CHAPTER 25
ELECTRIC POTENTIAL
Problem
2.
The potential difference between the two sides of an ordinary electrical outlet is 120 V. How much energy does an
electron gain when it moves from one side to the other?
Solution
Moving from the negative to the positive side (i.e., opposite to the electric field), an electron gains e ∆V = 120 eV =
(1.6 × 10 −19 C)(120 V ) = 1.92 × 10 −17 J of energy.
Problem
5.
Find the magnitude of the potential difference between two points located 1.4 m apart in a uniform 650 N/C electric
field, if a line between the points is parallel to the field.
Solution
For ℓ in the direction of a uniform electric field, Equation 25-2b gives ∆V = Eℓ = (650 N/C)(1.4 m) = 910 V. (See note in
solution to Problem 1. Since dV = −E ⋅ dℓℓ, the potential always decreases in the direction of the electric field.)
Problem
8.
Figure 25-37 shows a uniform electric field of magnitude E. Find expressions for (a) the potential difference ∆VAB and
(b) ∆VBC . (c) Use your result to determine ∆VAC .
C
d
E
45°
A
FIGURE
Solution
d
B
25-37 Problem 8.
z
z
(a) On the line A to B, dℓℓ is antiparallel to E, so Equation 25-2 gives VB − VA = − BA E ⋅ dℓ = E BA dℓ = Ed. (b) The line B
to C makes an angle of 45° with E, so VC − VB = − E CB cos 45° dℓ = − Ed= 2 . (c) Addition yields VC − VA =
VC − VB + VB − VA = Ed (1 − 1= 2 ) = 0.293Ed .
z
Problem
9.
A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential
difference of 100 V. Find the energy each gains.
Solution
The energy gained is q ∆V (see Example 25-1). The proton and singly-ionized helium atom have charge e, so they gain
100 eV = (1.6 × 10 −19 C)(100 V) = 1.6 × 10 −17 J, while the α-particle has charge 2e and gains twice this energy.
Problem
12. Electrons in a TV tube are accelerated from rest through a 25-kV potential difference. With what speed do they hit the
TV screen?
CHAPTER 25 595
Solution
The work done on an electron equals the change in its kinetic energy, W = e ∆V =
v=
2 e ∆V=m =
1
2
mv 2 (if it starts from rest). Thus,
2(1.6 × 10 −19 C)(25 × 10 3 V )
= 9.37 × 10 7 m/s.
(9.11 × 10 −31 kg)
Problem
15. Two large, flat metal plates are a distance d apart, where d is small compared with the plate size. If the plates carry
surface charge densities ±σ , show that the potential difference between them is V = σ d=ε 0 .
Solution
The electric field between the plates is uniform, with E = σ =ε 0 , directed from the positive to the negative plate (see last
paragraph of Section 24-6 and Fig. 24-35). Then Equation 25-2b gives V = V+ − V− = − (σ=ε 0 )( − d ) = σ d=ε 0 (the
displacement from the negative to the positive plate is opposite to the field direction).
Problem
17. A 5.0-g object carries a net charge of 3.8 µ C. It acquires a speed v when accelerated from rest through a potential
difference V. A 2.0-g object acquires twice the speed under the same circumstances. What is its charge?
Solution
The speed acquired by a charge q, starting from rest at point A and moving through a potential difference of V, is given by
2
1
2 mvB = q(VA − VB ) = qV , or vB = 2V ( q=m ). (This is the work-energy theorem for the electric force. A positive charge is
accelerated in the direction of decreasing potential.) If the second object acquires twice the speed of the first object, moving
through the same potential difference, it must have four times the charge to mass ratio, q=m. Thus,
q2 = 4( q1=m1 )m2 = 4(3.8 µ C)(2 g=5g) = 6.08 µ C.
Problem
22. What is the maximum potential allowable on a 5.0-cm-diameter metal sphere if the electric field at the sphere’s surface is
not to exceed the 3 MV/m breakdown field in air?
Solution
For an isolated metal sphere, the potential at the surface is V = kQ=R, while the electric field strength at the surface is
kQ=R 2 = V=R. Thus, V=R ≤ 3 MV/m implies V ≤ (3 MV/m )( 12 × 0.05 m ) = 75 kV.
