 is 0
2 2CQ and the
energy stored by the second capacitor is .4 0
2 CQ
(d) True
9 •
Determine the Concept True. The capacitance of a parallel-plate capacitor filled with a
dielectric of constant κ is given by
d
AC 0∈κ= or C ∝ κ.
*10 ••
Picture the Problem We can treat the configuration in (a) as two capacitors in parallel
and the configuration in (b) as two capacitors in series. Finding the equivalent
capacitance of each configuration and examining their ratio will allow us to decide
whether (a) or (b) has the greater capacitance. In both cases, we’ll let C1 be the
capacitance of the dielectric-filled capacitor and C2 be the capacitance of the air
capacitor.
In configuration (a) we have: 21 CCCa +=
Chapter 24
244
Express C1 and C2:
d
A
d
A
d
AC
2
02
1
0
1
10
1
∈=∈=∈= κκκ
and
d
A
d
A
d
AC
2
02
1
0
2
20
2
∈=∈=∈=
Substitute for C1 and C2 and
simplify to obtain:
( )1
222
000 +∈=∈+∈= κκ
d
A
d
A
d
ACa
In configuration (b) we have:
21
111
CCCb
+= ⇒
21
21
CC
CCCb +=
Express C1 and C2:
d
A
d
A
d
AC 0
2
1
0
1
10
1
2∈=∈=∈=
and
d
A
d
A
d
AC 0
2
1
0
2
20
2
2 ∈=∈=∈= κκκ
Substitute for C1 and C2 and
simplify to obtain:
( )
⎟⎠
⎞⎜⎝
⎛
+
∈=
+∈
⎟⎠
⎞⎜⎝
⎛ ∈⎟⎠
⎞⎜⎝
⎛ ∈
=
∈+∈
⎟⎠
⎞⎜⎝
⎛ ∈⎟⎠
⎞⎜⎝
⎛ ∈
=
1
2
12
22
22
22
0
0
00
00
00
κ
κ
κ
κ
κ
κ
d
A
d
A
d
A
d
A
d
A
d
A
d
A
d
A
Cb
Divide Cb by Ca:
( ) ( )20
0
1
4
1
2
1
2
+=+∈
⎟⎠
⎞⎜⎝
⎛
+
∈
= κ
κ
κ
κ
κ
d
A
d
A
C
C
a
b
Because ( ) 11
4
2 <+κ
κ
for κ > 1: ba CC >
Electrostatic Energy and Capacitance
245
11 •
(a) False. The capacitance of a parallel-plate capacitor is defined to be the ratio of the
charge on the capacitor to the potential difference across it.
(b) False. The capacitance of a parallel-plate capacitor depends on the area of its plates A,
their separation d, and the dielectric constant κ of the material between the plates
according to .0 dAC ∈=κ
(c) False. As in part (b), the capacitance of a parallel-plate capacitor depends on the area
of its plates A, their separation d, and the dielectric constant κ of the material between the
plates according to .0 dAC ∈=κ
12 ••
Picture the Problem We can use the expression 221 CVU = to express the ratio of the
energy stored in the single capacitor and in the identical-capacitors-in-series combination.
Express the energy stored in
capacitors when they are connected
to the 100-V battery:
2
eq2
1 VCU =
Express the equivalent capacitance
of the two identical capacitors
connected in series:
C
C
CC 21
2
eq 2
==
Substitute to obtain:
( ) 24122121 CVVCU ==
Express the energy stored in one
capacitor when it is connected to the
100-V battery:
2
2
1
0 CVU =
Express the ratio of U to U0:
2
1
2
2
1
2
4
1
0
==
CV
CV
U
U
or
02
1 UU = and correct is )(d
Estimation and Approximation
13 ••
Picture the Problem The outer diameter of a "typical" coaxial cable is about
5 mm, while the inner diameter is about 1 mm. From Table 24-1 we see that a reasonable
range of values for κ is 3-5. We can use the expression for the capacitance of a
Chapter 24
246
cylindrical capacitor to estimate the capacitance per unit length of a coaxial cable.
The capacitance of a cylindrical
dielectric-filled capacitor is given
by:
⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
∈=
1
2
0
ln
2
R
R
LC πκ
where L is the length of the capacitor, R1 is
the radius of the inner conductor, and R2 is
the radius of the second (outer) conductor.
Divide both sides by L to obtain an
expression for the capacitance per
unit length of the cable:
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈=
1
2
1
2
0
ln2ln
2
R
Rk
R
RL
C κπκ
If κ = 3:
( ) nF/m104.0
mm5.0
mm5.2lnC/mN1099.82
3
229
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×
=
L
C
If κ = 5:
( ) nF/m173.0
mm5.0
mm5.2lnC/mN1099.82
5
229
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛⋅×
=
L
C
A reasonable range of values for C/L,
corresponding to 3 ≤ κ ≤ 5, is: nF/m0.173nF/m104.0 ≤≤ L
C
*14 ••
Picture the Problem The energy stored in a capacitor is given by .22
1 CVU =
Relate the energy stored in a
capacitor to its capacitance and the
potential difference across it:
2
2
1 CVU =
Solve for C:
2
2
V
UC =
The potential difference across the
spark gap is related to the width of
the gap d and the electric field E in
the gap:
EdV =
Electrostatic Energy and Capacitance
247
Substitute for V in the expression for
C to obtain:
22
2
dE
UC =
Substitute numerical values and
evaluate C:
( )( ) ( )
F2.22
m001.0V/m103
J1002
226
µ=
×=C
15 ••
Picture the Problem Because ∆R << RE we can treat the atmosphere as a flat slab with
an area equal to the surface area of the earth. Then the energy stored in the atmosphere
can be estimated from U = uV, where u is the energy density of the atmosphere and V is
its volume.
Express the electric energy stored in
the atmosphere in terms of its energy
density and volume:
uVU =
Because ∆R << RE = 6370 km, we
can consider the volume: RR
RAV
∆=
∆=
2
E
earth theof area surface
4π
Express the energy density of the
Earth’s atmosphere in terms of the
average magnitude of its electric
field:
2
02
1 Eu ∈=
Substitute for V and u to obtain: ( )( )
k
RERRREU
2
4
22
E2
E
2
02
1 ∆=∆= π∈
Substitute numerical values and
evaluate U:
( ) ( ) ( )( )
J1003.9
/CmN1099.82
km1V/m200km6370
10
229
22
×=
⋅×=U
16 ••
Picture the Problem We’ll approximate the balloon by a sphere of radius R = 3 m and
use the expression for the capacitance of an isolated spherical conductor.
Relate the capacitance of an isolated
spherical conductor to its radius:
k
RRC == 04 ∈π
Chapter 24
248
Substitute numerical values and
evaluate C:
nF334.0
/CmN108.99
m3
229 =⋅×=C
Electrostatic Potential Energy
17 •
The electrostatic potential energy of this
system of three point charges is the work
needed to bring the charges from an
infinite separation to the final positions
shown in the diagram.
Express the work required to
assemble this system of charges:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
++=
3,2
32
3,1
31
2,1
21
3,2
32
3,1
31
2,1
21
r
qq
r
qq
r
qqk
r
qkq
r
qkq
r
qkqU
Find the distances r1,2, r1,3, and r2,3:
m6andm,3m,3 3,13,22,1 === rrr
(a) Evaluate U for q1 = q2 = q3 = 2 µC:
( ) ( )( ) ( )( ) ( )( )
mJ0.30
m3
C2C2
m6
C2C2
m3
C2C2/CmN1099.8 229
=
⎥⎦
⎤+⎢⎣
⎡ +⋅×= µµµµµµU
(b) Evaluate U for q1 = q2 = 2 µC and q3 = −2 µC:
( ) ( )( ) ( )( ) ( )( )
mJ99.5
m3
C2C2
m6
C2C2
m3
C2C2/CmN1099.8 229
−=
⎥⎦
⎤−+⎢⎣
⎡ −+⋅×= µµµµµµU
(c) Evaluate U for q1 = q3 = 2 µC and q2 = −2 µC:
( ) ( )( ) ( )( ) ( )( )
mJ0.18
m3
C2C2
m6
C2C2
m3
C2C2/CmN1099.8 229
−=
⎥⎦
⎤−+⎢⎣
⎡ +−⋅×= µµµµµµU
Electrostatic Energy and Capacitance
249
18 •
Picture the Problem The electrostatic
potential energy of this system of three
point charges is the work needed to bring
the charges from an infinite separation to
the final positions shown in the diagram.
