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:
1 1
1
1 qV 2eV K = =
Express the kinetic energy of the lithium nucleus when it has been accelerated through a potential difference V2:
2 2
2
2 qV 3eV
K = =
Equate the kinetic energies to obtain:
2 1 3
2eV = eV or
1 3 2 2
V
V
=
and( )
b iscorrect.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
l
ld
dV
E
=
−
provided V is known and differentiable or froml
l
∆
∆
−
=
V
5 •
Determine the Concept Because the field lines are always perpendicular to equipotential surfaces, you move always perpendicular to the field.
6 ••
Determine the ConceptV 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
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: 2 ˆ 2 ˆ 0 at at
= −
=
+
= + − +
i i
E E
E
a kQ a
kQ
a Q a Q
r r
r
Express the potential V at the origin:
a kQ a
kQ a kQ
V V
V Q a Q a
2
at at
= + =
+
= + − +
12 •
Picture the Problem We can use
E
i
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
i i
i i
E
ˆ 0 if 4
0 if 4
ˆ 0 if 1
0 if 1 4 ˆ 4
ˆ 4
ˆ
0
⎥ ⎦ ⎤ ⎢
⎣ ⎡
< ≥ −
=
⎥ ⎦ ⎤ ⎢
⎣ ⎡
< −
≥ −
= ∂
∂ − =
+ ∂
∂ − = ∂ ∂ − =
x x
x x x
x
V x x x
V
r
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. (c)iscorrect.
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: i i i E
E E
ˆ 2 ˆ ˆ
2 2
2
at at
a kQ a
kQ a
kQ
a Q a Q
= +
=
+
= r+ − r−
r
Express the potential V at the origin:
( )
0
at at
= − + =
+
= + − −
a Q k a kQ
V V
V Q a Q a
and (c)iscorrect
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.
(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. (c)iscorrent.
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:
Ed V =
Assuming that the thundercloud is at a distance of about 1 km above the surface of the earth, the potential difference is approximately:
(
)(
)
V
10
00
.
3
m
10
V/m
10
3
9 3 6
×
=
×
=
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
plug as a function of their separation d and electric field E between them: Substitute numerical values and
evaluate V:
(
)(
)
kV
0
.
20
m
10
V/m
10
2
7 3=
×
=
−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:
f f i
i U K U
K + = +
or, because Ui = Kf = 0, f
i U K =
Because each proton has kinetic
energy K:
r
e
K
0 2
4
2
∈
=
π
⇒r
e
K
0 2
8
∈
=
π
Substitute numerical values and evaluate K:
(
)
(
)(
)
MeV 719 . 0
J 10 6 . 1
eV 1 J 10 15 . 1 m 10 m N / C 10 85 . 8 8
C 10 6 . 1
19 13
15 2 2
12
2 19
=
× × ×
= ⋅
× ×
= − −
− −
−
π
K
(b) Express and evaluate the ratio of
the two energies:
938
MeV
0
.
0767
%
MeV
719
.
0
rest=
=
=
E
K
f
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:
(
3MV/m)(
2mm)
= 6000V=
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:
( ) ( )
(
)( )
kV 8.00
m 4 kN/C 2 0 m 4
m 4
0
− =
− =
− = ⋅ − =
−V
∫
r drl∫
EdlV
b
a
E
(b) By definition, ∆U is given by:
(
)(
)
mJ
0
.
24
kV
8
C
3
−
=
−
=
∆
=
∆
U
q
V
µ
(c) Use conservation of energy to relate ∆U and ∆K:
0
= ∆ +
∆K U
or
0
0m
4
−
K
+
∆
U
=
K
Because K0 = 0: K4m =−∆U = 24.0mJ Use the definition of finite potential
difference to obtain:
( ) ( )
(
)
(
)(
0)
0 0
kV/m
2
x
x
x
x
E
x
V
x
V
x−
−
=
−
−
=
−
(d) For V(0) = 0: V
( )
x −0=−(
2kV/m)(
x−0)
or( )
x(
)
xV = − 2kV/m
(e) For V(0) = 4 kV: V
( )
x −4kV=−(
2kV/m)(
x−0)
or( )
x(
)
xV = 4kV− 2kV/m
(f) For V(1m) = 0: V
( )
x −0=−(
2kV/m)(
x−1)
or
( )
x(
)
x22 •
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/m
5.00
m
0.1
V
500
=
=
∆
∆
=
x
V
E
potential.
higher
at the
is
plate
positive
the
plate,
negative
the
toward
and
plate
positive
the
from
away
is
charge
test
a
on
force
electric
the
Because
(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:
(
)
(
)
J
10
01
.