Problem
23. A 3.5-cm-diameter isolated metal sphere carries a net charge of 0.86 µ C. (a) What is the potential at the sphere’s
surface? (b) If a proton were released from rest at the sphere’s surface, what would be its speed far from the sphere?
Solution
(a) An isolated metal sphere has a uniform surface charge density, so Equation 25-4 gives the potential at its surface,
Vsurf = kQ=R = (9 GN ⋅ m 2 /C 2 )(0.86 µ C)=( 12 × 3.5 cm ) = 442 kV. (b) The work done by the repulsive electrostatic field (the
negative of the change in the proton’s potential energy) equals the proton’s kinetic energy at infinity, W =
− e(V∞ − Vsurf ) = eVsurf = 12 mv 2 . Then v = 2eVsurf =m = [2(1.6 × 10 −19 C)( 442 kV)=(1.67 × 10 −27 kg)]1=2 = 9.21 Mm/s.
596 CHAPTER 25
Problem
24. A sphere of radius R carries a negative charge of magnitude Q, distributed in a spherically symmetric way. Find the
“escape speed” for a proton at the sphere’s surface—that is, the speed that would enable the proton to escape to
arbitrarily large distances.
Solution
The work done by the electric field, when a proton escapes from the surface to an infinite distance, equals the change in
kinetic energy, or − e(V∞ − Vsurf ) = eVsurf = K ∞ − Ksurf = − 12 mv 2 . (We assumed zero kinetic energy for the proton at
infinity, and that the sphere is stationary.) For a uniformly negatively charged sphere, Vsurf = − kQ=R, so v = 2keQ=mR .
Problem
35. A hollow, spherical conducting shell of inner radius b and outer radius c surrounds, and is concentric with, a solid
conducting sphere of radius a, as shown in Fig. 25-39. The sphere carries a net charge −Q and the shell carries a net
charge +3Q. Both conductors are in electrostatic equilibrium. Find an expression for the potential difference from infinity
to the surface of the sphere.
FIGURE 25-39
Problem 35.
Solution
The electric field between the solid sphere and the shell is like that due to a point charge −Q located at their common center
(the origin), so the potential difference between the sphere and the shell is Vsph − Vshell = − kQ( a −1 − b −1 ). The electric field
outside the shell is like that due to a point charge 2Q at the origin (the total charge is 3Q − Q), so the potential difference
between the shell and infinity is Vshell − V∞ = 2kQc −1 . (See Examples 24-3 and 25-3.) The entire shell is at one potential, as
is the entire sphere, because each is a conductor in electrostatic equilibrium. Therefore, the potential difference between the
sphere and infinity is Vsph − V∞ = Vsph − Vshell + Vshell − V∞ = kQ(2c −1 + b −1 − a −1 ).
Problem
39. The annulus shown in Fig. 25-40 carries a uniform surface charge density σ . Find an expression for the potential at an
arbitrary point P on its axis.
CHAPTER 25 597
b
a
P
x
FIGURE 25-40
Problem 39.
Solution
The annulus can be considered to be composed of thin rings of radius r ( a ≤ r ≤ b ) and charge dq = 2πσ r dr (see Example
25-7 and Fig.’s 25-15 and 16). The element of potential from a ring on its axis, a distance x from the center, is
dV = k dq= x 2 + r 2 (see Example 25-6) so the potential from the whole annulus is:
V =
z
dV = 2πσ k
z
b
r dr
x2 + r2
a
= 2π kσ
x2 + r2
b
a
= 2π kσ
e
x 2 + b2 −
j
x 2 + a2 .
(Note: This reduces to the potential on the axis of a uniformly charged disk if a → 0.)
Problem
40. A thin rod of length ℓ carries a charge Q distributed uniformly over its length. (a) Show that the potential in the plane that
perpendicularly bisects the rod is given by
V (r) =
LM
MN
OP
PQ
2 kQ
ℓ
ℓ2
ln
+ 1+ 2 ,
ℓ
2r
4r
where r is the distance from the rod center. (b) Show that this expression reduces to an expected result when r À ℓ.
Hint: See Appendix A for a series expansion of the logarithm.