Express the work required to assemble
this system of charges:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
++=
3,2
32
3,1
31
2,1
21
3,2
32
3,1
31
2,1
21
r
qq
r
qq
r
qqk
r
qkq
r
qkq
r
qkqU
Find the distances r1,2, r1,3, and r2,3:
m5.23,13,22,1 === rrr
(a) Evaluate U for q1 = q2 = q3 = 4.2 µC:
( ) ( )( ) ( )( )
( )( )
J190.0
m5.2
C2.4C2.4
m5.2
C2.4C2.4
m5.2
C2.4C2.4/CmN1099.8 229
=
⎥⎦
⎤+
+⎢⎣
⎡⋅×=
µµ
µµµµU
(b) Evaluate U for q1 = q2 = 4.2 µC and q3 = −4.2 µC:
( ) ( )( ) ( )( )
( )( )
mJ4.63
m5.2
C2.4C2.4
m5.2
C2.4C2.4
m5.2
C2.4C2.4/CmN10988.8 229
−=
⎥⎦
⎤−+
−+⎢⎣
⎡⋅×=
µµ
µµµµU
(c) Evaluate U for q1 = q2 = −4.2 µC and q3 = +4.2 µC:
Chapter 24
250
( ) ( )( ) ( )( )
( )( )
mJ4.63
m5.2
C2.4C2.4
m5.2
C2.4C2.4
m5.2
C2.4C2.4/CmN1099.8 229
−=
⎥⎦
⎤−+
−+⎢⎣
⎡ −−⋅×=
µµ
µµµµU
*19 •
Picture the Problem The potential of an isolated spherical conductor is given by
rkQV = ,where Q is its charge and r its radius, and its electrostatic potential energy
by QVU 21= . We can combine these relationships to find the sphere’s electrostatic
potential energy.
Express the electrostatic potential
energy of the isolated spherical
conductor as a function of its charge
Q and potential V:
QVU 21=
Express the potential of the spherical
conductor:
r
kQV =
Solve for Q to obtain:
k
rVQ =
Substitute to obtain:
k
rVV
k
rVU
2
2
2
1 =⎟⎠
⎞⎜⎝
⎛=
Substitute numerical values and
evaluate U:
( )( )( )
J2.22
/CmN108.992
kV2m0.1
229
2
µ=
⋅×=U
20 ••
Picture the Problem The electrostatic
potential energy of this system of four
point charges is the work needed to bring
the charges from an infinite separation to
the final positions shown in the diagram. In
part (c), depending on the configuration of
the positive and negative charges, two
energies are possible.
Electrostatic Energy and Capacitance
251
Express the work required to assemble this system of charges:
⎟⎟⎠
⎞+++⎜⎜⎝
⎛ ++=
+++++=
4,3
43
4,2
42
3,2
32
4,1
41
3,1
31
2,1
21
4,3
43
4,2
42
3,2
32
4,1
41
3,1
31
2,1
21
r
qq
r
qq
r
qq
r
qq
r
qq
r
qqk
r
qkq
r
qkq
r
qkq
r
qkq
r
qkq
r
qkqU
Find the distances r1,2, r1,3, r1,4, r2,3,
r2,4, and r3,4,:
m44,14,33,22,1 ==== rrrr
and
m244,23,1 == rr
(a) Evaluate U for q1 = q2 = q3 = q4 = −2 µC:
( ) ( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
mJ7.48
m4
C2C2
m24
C2C2
m4
C2C2
m4
C2C2
m24
C2C2
m4
C2C2/CmN1099.8 229
=
⎥⎦
⎤−−+−−+−−+
−−+−−+⎢⎣
⎡ −−⋅×=
µµµµµµ
µµµµµµU
(b) Evaluate U for q1 = q2 = q3 = 2 µC and q4 = −2 µC:
( ) ( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
0
m4
C2C2
m24
C2C2
m4
C2C2
m4
C2C2
m24
C2C2
m4
C2C2/CmN1099.8 229
=
⎥⎦
⎤−+−++
−++⎢⎣
⎡⋅×=
µµµµµµ
µµµµµµU
(c) Let q1 = q2 = 2 µC and q3 = q4 = −2 µC:
( ) ( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
mJ7.12
m4
C2C2
m24
C2C2
m4
C2C2
m4
C2C2
m24
C2C2
m4
C2C2/CmN1099.8 229
−=
⎥⎦
⎤−−+−+−+
−+−+⎢⎣
⎡⋅×=
µµµµµµ
µµµµµµU
Let q1 = q3 = 2 µC and q2 = q4 = −2 µC:
Chapter 24
252
( ) ( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
mJ2.23
m4
C2C2
m24
C2C2
m4
C2C2
m4
C2C2
m24
C2C2
m4
C2C2/CmN1099.8 229
−=
⎥⎦
⎤−+−−+−+
−++⎢⎣
⎡ −⋅×=
µµµµµµ
µµµµµµU
21 ••
Picture the Problem The diagram shows
the four charges fixed at the corners of the
square and the fifth charge that is released
from rest at the origin. We can use
conservation of energy to relate the initial
potential energy of the fifth particle to its
kinetic energy when it is at a great distance
from the origin and the electrostatic
potential at the origin to express Ui.
Use conservation of energy to relate
the initial potential energy of the
particle to its kinetic energy when it
is at a great distance from the origin:
0=∆+∆ UK
or, because Ki = Uf = 0,
0if =−UK
Express the initial potential energy
of the particle to its charge and the
electrostatic potential at the origin:
( )0i qVU =
Substitute for Kf and Ui to obtain:
( ) 00221 =− qVmv
Solve for v: ( )
m
qVv 02=
Express the electrostatic potential at
the origin:
( )
a
kq
a
kq
a
kq
a
kq
a
kqV
2
6
2
6
2
3
2
2
2
0
=
+−++=
Substitute and simplify to obtain:
ma
kq
a
kq
m
qv 26
2
62 =⎟⎠
⎞⎜⎝
⎛=
Electrostatic Energy and Capacitance
253
Capacitance
*22 •
Picture the Problem The charge on the spherical conductor is related to its radius and
potential according to V = kQ/r and we can use the definition of capacitance to find the
capacitance of the sphere.
(a) Relate the potential V of the
spherical conductor to the charge on
it and to its radius:
r
kQV =
Solve for and evaluate Q:
( )( ) nC2.22
/CmN108.99
kV2m0.1
229 =⋅×=
=
k
rVQ
(b) Use the definition of capacitance
to relate the capacitance of the
sphere to its charge and potential:
pF11.1
kV2
nC22.2 ===
V
QC
(c) radius. its offunction a is sphere a of ecapacitanc The t.doesn'It
23 •
Picture the Problem We can use its definition to find the capacitance of this capacitor.
Use the definition of capacitance to
obtain:
nF0.75
V400
C30 === µ
V
QC
24 ••
Picture the Problem Let the separation of the spheres be d and their radii be R. Outside
the two spheres the electric field is approximately the field due to point charges of +Q
and −Q, each located at the centers of spheres, separated by distance d. We can derive an
expression for the potential at the surface of each sphere and then use the potential
difference between the spheres and the definition of capacitance and to find the
capacitance of the two-sphere system.
The capacitance of the two-sphere
system is given by:
V
QC ∆=
where ∆V is the potential difference
between the spheres.
Chapter 24
254
The potential at any point outside
the two spheres is:
( ) ( )
21 r
Qk
r
QkV −++=
where r1 and r2 are the distances from the
given point to the centers of the spheres.
For a point on the surface of the
sphere with charge +Q:
δ+== drRr 21 and
where R<δ
Substitute to obtain:
( ) ( )
δ+
−++=+ d
Qk
R
QkV Q
For δ << d:
d
kQ
R
kQV Q −=+
and
d
kQ
R
kQV Q +−=−
The potential difference between the
spheres is:
⎟⎠
⎞⎜⎝
⎛ −=
⎟⎠
⎞⎜⎝
⎛ +−−−=
−=∆ −
dR
kQ
d
kQ
R
kQ
d
kQ
R
kQ
VVV QQ
112
Substitute for ∆V in the expression
for C to obtain:
d
R
R
dRdR
kQ
QC
−
∈=
⎟⎠
⎞⎜⎝
⎛ −
∈=
⎟⎠
⎞⎜⎝
⎛ −
=
1
2
11
2
112
0
0
π
π
For d very large: RC 02 ∈= π
The Storage of Electrical Energy
25 •
Picture the Problem Of the three equivalent expressions for the energy stored in a
charged capacitor, the one that relates U to C and V is 221 CVU = .