8
V
0
0
5
C
10
6
.
1
17 19 −
−
×
=
×
=
∆
=
q
V
W
Convert 8.01×10−17 J to eV:
(
)
eV 500
J 10 1.6
eV 1 J 10
8.01 17 19
=
⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛
× ×
= − −
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:
eV 500
− = − =
∆U W
Apply conservation of energy to obtain:
eV 500
= ∆ − =
∆K U
23 •
(a) Express and evaluate the Coulomb potential of the charge:
(
)
(
)
kV 50 . 4
m 4
C 2 /C m N 10 99 .
8 9 2 2
=
⋅ × =
=
µ
r kq V
(b) Relate the work that must be done to the magnitude of the charge and the potential difference through which the charge is moved:
(
)(
)
mJ
5
.
13
kV
50
.
4
C
3
=
=
∆
=
q
V
µ
W
(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
q
kq
V
q
W
2 33 2
∆
=
=
Substitute numerical values and evaluate W:
(
)
(
)(
)
mJ
5
.
13
m
4
C
3
C
2
/C
m
N
10
99
.
8
9 2 2=
⋅
×
=
µ
µ
W
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
e e k r
q kq
U U
K W
2 i
ext 0
= − − = −
=
− = ∆ + ∆ =
+ −
Substitute numerical values and evaluate Wext:
(
)(
)
J 10 24 . 8 m
10 2.80
C 10 6 . 1 /C m N 10 99 .
8 19
10
2 19 2
2 9 ext
− −
−
× = ×
× ⋅
× =
Convert Wext to eV:
(
)
eV 14 . 5
J 10 1.6
eV 1 J 10 24 .
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:
f K K W =∆ =
or
2 2 1
mv
V
e
∆
=
Solve for v to obtain:m V e v= 2 ∆
Substitute numerical values and evaluate v:
(
)
(
)
m/s
10
10
.
3
kg
10
1.67
MV
5
C
10
1.6
2
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/m
2.50
m
2
MV
5
=
=
∆
∆
=
x
V
E
*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:
f K K W =∆ =
or
V e
Kf = ∆ (1)
(a) Substitute numerical values and evaluate Kf:
( )(
1 30kV)
3 104eVf = e = ×
(b) Convert this energy to eV:
(
)
J 10 80 . 4
eV J 10 1.6 eV 10 3
15
19 4
f
−
−
× =
⎟⎟ ⎠ ⎞ ⎜⎜
⎝
⎛ ×
× =
K
(c) From equation (1) we have:
mv
2=
e
∆
V
f 2 1 Solve for vf to obtain:
m V e vf = 2 ∆
Substitute numerical values and evaluate vf:
(
)
(
)
m/s
10
03
.
1
kg
10
11
.
9
kV
0
3
C
10
1.6
2
8 1 3 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
= ∆ +
∆K U
or
0
i f i
f −K +U −U = K
Because Kf = Ui = 0 and Ki = E: −E+Uf =0 Letting r be the separation of the
particles at closest approach, express Uf:
( )( )
r
kZe
r
e
Ze
k
r
q
kq
U
2 nucleus
f
2
2
=
=
=
αSubstitute to obtain:
0
2
2=
+
−
r
kZe
E
Solve for r to obtain:
E
kZe
r
2
2
=
(
)
( )
(
)
(
)
(
)
4.55 10 m 45.4fmJ/eV 10
6 . 1 MeV 5
C 10 1.6 79 /C m N 10 8.99
2 14
19
2 19 2
2 9
5 = × =
×
× ⋅
×
= −
−
− r
For a 9.0-MeV α particle and a gold nucleus:
(
)
( )
(
)
(
)
(
)
2.53 10 m 25.3fmJ/eV 10
6 . 1 MeV 9
C 10 1.6 79 /C m N 10 8.99
2 14
19
2 19 2
2 9
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
4 3 2 1
4 3 2 1
i i
q
r
k
q
q
q
q
r
k
r
kq
r
kq
r
kq
r
kq
V
(a) If the charges are positive:
( )(
)
kV
4
.