Solution
(a) The potential produced by an element of charge dq = λ dz = (Q=ℓ) dz, at a point in the perpendicular bisecting plane, is
dV = k dq= z 2 + r 2 , as shown in the sketch. Thus,
V =
kQ
ℓ
z
ℓ=2
− ℓ=2
dz
z2 + r2
=
F
GH
I
JK
ℓ=2
2kQ
2 kQ
ℓ
ℓ2
ln( z + z 2 + r 2 )
=
ln
+ 1+ 2 ,
0
ℓ
ℓ
2r
4r
where we used the symmetry of the rod about the bisecting plane ( z = 0). (b) For ℓ=2 r ¿ 1, ln[( ℓ=2 r ) + 1 + (ℓ=2 r ) 2 ] ¼
ln[(ℓ=2 r ) + 1] ¼ ℓ=2 r, so V ¼ (2 kQ=ℓ)(ℓ=2 r ) = kQ=r, as expected for a point charge.
598 CHAPTER 25
Problem 40 Solution.
Problem
44. Figure 25-42 shows some equipotentials in the x-y plane. (a) In what region is the electric field strongest? What are
(b) the direction and (c) the magnitude of the field in this region?
FIGURE
25-42 Problem 44 Solution.
Solution
(a) The equipotentials in Fig. 25-37 are most closely spaced along the x-axis between x = 2 m and x = 5 m. (b) The potential
decreases in the direction of the electric field, which, for 2 m ≤ x ≤ 5 m, is in the negative x direction. (c) The potential
drops by 10 V/m, which is the field strength in this region.
Problem
45. The potential in a certain region is given by V = axy, where a is a constant. (a) Determine the electric field in the region.
(b) Sketch some equipotentials and field lines.
Solution
(a) The x and y components of the electric field can be found from Equation 25-10: E x = −∂ V=∂ x = −∂=∂x ( axy) = − ay, and
E y = −∂ V=∂ y = − ax. Thus E = − a( yî + xɵj ). (The field has no z component.) (b) See sketch below. The field lines (dashed)
are perpendicular to the equipotentials (solid) in the direction of decreasing potential (arrows for a > 0, in this case). These
equipotentials and field lines are confocal hyperbolas, proportional to xy and 12 ( x 2 − y 2 ) respectively, and are sketched only
for x and y in the first quadrant.
CHAPTER 25 599
Problem 45 Solution.
Problem
50. A charge +4q is located at the origin and a charge −q is on the x-axis at x = a. (a) Write an expression for the potential
on the x-axis for x > a. (b) Find a point in this region where V = 0. (c) Use the result of (a) to find the electric field on
the x-axis for x > a and (d) find a point where E = 0.
Solution
(a) From Equation 25-6 on the x-axis, for x > a, V ( x ) = kq[ 4 x −1 − ( x − a) −1 ]. (b) V ( x ) = 0 when 4( x − a) = x, or
x = 4 a=3. (c) For points on the x-axis, the electric field only has an x component, so Equation 25-10 gives E = ( −dV=dx )î =
kq[ 4 x −2 − ( x − a) −2 ]î (see note to solution of previous problem). (d) E = 0 when x = 2( x − a), or x = 2 a (remember
x > a when taking square roots).
Problem
54. A large metal sphere has three times the diameter of a smaller sphere and carries three times as much charge. Both
spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials and (b) the electric field
strengths at their surfaces.
Solution
(a) The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R
has the same potential. (b) However, the electric field at the surface of the smaller sphere is σ=ε 0 = kQ=R 2 , so tripling Q and
R reduces the surface field by a factor of 13 .
Problem
57. Two conducting spheres are each 5.0 cm in diameter and each carries 0.12 µ C. They are 8.0 m apart. Determine
(a) the potential on each sphere; (b) the field strength at the surface of each sphere; (c) the potential midway between the
spheres; (d) the potential difference between the spheres.