(a) Express the energy stored in the
capacitor as a function of C and V:
2
2
1 CVU =
Electrostatic Energy and Capacitance
255
Substitute numerical values and
evaluate U:
( )( ) mJ0.15V100F3 221 == µU
(b) Express the additional energy
required as the difference between
the energy stored in the capacitor at
200 V and the energy stored at 100
V:
( ) ( )
( )( )
mJ0.45
mJ0.15V200F3
V100V200
2
2
1
=
−=
−=∆
µ
UUU
26 •
Picture the Problem Of the three equivalent expressions for the energy stored in a
charged capacitor, the one that relates U to Q and C is
C
QU
2
2
1= .
(a) Express the energy stored in the
capacitor as a function of C and Q:
C
QU
2
2
1=
Substitute numerical values and
evaluate U:
( ) J800.0
F10
C4
2
1 2 µµ
µ ==U
(b) Express the energy remaining
when half the charge is removed:
( ) ( ) J 0.200
F10
C2
2
1 2
2
1 µµ
µ ==QU
27 •
Picture the Problem Of the three equivalent expressions for the energy stored in a
charged capacitor, the one that relates U to Q and C is
C
QU
2
2
1= .
(a) Express the energy stored in the
capacitor as a function of C and Q:
C
QU
2
2
1=
Substitute numerical values and
evaluate U:
( ) ( ) J625.0
pF20
C5
2
1C5
2
== µµU
(b) Express the additional energy
required as the difference between
the energy stored in the capacitor
when its charge is 5 µC and when
its charge is 10 µC:
( ) ( )
( )
J .881
J 0.625 J 2.50
J625.0
pF20
C10
2
1
C5C10
2
=
−=
−=
−=∆
µ
µµ UUU
Chapter 24
256
*28 •
Picture the Problem The energy per unit volume in an electric field varies with the
square of the electric field according to 220 Eu ∈= .
Express the energy per unit volume
in an electric field:
2
02
1 Eu ∈=
Substitute numerical values and
evaluate u:
( )( )
3
22212
2
1
J/m8.39
MV/m3m/NC1085.8
=
⋅×= −u
29 •
Picture the
Problem Knowing the potential difference between the plates, we can use E
= V/d to find the electric field between them. The energy per unit volume is given by
2
02
1 Eu ∈= and we can find the capacitance of the parallel-plate capacitor using
.0 dAC =∈
(a) Express the electric field between
the plates in terms of their separation
and the potential difference between
them:
kV/m100
mm1
V100 ==
=
d
VE
(b) Express the energy per unit
volume in an electric field:
2
02
1 Eu ∈=
Substitute numerical values and
evaluate u:
( )( )
3
22212
2
1
mJ/m3.44
kV/m001m/NC1085.8
=
⋅×= −u
(c) The total energy is given by: ( )( )( )
J
uAduVU
µ6.88
mm1m2mJ/m3.44 23
=
=
==
(d) The capacitance of a parallel-plate
capacitor is given by:
( )( )
nF7.17
mm1
m2m/NC108.85 22212
0
=
⋅×=
∈=
−
d
AC
Electrostatic Energy and Capacitance
257
(e) The total energy is given by:
2
2
1 CVU =
Substitute numerical values and evaluate
U:
( )( )
).(with agreement in J,5.88
V100nF17.7 221
c
U
µ=
=
30 ••
Picture the Problem The total energy stored in the electric field is the product of the
energy density in the space between the spheres and the volume of this space.
(a) The total energy U stored in the
electric field is given by:
uVU =
where u is the energy density and V is the
volume between the spheres.
The energy density of the field is:
2
02
1 Eu ∈=
where E is the field between the spheres.
The volume between the spheres is
approximately:
( )12214 rrrV −≈ π
Substitute for u and V to obtain:
( )1221202 rrrEU −∈= π (1)
The magnitude of the electric field
between the concentric spheres is
the sum of the electric fields due to
each charge distribution:
QQ EEE −+=
Because the two surfaces are so
close together, the electric field
between them is approximately the
sum of the fields due to two plane
charge distributions:
000 22 ∈
=∈+∈=
− QQQE
σσσ
Substitute for σQ to obtain:
0
2
14 ∈
≈
r
QE π
Substitute for E in equation (1) and
simplify: ( )
2
1
12
0
2
12
2
1
2
0
2
1
0
8
4
2
r
rrQ
rrr
r
QU
−
∈=
−⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈∈=
π
ππ
Chapter 24
258
Substitute numerical values and evaluate U:
( ) ( )( )( ) nJ0.56cm0.10mN/C1085.88 cm0.10cm5.10nC5 22212
2
=⋅×
−= −πU
(b) The capacitance of the two-
sphere system is given by:
V
QC ∆=
where ∆V is the potential difference
between the two spheres.
The electric potentials at the
surfaces of the spheres are:
10
1 4 r
QV ∈= π and 202 4 r
QV ∈= π
Substitute for ∆V and simplify to
obtain: 12
21
0
2010
4
44
rr
rr
r
Q
r
Q
QC −∈=
∈−∈
= π
ππ
Substitute numerical values and evaluate C:
( )( )( ) nF234.0
cm0.10cm5.10
cm5.10cm0.10mN/C1085.84 2212 =−⋅×=
−πC
Use ½ Q2/C to find the total energy
stored in the electric field between
the spheres:
( ) nJ4.53
nF234.0
nC5
2
1 2 =⎥⎦
⎤⎢⎣
⎡=U
).(in obtained
resultexact our of 5% within is )(in result eapproximatour that Note
b
a
*31 ••
Picture the Problem We can relate the charge Q on the positive plate of the capacitor to
the charge density of the plate σ using its definition. The charge density, in turn, is
related to the electric field between the plates according to E0∈σ = and the electric field
can be found from E = ∆V/∆d. We can use VQU ∆=∆ 21 in part (b) to find the increase
in the energy stored due to the movement of the plates.
(a) Express the charge Q on the
positive plate of the capacitor in
terms of the plate’s charge density σ
and surface area A:
AQ σ=
Electrostatic Energy and Capacitance
259
Relate σ to the electric field E
between the plates of the capacitor:
E0∈σ =
Express E in terms of the change in
V as the plates are separated a
distance ∆d:
d
VE ∆
∆=
Substitute for σ and E to obtain:
d
VAEAQ ∆
∆== 00 ∈∈
Substitute numerical values and evaluate Q:
( )( ) nC1.11
cm0.4
V100cm500m/NC108.85 22212 =⋅×= −Q
(b) Express the change in the
electrostatic energy in terms of the
change in the potential difference:
VQU ∆=∆ 21
Substitute numerical values and
evaluate ∆U:
( )( ) J553.0V100nC11.121 µ==∆U
32 •••
Picture the Problem By symmetry, the electric field must be radial. In part (a) we can
find Er both inside and outside the ball by choosing a spherical Gaussian surface first
inside and then outside the surface of the ball and applying Gauss’s law.
(a) Relate the electrostatic energy
density at a distance r from the
center of the ball to the electric field
due to the uniformly distributed
charge Q:
2
02
1
e Eu ∈= (1)
Relate the flux through the Gaussian
surface to the electric field Er on the
Gaussian surface at r < R:
( )
0
inside24 ∈π
QrEr = (2)
Using the fact that the charge is
uniformly distributed, express the
ratio of the charge enclosed by the
Gaussian surface to the total charge
of the sphere:
3
3
3
3
4
3
3
4
ball
surfaceGaussian inside
R
r
R
r
V
V
Q
Q
==
=
π
π
ρ
ρ
Chapter 24
260
Solve for Qinside to obtain:
3
3
inside R
rQQ =
Substitute in equation (2): ( ) 3
0
3
24
R
QrrEr ∈π =
Solve for Er < R: r
R
kQ
R
QrE Rr 33
04
==< ∈π
Substitute in equation (1) to obtain: ( )
2
6
22
0
2
302
1
e
2
r
R
Qk
r
R
kQRru
∈
∈
=
⎟⎠
⎞⎜⎝
⎛=<
Relate the flux through the Gaussian
surface to the electric field Er on the
Gaussian surface at r > R:
( )
00
inside24 ∈∈π
QQrEr ==
Solve for Er > R: 2
0
24
−
> == kQrr
QE Rr ∈π
Substitute in equation (1) to obtain: ( ) ( )
422
02
1
22
02
1
e
−
−
=
=>
rQk
kQrRru
∈
∈
(b) Express the energy dU in a
spherical shell of thickness dr and
surface area 4π r2:
( )drrurdU 2shell 4π=
For r < R: ( )
drr
R
kQ
drr
R
QkrRrdU
4
6
2
2
6
22
02
shell
2
2
4
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=< ∈π
For r > R: ( ) ( )
drrkQ
drrQkrRrdU
22
2
1
422
02
12
shell 4
−
−
=
=> ∈π
(c) Express the total electrostatic energy:
( ) ( )RrURrUU >+<= (3)
Electrostatic Energy and Capacitance
261
Integrate Ushell(r < R) from 0 to R: ( )
R
kQ
drr
R
kQRrU
R
10
2
2
0
4
6
2
shell
=
=< ∫
Integrate Ushell(r > R) from R to ∞: ( )
R
kQdrrkQRrU
R 2
2
22
2
1
shell ==> ∫∞ −
Substitute in equation (3) to obtain:
R
kQ
R
kQ
R
kQU
5
3
210
222
=+=
sphere.