25
C
2
4
m
2
2
/C
m
N
10
8.99
9 2 2=
⋅
×
=
µ
V
(b) If three of the charges are positive
and one is negative:
( )(
)
kV
7
.
12
C
2
2
m
2
2
/C
m
N
10
8.99
9 2 2=
⋅
×
=
µ
V
(c) If two are positive and two are negative:
0
=
V
29 •
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 q k r q k V
i i i
(a) For q1 = q2 = q3 = 2 µC:
(
)
(
)
12.9kVm 5 3
1 m 2 3
1 m 3
1 C 2 /C m N 10 99 .
8 9 2 2 ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛
+ +
⋅ ×
=
µ
V
(b) For q1 = q2 = 2 µC and q3 = −2 µC:
(
)
(
)
7.55kVm 5 3
1 m 2 3
1 m 3
1 C 2 /C m N 10 99 .
8 9 2 2 ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛
− +
⋅ ×
=
µ
V
(c) For q1 = q3 = 2 µC and q2 = −2 µC:
(
)
(
)
4.44kVm 5 3
1 m 2 3
1 m 3
1 C 2 /C m N 10 99 .
8 9 2 2 ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛
+ −
⋅ ×
=
µ
V
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 q k V
Substitute numerical values and evaluate VC:
(
)
(
)
12.0kVm 3
1 m 3
1 C 2 /C m N 10 99 . 8 1
1 9 2 2
B A
C ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛
+ ⋅
× =
⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛
+
=
µ
r r kq V
(b) Express the required work in terms of the change in the potential
energy of the system:
(
)(
)
mJ
60.0
kV
12.0
µC
5
C 5
=
=
=
∆
=
U
q
V
W
(
)
0 m3 C 2 m 3
C 2 /C m N 10 99 .
8 9 2 2
B B A
A
C ⎟⎟=
⎠ ⎞ ⎜⎜
⎝
⎛ + −
⋅ × =
⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛
+
=
µ
µ
r q r q k V
and W =∆U =q5VC =
(
5µC)( )
0 = 031 •
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 q k r q k V
i i i
6
6 1
=
=
∑
=
Substitute numerical values and
evaluate V:
(
)
kV
270
m
6
.
0
C
3
/C
m
N
10
99
.
8
6
9 2 2=
⋅
×
=
µ
V
(b) Using geometry, find the distance from each charge to the north pole:
m
2
6
.
0
'
=
r
Proceed as in (a) with
r
'
=
0
.
6
2
m
:(
)
kV
191
m
2
6
.
0
C
3
/C
m
N
10
99
.
8
6
'
6
2 2 9 6
1 '
=
⋅
×
=
=
=
∑
=
µ
r
q
k
r
q
k
V
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
3
2
3
+
a
=
kq'
a
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, r= x−a and x+a 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:
⎟ ⎟ ⎠ ⎞ ⎜
⎜ ⎝ ⎛
+ + − =
a x a x kq
V 1 1
(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
dV
E
x*34 ••
a
x
=
21 on the x axis from the change in the potential energy of this third charge. Express the potential at x:( ) ( ) (
)
a x
e k x
e k x V
− − +
= 3 2
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: 0
2 3
= − −
a x x
For x > 0, V(x) = 0 when: x= 3a For 0 < x < a, V(x) = 0 when: x= 0.6a (c) Express the work that must be
done in terms of the change in potential energy of the charge:
( )
a
qV
U
W
=
∆
=
21Evaluate the potential at
x
=
21a
:( )
( ) (
)
a ke a
ke a
ke
a a
e k a
e k a V
2 4 6
2 3
2 1 2 1 2
1
= − =
Substitute to obtain:
a
ke
a
ke
e
W
2
2
2
=
⎟
⎠
⎞
⎜
⎝
⎛
=
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
a b V
V −
(b) Express E in terms of Vb−Va and
the separation of points a and b:
x
V
V
x
V
E
b ax
∆
−
=
∆
∆
=
Substitute numerical values and evaluate Ex:
kV/m
25.0
m
4
V
10
5=
=
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:
m 5 . 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 2
2
y
x
r
=
+
.(a) Express the potential on the x axis
as a function of x and q:
( )
x
kq
x
V
=
Evaluate V at x = 3 m:
( )
(
)
(
)
kV
99
.
8
m
3
C
3
/C
m
N
10
99
.