Solution
Since the spheres are small and widely separated, at small distances (¿ 8 m ) they behave like isolated spheres, and at large
distances (À 5 cm ) they behave like two point charges. (a) At the surface of either sphere, Equation 25-11 gives
Vsurf = kq=R = (9 × 10 9 N ⋅ m 2 /C 2 )(1.2 × 10 −7 C)=(0.025 m ) = 43.2 kV. (b) Then Esurf = σ=ε 0 = kq=R 2 = Vsurf =R =
( 43.2 kV)=(2.5 cm) = 1.73 × 10 6 V/m. (c) Midway between the spheres, the potential from each one is the same, so
600 CHAPTER 25
Vmid - pt. = 2 kq=r = 2(9 × 10 9 V ⋅ m/C)(1.2 × 10 −7 C)=( 4 m ) = 540 V. (d) The spheres are at the same potential, so the
difference is zero.
Problem
58. Two small metal spheres are located 2.0 m apart. One has radius 0.50 cm and carries 0.20 µ C. The other has radius
1.0 cm and carries 0.080 µ C. (a) What is the potential difference between the spheres? (b) If they were connected by a
thin wire, how much change would move along it, and in which direction?
Solution
(a) Since the spheres are approximately isolated, their potentials are V1 = kq1=R1 = (9 × 10 9 V ⋅ m/C)(2 × 10 −7 C) ÷
(5 × 10 −3 m ) = 3.6 × 10 5 V, and V2 = (9 × 10 9 V ⋅ m /C)(8 × 10 −8 C)=(10 −2 m) = 0.72 × 10 5 V. The difference is
V1 − V2 = 2.88 × 10 5 V. (b) The total charge on the spheres is 0.28 µ C = q1′ + q2′ , where primes refer to the spheres
after they are connected. (We assume that the wire has a negligible effect, other than bringing both spheres to the same
potential.) Since V1′ = kq1′=R1 = V2′ = kq2′ =R2 , R2 q1′ − R1q2′ = 0, or 2 q1′ − q2′ = 0. Solving for q1′ , we find
q1′ = 13 (0.28 µ C) = 0.0933 µ C, therefore a charge of q1 − q1′ = 0.107 µ C was transferred from the first sphere to the second.
Problem
74. For the dipole of Example 25-5, show that the electric field at an arbitrary point far from the dipole can be written
E=
kp
[(3 cos 2 θ − 1)î + 3 sin θ cos θ ɵj ].
r3
Solution
In the solution to Problem 48, we found Er = 2 kp cos θ=r 3 and Eθ = kp sin θ=r 3 . In Fig. 25-10 (in which a ¿ r for a
point dipole) p is parallel to the x-axis, so the unit vectors are rɵ = î cos θ + ɵj sin θ and θɵ = − î sin θ + ɵj cos θ . (θɵ is in the
direction of increasing θ . ) Thus, E dip = Er rɵ + Eθ θɵ = ( kp=r 3 )[2 cos θ ( î cos θ + ɵj sin θ ) + sin θ ( − î sin θ + ɵj cos θ )] =
( kp=r 3 )[(3 cos 2 θ − 1)î + 3 sin θ cos θ ɵj ], where sin 2 θ + cos 2 θ = 1 was used to express the x component.
Problem
75. A thin rod of length ℓ lies on the x-axis with its center at the origin. It carries a line charge density given by
λ = λ 0 ( x=ℓ) 2 , where λ 0 is a constant. (a) Find an expression for the potential on the x-axis for x > ℓ=2. (b) Integrate the
charge density to find the total charge on the rod. (c) Show that your answer for (a) reduces to the potential of a point
charge whose charge is the answer to (b), for x À ℓ.
Solution
(a) For points P on the x-axis at x > ℓ=2, the potential from a charge element dq = λ dx ′ at x ′ along the rod
( − ℓ=2 ≤ x ′ ≤ ℓ=2) is dV = k dq= x − x ′ = kλ dx ′=( x − x ′). (We used x ′ for the variable position of dq, and took the potential
relative to zero at infinity.) The potential due to the entire rod, for λ = λ 0 ( x ′=ℓ) 2 , is
V ( x) =
z
ℓ=2
− ℓ=2
dV =
kλ 0
ℓ2
z
ℓ=2
− ℓ=2
kλ
x ′ 2 dx ′
x ′2
= 20 − x 2 ln( x − x ′ ) − xx ′ −
( x − x ′)
2
ℓ
ℓ=2
=
− ℓ=2
LM
MN
FG
H
IJ
K
OP
PQ
kλ 0 2
x + ℓ=2
x ln
− xℓ .