hin theenergy wit field theincludesit because sphere for thegreater is
result The vanishes.integralfirst theso zero, is shell theinside field The
Combinations of Capacitors
33 •
Picture the Problem We can apply the properties of capacitors connected in parallel to
determine the number of 1.0-µF capacitors connected in parallel it would take to store a
total charge of 1 mC with a potential difference of 10 V across each capacitor. Knowing
that the capacitors are connected in parallel (parts (a) and (b)) we determine the potential
difference across the combination. In part (c) we can use our knowledge of how potential
differences add in
a series circuit to find the potential difference across the combination
and the definition of capacitance to find the charge on each capacitor.
(a) Express the number of
capacitors n in terms of the charge q
on each and the total charge Q:
q
Qn =
Relate the charge q on one capacitor
to its capacitance C and the potential
difference across it:
CVq =
Substitute to obtain:
CV
Qn =
Substitute numerical values and
evaluate n: ( )( ) 100V10F1
mC1 == µn
Chapter 24
262
(b) Because the capacitors are
connected in parallel the potential
difference across the combination is
the same as the potential difference
across each of them:
V10ncombinatio parallel ==VV
(c) With the capacitors connected in
series, the potential difference
across the combination will be the
sum of the potential differences
across the 100 capacitors:
( )
kV00.1
V10100
100ncombinatio series
=
=
= VV
Use the definition of capacitance to
find the charge on each capacitor:
( )( ) C0.10V10F1 µµ === CVq
34 •
Picture the Problem The capacitor array
is shown in the diagram. We can find the
equivalent capacitance of this combination
by first finding the equivalent capacitance
of the 3.0-µF and 6.0-µF capacitors in
series and then the equivalent capacitance
of this capacitor with the 8.0-µF capacitor
in parallel.
Express the equivalent capacitance
for the 3.0-µF and 6.0-µF capacitors
in series:
F6
1
F3
11
63 µµ +=+C
Solve for C3+6: F263 µ=+C
Find the equivalent capacitance of a
2-µF capacitor in parallel with an 8-
µF capacitor:
F10F8F282 µµµ =+=+C
*35 •
Picture the Problem Because we’re interested in the equivalent capacitance across
terminals a and c, we need to recognize that capacitors C1 and C3 are in series with each
other and in parallel with capacitor C2.
Find the equivalent capacitance of
C1 and C3 in series:
3131
111
CCC
+=
+
Electrostatic Energy and Capacitance
263
Solve for C1+3:
31
31
31 CC
CCC +=+
Find the equivalent capacitance of
C1+3 and C2 in parallel:
31
31
2312eq CC
CCCCCC ++=+= +
36 •
Picture the Problem Because the capacitors are connected in parallel we can add their
capacitances to find the equivalent capacitance of the combination. Also, because they
are in parallel, they have a common potential difference across them. We can use the
definition of capacitance to find the charge on each capacitor.
(a) Find the equivalent capacitance
of the two capacitors in parallel:
F30F20F0.10eq µµµ =+=C
(b) Because capacitors in parallel
have a common potential difference
across them:
V00.62010 =+= VVV
(c) Use the definition of capacitance
to find the charge on each capacitor:
( )( ) C0.60V6F101010 µµ === VCQ
and
( )( ) C120V6F202020 µµ === VCQ
37 ••
Picture the Problem We can use the properties of capacitors in series to find the
equivalent capacitance and the charge on each capacitor. We can then apply the definition
of capacitance to find the potential difference across each capacitor.
(a) Because the capacitors are
connected in series they have equal
charges:
VCQQ eq2010 ==
Express the equivalent capacitance
of the two capacitors in series:
F20
1
F10
11
eq µµ +=C
Solve for Ceq to obtain:
( )( ) F67.6
F20F10
F20F10
eq µµµ
µµ =+=C
Substitute to obtain: ( )( ) C0.40V6F67.62010 µµ === QQ
Chapter 24
264
(b) Apply the definition of
capacitance to find the potential
difference across each capacitor:
V00.4
F10
C0.40
10
10
10 === µ
µ
C
QV
and
V00.2
F20
C0.40
20
20
20 === µ
µ
C
QV
*38 ••
Picture the Problem We can use the properties of capacitors connected in series and in
parallel to find the equivalent capacitances for various connection combinations.
(a) parallel.in connected bemust they maximum, a be tois ecapacitanc their If
Find the capacitance of each
capacitor:
F153eq µ== CC
and
F5µ=C
(b) (1) Connect the three capacitors
in series:
F5
31
eq µ=C and F67.1eq µ=C
(2) Connect two in parallel, with the
third in series with that
combination:
( ) F10F52parallelin twoeq, µµ ==C
and
F5
1
F10
11
eq µµ +=C
Solve for Ceq: ( )( ) F33.3
F5F10
F5F10
eq µµµ
µµ =+=C
(3) Connect two in series, with the
third in parallel with that
combination:
F5
21
seriesin twoeq, µ=C
or
F5.2seriesin twoeq, µ=C
Find the capacitance equivalent to
2.5 µF and 5 µF in parallel:
F50.7F5F5.2eq µµµ =+=C
39 ••
Picture the Problem We can use the properties of capacitors connected in series and in
parallel to find the equivalent capacitance between the terminals and these properties and
the definition of capacitance to find the charge on each capacitor.
Electrostatic Energy and Capacitance
265
(a) Relate the equivalent
capacitance of the two capacitors in
series to their individual
capacitances:
F15
1
F4
11
154 µµ +=+C
Solve for C4+15: ( )( ) F16.3
F15F4
F15F4
154 µµµ
µµ =+=+C
Find the equivalent capacitance of
C4+15 in parallel with the
12-µF capacitor:
F2.15F12F16.3eq µµµ =+=C
(b) Using the definition of
capacitance, express and evaluate
the charge stored on the 12-µF
capacitor:
( )( )
mC40.2
V200F12
12121212
=
=
==
µ
VCVCQ
Because the capacitors in series
have the same charge:
( )( )
mC632.0
V200F16.3
154154
=
=
== +
µ
VCQQ
(c) The total energy stored is given
by:
2
eqtotal 2
1 VCU =
Substitute numerical values and
evaluate Utotal:
( )( ) J304.0V200F2.15
2
1 2
total == µU
40 ••
Picture the Problem We can use the properties of capacitors in series to establish the
results called for in this problem.
(a) Express the equivalent
capacitance of two capacitors in
series:
21
12
21eq
111
CC
CC
CCC
+=+=
Solve for Ceq by taking the
reciprocal of both sides of the
equation to obtain:
21
21
eq CC
CCC +=
Chapter 24
266
(b) Divide numerator and
denominator of this expression by
C1 to obtain:
2
1
2
2
eq
1
C
C
C
CC <
+
=
because 11
1
2 >+
C
C
.
Divide numerator and denominator
of this expression by C2 to obtain:
1
2
1
1
eq
1
C
C
C
CC <
+
=
because 11
2
1 >+
C
C
.