8
m
3
2 2 9
=
⋅
×
=
µ
Evaluate V at x = 3.01 m:
(
)
(
)
(
)
kV
96
.
8
m
01
.
3
C
3
/C
m
N
10
99
.
8
m
01
.
3
2 2 9
=
⋅
×
=
µ
V
(b)The potential decreases as x increases and:
kV/m
00
.
3
m
3.00
m
3.01
kV
8.99
kV
8.96
=
−
−
−
=
∆
∆
−
x
V
(c) Express the Coulomb field as a
function of x:
( )
x
2kq
x
E
=
Evaluate this expression at
x = 3.00 m to obtain:
( )
(
)
(
)
( )
kV/m
00
.
3
m
3
C
3
/C
m
N
10
99
.
8
m
3
22 2 9
=
⋅
×
=
µ
E
in agreement with our result in (b). (d) Express the potential at (x, y)
due to a point charge q at the origin:
( )
,
2 2y
x
kq
y
x
V
+
=
Evaluate this expression at (3.00 m, 0.01 m):
(
)
(
)
(
)
(
3
.
00
m
) (
0
.
01
m
)
8
.
99
kV
C
3
/C
m
N
10
99
.
8
m
m,0.01
.00
3
2 2
2 2 9
=
+
⋅
×
=
µ
V
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 q1and q2located 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 q1and q2located at the origin and at
x = 6 m:
( )
⎟⎟⎠ ⎞ ⎜⎜
⎝ ⎛
+ =
2 2 1 1
Substitute numerical values and evaluate V(3 m):
( )
(
)
0
m
3
C
3
m
3
C
3
/C
m
N
10
99
.
8
9 2 2=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
×
⋅
×
=
µ
µ
x
V
(b) Express the Coulomb field on the x axis as the sum of the fields due to the charges q1and q2located at the origin and at
x = 6 m:
⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = + = 2 2 2 2 1 1 2 2 2 2 1 1 r q r q k r kq r kq Ex
Substitute numerical values and evaluate E(3 m):
(
)
( ) ( )
kV/m
99
.
5
m
3
C
3
m
3
C
3
/C
m
N
10
99
.
8
2 2 2 2 9=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
×
⋅
×
=
µ
µ
xE
(c) Express the potential on the x axis as the sum of the potentials due to the charges q1and q2located at the origin and at
x = 6 m:
( )
⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = 2 2 1 1 r q r q k x VSubstitute numerical values and evaluate V(3.01 m):
(
)
(
)
V
9
.
59
m
99
.
2
C
3
m
01
.
3
C
3
/C
m
N
10
99
.
8
m
01
.
3
9 2 2−
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
×
⋅
×
=
µ
µ
V
Compute −∆V/∆x:
(
3
.
00
m
)
kV/m
99
.
5
m
3.00
m
3.01
0
V
59.9
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
a b V
(b) Express E in terms of Vb−Va and
the separation of points a and b:
y
V
V
y
V
E
b ay
∆
−
=
∆
∆
=
Substitute numerical values and evaluate Ey:
kV/m
00
.
5
m
4
V
10
2
4=
×
=
yE
40 •
Picture the Problem Given V(x), we can find Ex from −dV/dx.
(a) Find Ex from −dV/dx:
[
]
kV/m 00 . 3
3000 2000
− =
+ −
= x
dx d Ex
(b) Find Ex from −dV/dx:
[
]
kV/m 00 . 3
3000 4000
− =
+ −
= x
dx d Ex
(c) Find Ex from −dV/dx:
[
]
kV/m 00 . 3
3000 2000
=
− −
= x
dx d Ex
(d) Find Ex from −dV/dx:
[
]
0
2000
=
−
−
=
dx
d
E
x41 ••
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 q k
x q k x kq x V
(b) Set V(x) = 0:
0
1
3
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
x
q
x
q
k
0 1 3 1 = − − x x
For x < 0:
(
)
0
0
.
500
m
1
3
1
=
⇒
=
−
−
−
−
−
x
x
x
For 0 < x < 1:
(
)
0
0
.