2
x − ℓ=2
ℓ
(Use partial fractions, x ′ 2=( x − x ′) = − x − x ′ + x 2=( x − x ′), or standard tables to evaluate the integral.) (b) The total charge
z
on the rod is Q = dq =
ln(1 ± ℓ=2 x ) = ±( ℓ=2 x ) −
z
ℓ=2
2
2
2 2
3
−ℓ=2 ( λ 0 =ℓ ) x dx = ( λ 0 =ℓ )( 3 )( ℓ=2) = λ 0 ℓ=12.
2
3
1
1
. . . . .Therefore:
2 ( ℓ=2 x ) ± 3 ( ℓ=2 x ) −
(c) For x À ℓ, we can expand the logarithms,
CHAPTER 25 601
V ( x ) = (kλ 0 =ℓ 2 )[ x 2 ln(1 + ℓ=2 x ) − x 2 ln(1 − ℓ=2 x ) − xℓ]
FG kλ
Hℓ
F kλ
=G
Hℓ
=
IJ LMx FG ℓ − 1 F ℓ I + 1 F ℓ I − . . .IJ − x FG − ℓ − 1 F ℓ I
K MN H 2 x 2 H 2 x K 3 H 2 x K
K H 2x 2 H 2x K
IJ LMx FG ℓ + 2 F ℓ I + . . .IJ − x ℓOP = kλ ℓ = kQ=x,
K MN H x 3 H 2 x K
K PQ 12 x
2
0
2
0
2
3
2
2
2
−
F I
H K
1 ℓ
3 2x
3
I
JK
OP
PQ
− . . . − xℓ
3
2
0
as expected.
Problem 75 Solution.
Problem
78. A disk of radius a carries a nonuniform surface charge density given by σ = σ 0 (r=a), where σ 0 is a constant. (a) Find
the potential at an arbitrary point on the disk axis, a distance x from the disk center. (b) Use the result of (a) to find the
electric field on the disk axis, and (c) show that the field reduces to an expected form for x À a.
Solution
(a) As in Example 25-7, V ( x ) =
integral tables gives
V ( x) =
z
a
0
k dq= x 2 + r 2 , where dq = 2πσ r dr , but now σ = σ 0 r=a. Reference to standard
2π kσ 0
a
= π kσ 0 a
z
a
r 2 dr
0
x2 + r2
=
LM
MN
2π kσ 0 a
a
2
x 2 + a2 −
F
GH
x2
a+
ln
2
x 2 + a2
x
I OP
JK P
Q
1 + ( x=a) 2 − ( x=a) 2 ln( a=x + 1 + (a=x ) 2 ) .
(b) As in Example 25-9, E x = − dV=dx results in
E x = π kσ 0
LM 2 x lnF a +
MN a GH
x 2 + a2
x
I−
JK
x
x 2 + a2
+
x2
a
F
GH a +
x
x 2 + a2
IF
JK GH
1
x 2 + a2
−
a+
x 2 + a2
x2
I OP
JK P
Q
= (2π kσ 0 x=a) ln( a=x + 1 + ( a=x ) 2 ) − (a=x )(1 + ( a=x ) 2 ) −1=2 .
(c) The logarithm has to be expanded carefully, up to order ( a=x )3 , to evaluate E x for x À a. Thus,
ln
Fa +
GH x
1+
F a I IJ = lnFG1 + a + a + . . .IJ
H x K K H x 2x
K
Fa a I − 1 Fa + a I
¼G +
H x 2 x JK 2 GH x 2 x JK
2
2
2
2
2
2
2
2
+
F
GH
1 a
a2
+
3 x 2x2
I
JK
3
+ . . .¼
a
a3
− 3.
x 6x
602 CHAPTER 25
Also,
F
GH
a
a2
1+ 2
x
x
Then,
Ex ¼
This is like a point charge field for Q =
z
I
JK
−1=2
¼
F
GH
a
a2
1−
x
2x2
I=a− a
JK x 2 x
LM
N
3
OP
Q
3
.
2π kσ 0 x a
2π kσ 0 a 2
a3
a
a3
−
− +
=
.
3
3
a
x 6x
x 2x
3x 2
a
2
0 ( 2πσ 0 =a )r
dr = 2πσ 0 a 2 =3.