Using our result from part (a) for
two of the capacitors, add a third
capacitor C3 in series to obtain:
321
213231
321
21
eq
11
CCC
CCCCCC
CCC
CC
C
++=
++=
Take the reciprocal of both sides of
the equation to obtain: 313221
321
eq CCCCCC
CCCC ++=
41 ••
Picture the Problem Let Ceq1 represent the equivalent capacitance of the parallel
combination and Ceq the total equivalent capacitance between the terminals. We can use
the equations for capacitors in parallel and then in series to find Ceq. Because the charge
on Ceq is the same as on the 0.3-µF capacitor and Ceq1, we’ll know the charge on the 0.3-
µF capacitor when we have found the total charge Qeq stored by the circuit. We can find
the charges on the 1.0-µF and 0.25-µF capacitors by first finding the potential difference
across them and then using the definition of capacitance.
(a) Find the equivalent capacitance
for the parallel combination:
F1.25F0.25F1eq1 µµµ =+=C
Electrostatic Energy and Capacitance
267
The 0.30-µF capacitor is in series
with Ceq1 … find their equivalent
capacitance Ceq:
F242.0
and
F25.1
1
F3.0
11
eq
eq
µ
µµ
=
+=
C
C
(b) Express the total charge stored
by the circuit Qeq:
( )( )
C42.2
V10 F242.0
eq25.13.0eq
µ
µ
=
=
=== VCQQQ
The 1-µF and 0.25-µF capacitors,
being in parallel, have a common
potential difference. Express this
potential difference in terms of the
10 V across the system and the
potential difference across the 0.3-
µF capacitor: V93.1
F3.0
C42.2V10
V10
V10
3.0
3.0
3.025.1
=
−=
−=
−=
µ
µ
C
Q
VV
Using the definition of capacitance,
find the charge on the 1-µF and 0.25-
µF capacitors:
( )( ) C93.1V93.1F1111 µµ === VCQ
and ( )( )
C483.0
V93.1F25.025.025.025.0
µ
µ
=
== VCQ
(c) The total stored energy is given
by:
2
eq2
1 VCU =
Substitute numerical values and
evaluate U:
( )( ) J1.12V10F242.0 221 µµ ==U
42 ••
Picture the Problem Note that there are three parallel paths between a and b. We can
find the equivalent capacitance of the capacitors connected in series in the upper and
lower branches and then find the equivalent capacitance of three capacitors in parallel.
(a) Find the equivalent capacitance
of the series combination of
capacitors in the upper and lower
branch:
021
0
2
0
eq
00eq
2
C
or
111
C
C
C
CCC
==
+=
Chapter 24
268
Now we have two capacitors with
capacitance C0/2 in parallel with a
capacitor whose capacitance is C0.
Find their equivalent capacitance:
002
1
002
1
eq 2CCCCC' =++=
(b) If the central capacitance is
10C0, then:
002
1
002
1
eq 1110 CCCCC' =++=
43 ••
Picture the Problem Place four of the capacitors in series. Then the potential across each
is 100 V when the potential across the combination is 400 V. The equivalent capacitance
of the series is 2/4 µF = 0.5 µF. If we place four such series combinations in parallel, as
shown in the circuit diagram, the total capacitance between the terminals is 2 µF.
*44 ••
Picture the Problem We can connect two capacitors in parallel, all three in parallel, two
in series, three in series, two in parallel in series with the third, and two in series in
parallel with the third.
Connect 2 in parallel to obtain: F3F2F1eq µµµ =+=C
or
F5F4F1eq µµµ =+=C
or
F6F4F2eq µµµ =+=C
Connect all three in parallel to
obtain:
F7F4F2F1eq µµµµ =++=C
Connect two in series: ( )( ) F
3
2
F2F1
F2F1
eq µµµ
µµ =+=C
or
( )( ) F
5
4
F4F1
F4F1
eq µµµ
µµ =+=C
or
Electrostatic Energy and Capacitance
269
( )( ) F
3
4
F4F2
F4F2
eq µµµ
µµ =+=C
Connect all three in series:
( )( )( )
( )( ) ( )( ) ( )( ) F7
4
F4F1F4F2F2F1
F4F2F1
eq µµµµµµµ
µµµ =++=C
Connect two in parallel, in series
with the third:
( )( ) F
7
12
F4F2F1
F2F1F4
eq µµµµ
µµµ =++
+=C
or
( )( ) F
7
6
F4F2F1
F2F4F1
eq µµµµ
µµµ =++
+=C
or
( )( ) F
7
10
F4F2F1
F1F4F2
eq µµµµ
µµµ =++
+=C
Connect two in series, in parallel
with the third:
( )( ) F
3
14F4
F2F1
F2F1
eq µµµµ
µµ =++=C
or
( )( ) F
3
7F1
F2F4
F2F4
eq µµµµ
µµ =++=C
or
( )( ) F
5
14F2
F4F1
F4F1
eq µµµµ
µµ =++=C
45 •••
Picture the Problem Let C be the
capacitance of each capacitor in the ladder
and let Ceq be the equivalent capacitance of
the infinite ladder less the series capacitor
in the first rung. Because the capacitance is
finite and non-zero, adding one more stage
to the ladder will not change the
capacitance of the network. The
capacitance of the two capacitor
combination shown to the right is the
equivalent of the infinite ladder, so it has
capacitance Ceq also.
Chapter 24
270
(a) The equivalent capacitance of
the parallel combination of C and
Ceq is:
C + Ceq
The equivalent capacitance of the
series combination of C and
(C + Ceq) is Ceq, so:
eqeq
111
CCCC ++=
Simply this expression to obtain a
quadratic equation in Ceq:
02eq
2
eq =−+ CCCC
Solve for the positive value of Ceq to
obtain:
CCC 618.0
2
15
eq =⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
Because C = 1 µF:
F618.0eq µ=C
(b) The capacitance C′ required so
that the combination has the same
capacitance as the infinite ladder is:
eqCCC' +=
Substitute for Ceq and evaluate C′:
CCCC' 618.10.618 =+=
Because C = 1 µF:
F618.1 µ=C'
Parallel-Plate Capacitors
46 •
Picture the Problem The potential difference V across a parallel-plate capacitor, the
electric field E between its plates, and the separation d of the plates are related according
to V = Ed. We can use this relationship to find Vmax corresponding to dielectric
breakdown and the definition of capacitance to find the maximum charge on the
capacitor.
(a) Express the potential difference
V across the plates of the capacitor
in terms of the electric field between
the plates E and their separation d:
EdV =
Vmax corresponds to Emax: ( )( ) kV4.80mm1.6MV/m3max ==V
(b) Using the definition of
capacitance, find the charge Q ( )( ) mC60.9kV80.4F0.2
max
==
=
µ
CVQ
Electrostatic Energy and Capacitance
271
stored at this maximum potential
difference:
47 •
Picture the Problem The potential difference V across a parallel-plate capacitor, the
electric field E between its plates, and the separation d of the plates are related according
to V = Ed. In part (b) we can use the definition of capacitance and the expression for the
capacitance of a parallel-plate capacitor to find the required plate radius.
(a) Express the potential difference
V across the plates of the capacitor
in terms of the electric field between
the plates E and their separation d:
EdV =
Substitute numerical values and
evaluate V:
( )( ) V40.0mm2V/m102 4 =×=V
(b) Use the definition of capacitance
to relate the capacitance of the
capacitor to its charge and the
potential difference across it:
V
QC =
Express the capacitance of a
parallel-plate capacitor:
d
R
d
AC
2
00 π∈∈ ==
where R is the radius of the circular plates.
Equate these two expressions for C:
V
Q
d
R =
2
0 π∈
Solve for R to obtain:
V
QdR π∈0=
Substitute numerical values and
evaluate R:
( )( )( )( )
m24.4
V40m/NC1085.8
mm2C10
2212
=
⋅×= −π
µR
48 ••
Picture the Problem We can use the expression for the capacitance of a parallel-plate
capacitor to find the area of each plate and the definition of capacitance to find the
potential difference when the capacitor is charged to 3.2 µC. We can find the stored
energy using 221 CVU = and the definition of capacitance and the relationship between
Chapter 24
272
the potential difference across a parallel-plate capacitor and the electric field between its
plates to find the charge at which dielectric breakdown occurs. Recall that Emax, air = 3
MV/m.