250
m
1
3
1
=
⇒
=
−
−
−
x
x
x
Note also that: V
( )
x →0asx→±∞ (c) Evaluate V(x) for 0 < x < 1:(
)
⎟
⎠
⎞
⎜
⎝
⎛
−
+
=
<
<
1
3
1
0
x
q
x
q
k
x
V
Apply Ex = −dV/dx to find Ex in this
region:
(
)
(
)
⎥
⎦
⎤
⎢
⎣
⎡
−
+
=
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−
+
−
=
<
<
2 21
3
1
1
3
1
0
x
x
kq
x
q
x
q
k
dx
d
x
E
xEvaluate this expression at
x = 0.25 m to obtain:
(
)
(
) (
)
(
)
kq
kq
E
x 2 2 2m
3
.
21
m
75
.
0
3
m
25
.
0
1
m
25
.
0
−=
⎥
⎦
⎤
⎢
⎣
⎡
+
=
Evaluate V(x) for x < 0:
(
)
⎟
⎠
⎞
⎜
⎝
⎛
−
+
−
=
<
x
x
kq
x
V
1
3
1
0
Apply Ex = −dV/dx to find Ex in this
region:
(
)
(
)
⎥⎦ ⎤ ⎢ ⎣ ⎡ − + − = ⎥⎦ ⎤ ⎢⎣ ⎡ − + − = < 2 2 1 3 1 1 3 1 0 x x kq x x dx d kq x ExEvaluate this expression at x = −0.5 m to obtain:
(
)
kq(
) (
)
(
)
kqEx 2 2 2.67m 2
m 5 . 1 3 m 5 . 0 1 m 5 .
0 ⎥= − −
⎦ ⎤ ⎢ ⎣ ⎡ + − − = −
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:
dV
=
−
E
xdx
=
−
(
2
.
0
x
3kN/C
)
dx
Integrate V from V1 to V2 and x from1 m to 2 m:
(
)
(
)
[ ]
2mm 1 4 4 1
3
kN/C
0
.
2
kN/C
0
.
2
2
1 2
1
x
dx
x
dV
x
x V
V
−
=
−
=
∫
∫
Simplify to obtain: 7.50kV
1 2 −V = − V
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
kq
x
V
=
+
+
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
As x→∞, i.e., for x >> a, the three charges appear as a point charge 3q, so Ex = 3kq/x
2
; this is also the result one obtains from (b) for x >> a. Substitute for the rito obtain:
( )
⎟⎟⎠ ⎞ ⎜
⎜ ⎝ ⎛
− + + =
⎟ ⎟ ⎠ ⎞ ⎜
⎜ ⎝ ⎛
− + + + + =
a x a x kq a
x a x a
x kq x
V 1 1 1 2 1
2 2 2
2 2
2
(b) For x > a, x−a > 0 and: x−a =x−a
Use Ex = −dV/dx to find Ex:
(
)
(
2 2)
32(
)
2 22
2
1
2
a
x
kq
a
x
kqx
a
x
a
x
kq
dx
d
a
x
E
x−
+
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
+
+
−
=
>
For x < a, x−a < 0 and: x−a =−
(
x−a)
=a−xUse Ex = −dV/dx to find Ex:
(
)
2 2(
2 2)
32(
)
22
1
2
x
a
kq
a
x
kqx
x
a
a
x
kq
dx
d
a
x
E
x−
−
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
+
+
−
=
<
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
= =
⋅
∫
Er dAr Q∈
because the charge resides on the outer surface of the spherical surface. Hence
(
r<12cm)
= 0Er
Apply Gauss’s law to a spherical
Gaussian surface of radius
(
)
0 24
∈
π
r
q
r > 12 cm: and
(
)
20 2
4
cm
12
r
kq
r
q
r
E
>
=
=
∈
π
Substitute numerical values and evaluate E
(
r >12cm)
:(
)
(
)(
)
(
0.12
m
)
6
.
24
kV/m
C
10
/C
m
N
10
8.99
cm
12
28 2 2 9
=
⋅
×
=
>
−r
E
(b) Express and evaluate the potential just inside the spherical shell:
(
)
(
)(
)
749
V
m
0.12
C
10
/C
m
N
10
8.99
9 2 2 8=
⋅
×
=
=
≤
−R
kq
R
r
V
Express and evaluate the potential just outside the spherical shell:
(
)
(
)(
)
749
V
m
0.12
C
10
/C
m
N
10
8.99
9 2 2 8=
⋅
×
=
=
≥
−r
kq
R
r
V
(c) The electric potential inside a uniformly charged spherical shell is constant and given by:
(
)
(
)(
)
749
V
m
0.12
C
10
/C
m
N
10
8.99
×
9⋅
2 2 8=
=
=
≤
−R
kq
R
r
V
In part (a) we showed that: Er
(
r<12cm)
= 045 •
Picture the Problem We can use the expression for the potential due to a line charge
a
r
k
V
=
−
2
λ
ln
, 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
r
k
V
=
−
2
λ
ln
Because V = 0 at r = 2.5 m:
a
k
ln
2
.
