# Selected Solutions

```Halliday ♦ Resnick ♦Walker
FUNDAMENTALS OF PHYSICS
SIXTH EDITION
Selected Solutions
Chapter 26
26.15
26.19
26.31
26.45
15. Let x be the separation of the plates in the lower capacitor. Then the plate separation in the upper
capacitor is a − b − x. The capacitance of the lower capacitor is C = ε0 A/x and the capacitance of
the upper capacitor is Cu = ε0 A/(a − b − x), where A is the plate area. Since the two capacitors are in
series, the equivalent capacitance is determined from
1
a−b−x
a−b
1
1
x
+
=
.
=
+
=
Ceq
C
Cu
ε0 A
ε0 A
ε0 A
Thus, the equivalent capacitance is given by Ceq = ε0 A/(a − b) and is independent of x.
19. (a) After the switches are closed, the potential diﬀerences across the capacitors are the same and the
two capacitors are in parallel. The potential diﬀerence from a to b is given by Vab = Q/Ceq , where
Q is the net charge on the combination and Ceq is the equivalent capacitance. The equivalent
capacitance is Ceq = C1 + C2 = 4.0 &times; 10−6 F. The total charge on the combination is the net charge
on either pair of connected plates. The charge on capacitor 1 is
q1 = C1 V = (1.0 &times; 10−6 F)(100 V) = 1.0 &times; 10−4 C
and the charge on capacitor 2 is
q2 = C2 V = (3.0 &times; 10−6 F)(100 V) = 3.0 &times; 10−4 C ,
so the net charge on the combination is 3.0 &times; 10−4 C − 1.0 &times; 10−4 C = 2.0 &times; 10−4 C. The potential
diﬀerence is
2.0 &times; 10−4 C
Vab =
= 50 V .
4.0 &times; 10−6 F
(b) The charge on capacitor 1 is now q1 = C1 Vab = (1.0 &times; 10−6 F)(50 V) = 5.0 &times; 10−5 C.
(c) The charge on capacitor 2 is now q2 = C2 Vab = (3.0 &times; 10−6 F)(50 V) = 1.5 &times; 10−4 C.
31. We ﬁrst need to ﬁnd an expression for the energy stored in a cylinder of radius R and length L, whose
surface lies between the inner and outer cylinders of the capacitor (a &lt; R &lt; b). The energy density at
any point is given by u = 12 ε0 E 2 , where E is the magnitude of the electric ﬁeld at that point. If q is the
charge on the surface of the inner cylinder, then the magnitude of the electric ﬁeld at a point a distance
r from the cylinder axis is given by
q
E=
2πε0 Lr
(see Eq. 26-12), and the energy density at that point is given by
u=
1
q2
.
ε0 E 2 =
2
2
8π ε0 L2 r2
The energy in the cylinder is the volume integral
UR =
u dV .
Now, dV = 2πrL dr, so
UR =
a
R
q2
q2
2πrL dr =
2
2
2
8π ε0 L r
4πε0 L
a
R
q2
R
dr
=
ln
.
r
4πε0 L
a
To ﬁnd an expression for the total energy stored in the capacitor, we replace R with b:
Ub =
q2
b
ln .
4πε0 L
a
We want the ratio UR /Ub to be 1/2, so
R
1 b
ln = ln
a
2 a
√
1
or, since 2 ln(b/a) = ln( b/a), ln(R/a) = ln( b/a). This means R/a = b/a or R = ab.
45. (a) According to Eq. 26-17 the capacitance of an air-ﬁlled spherical capacitor is given by
C0 = 4πε0
ab
.
b−a
When the dielectric is inserted between the plates the capacitance is greater by a factor of the
dielectric constant κ. Consequently, the new capacitance is
C = 4πκε0
ab
.
b−a
(b) The charge on the positive plate is
q = CV = 4πκε0
ab
V .
b−a
(c) Let the charge on the inner conductor to be −q. Immediately adjacent to it is the induced charge
q . Since the electric ﬁeld is less by a factor 1/κ than the ﬁeld when no dielectric is present, then
−q + q = −q/κ. Thus,
κ−1
ab
q =
q = 4π(κ − 1)ε0
V .
κ
b−a
```