(a) Relate the capacitance of a
parallel-plate capacitor to the area A
of its plates and their separation d:
d
AC 0∈=
Solve for A:
0∈
CdA =
Substitute numerical values and
evaluate A:
( )( ) 2
2212 m91.7m/NC108.85
mm5.0F14.0 =⋅×= −
µA
(b) Using the definition of
capacitance, express and evaluate
the potential difference across the
capacitor when it is charged to 3.2
µC:
V9.22
F0.14
C2.3 === µ
µ
C
QV
(c) Express the stored energy as a
function of the capacitor’s
capacitance and the potential
difference across it:
2
2
1 CVU =
Substitute numerical values and
evaluate U:
( )( ) J7.36V9.22F14.0 221 µµ ==U
(d) Using the definition of
capacitance, relate the charge on the
capacitor to breakdown potential
difference:
maxmax CVQ =
Relate the maximum potential
difference to the maximum electric
field between the plates:
dEV maxmax =
Substitute to obtain: dCEQ maxmax =
Substitute numerical values and
evaluate Qmax:
( )( )( )
C210
mm0.5MV/m3F14.0max
µ
µ
=
=Q
Electrostatic Energy and Capacitance
273
*49 ••
Picture the Problem The potential difference across the capacitor plates V is related to
their separation d and the electric field between them according to
V = Ed. We can use this equation with Emax = 3 MV/m to find dmin. In part (b) we can use
the expression for the capacitance of a parallel-plate capacitor to find the required area of
the plates.
(a) Use the relationship between the
potential difference across the plates
and the electric field between them
to find the minimum separation of
the plates:
mm333.0
MV/m3
V1000
max
min === E
Vd
(b) Use the expression for the
capacitance of a parallel-plate
capacitor to relate the capacitance to
the area of a plate:
d
AC 0∈=
Solve for A:
0∈
= CdA
Substitute numerical values and
evaluate A:
( )( ) 2
2212- m76.3m/NC108.85
mm333.0F1.0 =⋅×=
µA
Cylindrical Capacitors
50 •
Picture the Problem The capacitance of a cylindrical capacitor is given by ( )120 ln2 rrLC ∈= πκ where L is its length and r1 and r2 the radii of the inner and
outer conductors.
(a) Express the capacitance of the
coaxial cylindrical shell:
⎟⎠
⎞⎜⎝
⎛
∈=
R
r
LC
ln
2 0πκ
Substitute numerical values and
evaluate C:
( )( )( )
pF55.1
mm0.2
cm5.1ln
m12.0m/NC1085.812 2212
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅×=
−πC
Chapter 24
274
(b) Use the definition of
capacitance to express the charge
per unit length:
L
CV
L
Q ==λ
Substitute numerical values and
evaluate λ:
( )( ) nC/m5.15
m0.12
kV2.1pF55.1 ==λ
51 ••
Picture the Problem The diagram shows a
partial cross-sectional view of the inner
wire and the outer cylindrical shell. By
symmetry, the electric field is radial in the
space between the wire and the concentric
cylindrical shell. We can apply Gauss’s
law to cylindrical surfaces of radii r < R1,
R1 < r < R2, and r > R2 to find the electric
field and, hence, the energy density in
these regions.
(a) Apply Gauss’s law to a
cylindrical surface of radius r < R1
and length L to obtain:
( ) 02
0
inside =∈=
QrLEr π
and
0
1
=<RrE
Because E = 0 for r < R1: 0
1
=<Rru
Apply Gauss’s law to a cylindrical
surface of radius
R1 < r < R2 and length L to obtain:
( )
00
inside2 ∈=∈=
LQrLEr
λπ
where λ is the linear charge density.
Solve for Er to obtain:
r
k
r
LEr
λ
π
λ 2
2 0
=∈=
Express the energy density in the
region R1 < r < R2:
22
2
0
22
02
1
2
02
12
02
1
22
2
Lr
Qk
rL
kQ
r
kEu r
∈=⎟⎠
⎞⎜⎝
⎛∈=
⎟⎠
⎞⎜⎝
⎛∈=∈= λ
Electrostatic Energy and Capacitance
275
Apply Gauss’s law to a cylindrical
surface of radius
r > R2 and length L to obtain:
( ) 02
0
inside =∈=
QrLEr π
and
0
2
=>RrE
Because E = 0 for r > R2: 0
2
=>Rru
(b) Express the energy residing in a
cylindrical shell between the
conductors of radius r, thickness dr,
and volume 2π rL dr:
( )
dr
rL
kQdr
Lr
QkrL
drrrLudU
2
22
2
0
222
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ∈=
=
π
π
(c) Integrate dU from r = R1 to R2 to
obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛== ∫
1
2
22
ln
2
1
R
R
L
kQ
r
dr
L
kQU
R
R
Use 221 CVU = and the expression
for the capacitance of a cylindrical
capacitor to obtain:
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
∈
==
=
1
2
2
1
2
0
22
2
1
2
2
1
ln
ln
22
R
R
L
kQ
R
R
L
Q
C
Q
CVU
π
in agreement with the result from part (b).
52 •••
Picture the Problem Note that with the innermost and outermost cylinders connected
together the system corresponds to two cylindrical capacitors connected in parallel. We
can use ( )io
0
ln
2
RR
LC κπ ∈= to express the capacitance per unit length and then calculate and
add the capacitances per unit length of each of the cylindrical shell capacitors.
Relate the capacitance of a
cylindrical capacitor to its length L
and inner and outer radii Ri and Ro:
( )io
0
ln
2
RR
LC κπ ∈=
Divide both sides of the equation by
L to express the capacitance per unit ( )io
0
ln
2
RRL
C κπ ∈=
Chapter 24
276
length:
Express the capacitance per unit
length of the cylindrical system:
innerouter
⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛=
L
C
L
C
L
C
(1)
Find the capacitance per unit length
of the outer cylindrical shell
combination:
( )( )
( )
pF/m3.118
cm0.5cm0.8ln
1m/NC1085.82 2212
outer
=
⋅×=⎟⎠
⎞⎜⎝
⎛ −π
L
C
Find the capacitance per unit length
of the inner cylindrical shell
combination:
( )( )
( )
pF/m7.60
cm0.2cm0.5ln
1m/NC1085.82 2212
inner
=
⋅×=⎟⎠
⎞⎜⎝
⎛ −π
L
C
Substitute in equation (1) to obtain:
pF/m179
pF/m7.60pF/m3.118
=
+=
L
C
*53 ••
Picture the Problem We can use the
expression for the capacitance of a parallel-
plate capacitor of variable area and the
geometry of the figure to express the
capacitance of the goniometer.
The capacitance of the parallel-plate
capacitor is given by:
( )
d
AAC ∆−∈= 0
The area of the plates is:
( ) ( )22 21222122 θπθπ RRRRA −=−=
If the top plate rotates through an
angle ∆θ, then the area is reduced
by:
( ) ( )
22
2
1
2
2
2
1
2
2
θ
π
θπ ∆−=∆−=∆ RRRRA
Substitute for A and ∆A in the
expression for C to obtain: ( ) ( )
( )( )θθ
θθ
∆−−∈=
⎥⎦
⎤⎢⎣
⎡ ∆−−−∈=
d
RR
RRRR
d
C
2
22
2
1
2
20
2
1
2
2
2
1
2
2
0
Electrostatic Energy and Capacitance
277
54 ••
Picture the Problem Let C be the capacitance of the capacitor when the pressure is P
and C′ be the capacitance when the pressure is P + ∆P. We’ll assume that (a) the change
in the thickness of the plates is small, and (b) the total volume of material between the
plates is conserved. We can use the expression for the capacitance of a dielectric-filled
parallel-plate capacitor and the definition of Young’s modulus to express the change in
the capacitance ∆C of the given capacitor when the pressure on its plates is increased by
∆P.