5
m
2
a
m
5
.
2
ln
0
=
,and
1
0
ln
m
5
.
2
=
−1=
a
Thus we have a = 2.5 m and:
(
)
(
)
(
)
⎟⎟⎠ ⎞ ⎜⎜
⎝ ⎛ ⋅
× − = ⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛ ⋅
× −
=
m 5 . 2 ln m/C N 10 70 . 2 m
5 . 2 ln C/m 5 . 1 /C m N 10 99 . 8
2 9 2 2 r 4 r
V
µ
(a) Evaluate V at r = 2.0 m:
(
)
kV 02 . 6
m 5 . 2
m 2 ln m/C N 10 70 .
2 4
=
⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛ ⋅
× − =
V
(b) Evaluate V at r = 4.0 m:
(
)
kV 7 . 12
m 5 . 2
m 4 ln m/C N 10 70 .
2 4
− =
⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛ ⋅
× − =
V
(c) Evaluate V at r = 12.0 m:
(
)
kV 3 . 42
m 5 . 2
m 12 ln m/C N 10 70 .
2 4
− =
⎟⎟ ⎠ ⎞ ⎜⎜
⎝ ⎛ ⋅
× − =
V
46 ••
Picture the Problem The electric field on the x axis of a disk charge of radius R is given
by
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
=
2 2
1
2
R
x
x
k
E
xπ
σ
. 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:
dx
E
dV
=
−
xExpress the electric field on the x
axis of a disk charge:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
=
2 2
1
2
R
x
x
k
Substitute to obtain:
dx
R
x
x
k
dV
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
−
=
2 21
2
π
σ
Let V(∞) = 0 and integrate from x′=
∞ to x′ = x:
(
)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
=
−
+
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
−
=
∫
∞1
1
2
2
'
'
1
2
2 2 2 2 2 2x
R
x
k
x
R
x
k
dx'
R
x
x
k
V
xσ
π
σ
π
σ
π
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:
r
dy
k
dV
=
λ
where
r
=
x
2+
y
2 Integrate dV from y = −L/2 toy = L/2:
( )
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
+
+
=
+
=
∫
−2
4
2
4
ln
0
,
2 2 2 2 22 2 2
L
L
x
L
L
x
L
kQ
y
x
dy
L
kQ
x
V
L L(b) Factor x from the numerator and denominator within the parentheses to
obtain:
( )
⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − + + + = x L x L x L x L L kQ x V 2 4 1 2 4 1 ln 0 , 2 2 2 2
Use
a
b
b
a
ln
ln
ln
=
−
to obtain:Let 2 2
4
x
L
=
ε
and use(
1)
1 81 2 ...2 1 2
1 = + − +
+
ε
ε
ε
to expand 22
4
1
x
L
+
:1
...
4
8
1
4
2
1
1
4
1
2 2 2 2 2 2 1 2 2≈
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
+
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
x
L
x
L
x
L
for x >> L.
Substitute to obtain:
( )
⎭
⎬
⎫
⎟
⎠
⎞
⎜
⎝
⎛ −
−
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛ +
=
x
L
x
L
L
kQ
x
V
2
1
ln
2
1
ln
0
,
Letx
L
2
=
δ
and useln
(
1
+
δ
)
=
δ
−
12δ
2+
...
to expand⎟
⎠
⎞
⎜
⎝
⎛ ±
x
L
2
1
ln
: 2 2 4 2 2 1 ln x L x L xL ≈ −
⎟ ⎠ ⎞ ⎜
⎝
⎛ + and 2
2 4 2 2 1 ln 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
kQ
x
V
=
⎭
⎬
⎫
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
−
⎩
⎨
⎧
−
=
22 224
2
4
2
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
R
r
dr
r
dq
0 02
2
2
πσ
σ
π
σ
π
=
⎟
⎠
⎞
⎜
⎝
⎛
=
=
Integrate from r = 0 to r = R to obtain: 2 0 0
0 2
2 R dr R
Q
R