Express the change in capacitance
resulting from the decrease in
separation of the capacitor plates by
∆d:
d
A
dd
A'CC'C 00 ∈−∆−
∈=−=∆ κκ
Because the volume is constant: AdA'd' =
or
A
dd
dA
d
dA' ⎟⎠
⎞⎜⎝
⎛
∆−=⎟⎠
⎞⎜⎝
⎛=
'
Substitute for A′ in the expression
for ∆C and simplify to obtain:
( )
( )
( ) ⎥⎦
⎤⎢⎣
⎡ −∆−=
⎥⎦
⎤⎢⎣
⎡ −∆−
∈=
∈−∆−
∈=
∈−⎟⎠
⎞⎜⎝
⎛
∆−∆−
∈=∆
1
1
2
2
2
2
0
02
2
0
00
dd
dC
dd
d
d
A
d
Ad
ddd
A
d
A
dd
d
dd
AC
κ
κκ
κκ
From the definition of Young’s
modulus:
Y
P
d
d −=∆ ⇒ d
Y
Pd ⎟⎠
⎞⎜⎝
⎛−=∆
Substitute for ∆d in the expression
for ∆C to obtain:
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −⎭⎬
⎫
⎩⎨
⎧ ⎟⎠
⎞⎜⎝
⎛+=
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢
⎣
⎡
−
⎭⎬
⎫
⎩⎨
⎧ ⎟⎠
⎞⎜⎝
⎛+
∈=∆
−
11
1
2
2
2
0
Y
PC
d
Y
Pd
d
d
AC κ
Expand
2
1
−
⎟⎠
⎞⎜⎝
⎛ −
Y
P
binomially to
obtain:
...3211
22
+⎟⎠
⎞⎜⎝
⎛+−=⎟⎠
⎞⎜⎝
⎛ −
−
Y
P
Y
P
Y
P
Chapter 24
278
Provided P << Y:
Y
P
Y
P 211
2
−≈⎟⎠
⎞⎜⎝
⎛ −
−
Substitute in the expression for ∆C
and simplify to obtain:
C
Y
P
Y
PCC 2121 −=⎥⎦
⎤⎢⎣
⎡ −−=∆
Spherical Capacitors
*55 ••
Picture the Problem We can use the definition of capacitance and the expression for the
potential difference between charged concentric spherical shells to show that ( ).4 12210 RRRRC −∈= π
(a) Using its definition, relate the
capacitance of the concentric
spherical shells to their charge Q
and the potential difference V
between their surfaces:
V
QC =
Express the potential difference
between the conductors:
21
12
21
11
RR
RRkQ
RR
kQV −=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
Substitute to obtain: ( )
12
210
12
21
21
12
4
RR
RR
RRk
RR
RR
RRkQ
QC
−
∈=
−=−=
π
(b) Because R2 = R1 + d: ( )
22
1
1
2
1
1121
RR
dRR
dRRRR
=≈
+=
+=
because d is small.
Substitute to obtain:
d
A
d
RC 0
2
04 ∈=∈≈ π
Electrostatic Energy and Capacitance
279
56 ••
Picture the Problem The diagram shows a
partial cross-sectional view of the inner and
outer spherical shells. By symmetry, the
electric field is radial. We can apply
Gauss’s law to spherical surfaces of radii r
< R1, R1 < r < R2, and r > R2 to find the
electric field and, hence, the energy density
in these regions.
(a) Apply Gauss’s law to a spherical
surface of radius r < R1 to obtain:
( ) 04
0
inside2 =∈=
QrEr π
and
0
1
=<RrE
Because E = 0 for r < R1: 0
1
=<Rru
Apply Gauss’s law to a spherical
surface of radius R1 < r < R2 to
obtain:
( )
00
inside24 ∈=∈=
QQrEr π
Solve for Er to obtain:
22
04 r
kQ
r
QEr =∈= π
Express the energy density in the
region R1 < r < R2:
4
2
0
2
2
202
12
02
1
2r
Qk
r
kQEu r
∈=
⎟⎠
⎞⎜⎝
⎛∈=∈=
Apply Gauss’s law to a cylindrical
surface of radius
r > R2 to obtain:
( ) 04
0
inside2 =∈=
QrEr π
and
0
2
=>RrE
Because E = 0 for r > R2: 0
2
=>Rru
Chapter 24
280
(b) Express the energy in the
electrostatic field in a spherical shell
of radius r, thickness dr, and volume
4π r2dr between the conductors:
( )
dr
r
kQ
dr
r
QkrdrrurdU
2
2
4
2
0
2
22
2
2
44
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ∈== ππ
(c) Integrate dU from r = R1 to R2 to
obtain:
( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∈
−=
−== ∫
210
122
21
12
2
2
2
42
1
22
2
1
RR
RRQ
RR
RRkQ
r
drkQU
R
R
π
Note that the quantity in parentheses is 1/C , so we have .221 CQU =
57 •••
Picture the Problem We know, from Gauss’s law, that the field inside the shell is zero.
Applying Gauss’s law to a spherical surface of radius R > r will allow us to find the
energy density in this region. We can then express the energy in the electrostatic field in a
spherical shell of radius R, thickness dR, and volume 4π R2dR outside the spherical shell
and find the total energy in the electric field by integrating from r to ∞. If we then
integrate the same expression from r to R we can find the radius R of the sphere such that
half the total electrostatic field energy of the system is contained within that sphere.
Apply Gauss’s law to a spherical shell
of radius R > r to obtain:
( )
00
inside24 ∈=∈=
QQREr π
Solve for Er outside the spherical
shell: 2R
kQEr =
Express the energy density in the
region R > r: 4
2
0
22
202
12
02
1
2R
Qk
R
kQEu R
∈=⎟⎠
⎞⎜⎝
⎛∈=∈=
Express the energy in the
electrostatic field in a spherical shell
of radius R, thickness dR, and
volume 4πR2dR outside the spherical
shell:
( )
dR
R
kQ
dR
R
QkR
dRRuRdU
2
2
4
2
0
2
2
2
2
2
4
4
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ∈=
=
π
π
Integrate dU from r to ∞ to obtain:
r
kQ
R
dRkQU
r 22
2
2
2
tot == ∫∞
Electrostatic Energy and Capacitance
281
Integrate dU from r to R to obtain:
⎟⎠
⎞⎜⎝
⎛ −== ∫ RrkQR'dR'kQU
R
r
11
22
2
2
2
Set tot21 UU = to obtain:
r
kQ
Rr
kQ
4
11
2
22
=⎟⎠
⎞⎜⎝
⎛ −
Solve for R: rR 2=
Disconnected and Reconnected Capacitors
58 ••
Picture the Problem Let C1 represent the capacitance of the 2.0-µF capacitor and C2 the
capacitance of the 2nd capacitor. Note that when they are connected as described in the
problem statement they are in parallel and, hence, share a common potential difference.
We can use the equation for the equivalent capacitance of two capacitors in parallel and
the definition of capacitance to relate C2 to C1 and to the charge stored in and the
potential difference across the equivalent capacitor.
Using the definition of capacitance,
find the charge on capacitor C1:
( )( ) C24V12F211 µµ === VCQ
Express the equivalent capacitance
of the two-capacitor system and
solve for C2:
21eq CCC +=
and
1eq2 CCC −=
Using the definition of capacitance,
express Ceq in terms of Q2 and V2:
2
1
2
2
eq V
Q
V
QC ==
where V2 is the common potential
difference (they are in parallel) across the
two capacitors and Q1 and Q2 are the
(equal) charges on the two capacitors.
Substitute to obtain:
1
2
1
2 CV
QC −=
Substitute numerical values and
evaluate C2:
F00.4F2
V4
C24
2 µµµ =−=C
Chapter 24
282
59 ••
Picture the Problem Because, when the capacitors are connected as described in the
problem statement, they are in parallel, they will have the same potential difference
across them. In part (b) we can find the energy lost when the connections are made by
comparing the energies stored in the capacitors before and after the connections.
(a) Because the capacitors are in parallel: kV00.2400100 ==VV
(b) Express the energy lost when the
connections are made in terms of the
energy stored in the capacitors before
and after their connection:
afterbefore UUU −=∆
Express and evaluate Ubefore:
( )
( ) ( )
mJ00.1
pF500kV2 221
400100
2
2
1
2
4004002
12
1001002
1
400100before
=
=
+=
+=
+=
CCV
VCVC
UUU
Express and evaluate Uafter:
( )
( ) ( )
mJ00.1
pF500kV2 221
400100
2
2
1
2
4004002
12
1001002
1
400100after
=
=
+=
+=
+=
CCV
VCVC
UUU
Substitute to obtain: 0mJ1.00mJ1.00 =−=∆U
*60 ••
Picture the Problem When the capacitors are reconnected, each will have the charge it
acquired while they were connected in series across the 12-V battery and we can use the
definition of capacitance and their equivalent capacitance to find the common potential
difference across them. In part (b) we can use 221 CVU = to find the initial and final
energy stored in the capacitors.
(a) Using the definition of
capacitance, express the potential
difference across each capacitor when
they are reconnected:
eq
2
C
QV = (1)
where Q is the charge on each capacitor
before they are disconnected.
Electrostatic Energy and Capacitance
283
Find the equivalent capacitance of
the two capacitors after they are
connected in parallel:
F16
F12F4
21eq
µ
µµ
=
+=
+= CCC
Express
the charge Q on each
capacitor before they are
disconnected:
VC'Q eq=
Express the equivalent capacitance
of the two capacitors connected in
series:
( )( ) F3
F12F4
F12F4
21
21
eq µµµ
µµ =+=+= CC
CCC'
Substitute to find Q: ( )( ) C36V12F3 µµ ==Q
Substitute in equation (1) and
evaluate V:
( ) V50.4
F16
C362 == µ
µV
(b) Express and evaluate the energy
stored in the capacitors initially:
( )( )
J216
V12F3 221
2
ieq2
1
i
µ
µ
=
== VC'U
Express and evaluate the energy
stored in the capacitors when they
have been reconnected:
( )( )
J162
V5.4F16 221
2
feq2
1
f
µ
µ
=
== VCU
61 ••
Picture the Problem Let C1 represent the capacitance of the 1.2-µF capacitor and C2 the
capacitance of the 2nd capacitor. Note that when they are connected as described in the
problem statement they are in parallel and, hence, share a common potential difference.
We can use the equation for the equivalent capacitance of two capacitors in parallel and
the definition of capacitance to relate C2 to C1 and to the charge stored in and the
potential difference across the equivalent capacitor. In part (b) we can use 221 CVU = to
find the energy before and after the connection was made and, hence, the energy lost
when the connection was made.
(a) Using the definition of
capacitance, find the charge on
capacitor C1:
( )( ) C36V30F2.111 µµ === VCQ
Express the equivalent capacitance
of the two-capacitor system and
solve for C2:
21eq CCC +=
and
Chapter 24
284
1eq2 CCC −=
Using the definition of capacitance,
express Ceq in terms of Q2 and V2:
2
1
2
2
eq V
Q
V
QC ==
where V2 is the common potential
difference (they are in parallel) across the
two capacitors.
Substitute to obtain:
1
2
1
2 CV
QC −=
Substitute numerical values and
evaluate C2:
F40.2F2.1
V10
C36
2 µµµ =−=C
(b) Express the energy lost when the
connections are made in terms of the
energy stored in the capacitors
before and after their connection:
( )2feq21121
2
feq2
12
112
1
afterbefore
VCVC
VCVC
UUU
−=
−=
−=∆
Substitute numerical values and evaluate ∆U:
( )( )[ ( )( ) ] J360V10F6.3V30F2.1 2221 µµµ =−=∆U
62 ••
Picture the Problem Because, when the capacitors are connected as described in the
problem statement, they are in parallel, they will have the same potential difference
across them. In part (b) we can find the energy lost when the connections are made by
comparing the energies stored in the capacitors before and after the connections.
(a) Using the definition of
capacitance, express the charge Q
on the capacitors when they have
been reconnected:
( )VCC
VCVC
QQQ
100400
100100400400
100400
−=
−=
−=
where V is the common potential difference
to which the capacitors have been charged.
Substitute numerical values to obtain:
( )( ) nC600kV2pF100pF400 =−=Q
Using the definition of capacitance,
relate the equivalent capacitance,
charge, and final potential difference
for the parallel connection:
( ) f21 VCCQ +=
Electrostatic Energy and Capacitance
285
Solve for and evaluate Vf:
kV20.1
pF400pF100
C600
21
f
=
+=+=
n
CC
QV
across both capacitors.
(b) Express the energy lost when the
connections are made in terms of the
energy stored in the capacitors before
and after their connection:
( )2feq22221121
2
feq2
12
222
12
112
1
afterbefore
VCVCVC
VCVCVC
UUU
−+=
−+=
−=∆
Substitute numerical values and evaluate ∆U:
( )( )[ ( )( ) ( )( ) ] mJ640.0kV2.1pF500kV2pF400kV2pF100 22221 =−+=∆U
63 ••
Picture the Problem When the capacitors are reconnected, each will have a charge equal
to the difference between the charges they acquired while they were connected in parallel
across the 12-V battery. We can use the definition of capacitance and their equivalent
capacitance to find the common potential difference across them. In part (b) we can use
2
2
1 CVU = to find the initial and final energy stored in the capacitors.
(a) Using the definition of
capacitance, express the potential
difference across the capacitors
when they are reconnected:
21
f
eq
f
f CC
Q
C
Q
V +== (1)
where Qf is the common charge on the
capacitors after they are reconnected.
Express the final charge Qf on each
capacitor:
12f QQQ −=
Use the definition of capacitance to
substitute for Q2 and Q1:
( )VCCVCVCQ 1212f −=−=
Substitute in equation (1) to obtain:
V
CC
CCV
21
12
f +
−=
Substitute numerical values and
evaluate Vf:
( ) V00.6V12
F4F12
F4F12
f =+
−= µµ
µµV
Chapter 24
286
(b) Express and evaluate the energy
stored in the capacitors initially:
( )
( ) ( )
mJ15.1
F4F12V12 221
21
2
2
1
2
22
12
12
1
i
=
+=
+=
+=
µµ
CCV
VCVCU
Express and evaluate the energy
stored in the capacitors when they
have been reconnected:
( )
( ) ( )
mJ288.0
F4F12V6 221
21
2
f2
1
2
f22
12
f12
1
f
=
+=
+=
+=
µµ
CCV
VCVCU
*64 ••
Picture the Problem Let the numeral 1 refer to the 20-pF capacitor and the numeral 2 to
the 50-pF capacitor. We can use conservation of charge and the fact that the connected
capacitors will have the same potential difference across them to find the charge on each
capacitor. We can decide whether electrostatic potential energy is gained or lost when the
two capacitors are connected by calculating the change ∆U in the electrostatic energy
during this process.
(a) Using the fact that no charge is
lost in connecting the capacitors,
relate the charge Q initially on the 20-
pF capacitor to the charges on the two
capacitors when they have been
connected:
21 QQQ += (1)
Because the capacitors are in parallel,
the potential difference across them is
the same:
21 VV = ⇒
2
2
1
1
C
Q
C
Q =
Solve for Q1 to obtain:
22
1
1 QC
CQ =
Substitute in equation (1) and solve
for Q2 to obtain:
21
2 1 CC
QQ += (2)
Use the definition of capacitance to
find the charge Q initially on the 20-
pF capacitor:
( )( ) nC60kV3pF201 === VCQ
Electrostatic Energy and Capacitance
287
Substitute in equation (2) and
evaluate Q2:
nC9.42
pF50pF201
nC60
2 =+=Q
Substitute in equation (1) to obtain:
nC17.1nC42.960nC
21
=−=
−= QQQ
(b) Express the change in the
electrostatic potential energy of the
system when the two capacitors are
connected:
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
−=
−=∆
1eq
2
1
2
eq
2
if
11
2
22
CC
Q
C
Q
C
Q
UUU
Substitute numerical values and
evaluate ∆U:
( )
J3.64
pF20
1
pF70
1
2
nC60 2
µ−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=∆U
connected.
are capacitors twothelost when isenergy ticelectrosta 0, Because <∆U
65 •••
Picture the Problem Let upper case Qs refer to the charges before S3 is closed and lower
case qs refer to the charges after this switch is closed. We can use conservation of charge
to relate the charges on the capacitors before S3 is closed to their charges when this
switch is closed. We also know that the sum of the potential differences around the circuit
when S3 is closed must be zero and can use this to obtain a fourth equation relating the
charges on the capacitors after the switch is closed to their capacitances. Solving these
equations simultaneously will yieldMais conteúdos dessa disciplina
Introdução ao Direito Eleitoral- Alistamento Eleitoral: Regras e Procedimentos
- Edição de material via content-update-api
- arquivo_225
- arquivo_101
- arquivo_18
- arquivo_18
- arquivo_31
- campanha_teste_jul17_13_06_97.txt
- campanha_teste_jul17_12_15_72
- updateOrder 0 0 1 2 0 0 1 2 0 1 0 0 1 2 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0
- Psicología educativa
- Intervenciones cognitivo conductuales para diabéticos en México
- youdontsay
- KoalaPerfil
Mais conteúdos dessa disciplina
- Introdução ao Direito Eleitoral
- Alistamento Eleitoral: Regras e Procedimentos
- Edição de material via content-update-api
- arquivo_225
- arquivo_101
- arquivo_18
- arquivo_18
- arquivo_31
- campanha_teste_jul17_13_06_97.txt
- campanha_teste_jul17_12_15_72
- updateOrder 0 0 1 2 0 0 1 2 0 1 0 0 1 2 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0
- Psicología educativa
- Intervenciones cognitivo conductuales para diabéticos en México
- youdontsay
- KoalaPerfil