Chap. 24, Exercises
1, 2, 3,a, 3, b, 9, a. 32, a, b, c 38, a, 15
24.9. A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The
inner cylinder is negatively charged and the outer is positively charged; the magnitude of the
charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00
mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance?
24.1. A capacitor has a capacitance of 728 F What amount of charge must be placed on each
of its plates to make the potential difference between its plates equal to 25.0 V?
24.2. The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of
122 cm2 Each plate carries a charge of magnitude 435 102 8 C The plates are in vacuum. (a)
What is the capacitance? (b) What is the potential difference between the plates? (c) What is the
magnitude of the electric field between the plates?
24.3. A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0148 C
on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the
plates? (b) What is the area of each plate?
24.32. For the capacitor network shown in Fig. 24.28, the potential difference across ab is 36 V.
Find (a) the total charge stored in this network; (b) the charge on each capacitor; (c) the total
energy stored in the network; (d) the energy stored in each capacitor; (e) the potential differences
across each capacitor.
24.38. A parallel-plate capacitor has capacitance C0 500 pF when there is air between the
plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of
charge Q that can be placed on each plate if the electric field in the region between the plates is
not to exceed 300 104 V/m? (b) A dielectric with K 270 is inserted between the plates of
the capacitor, completely filling the volume between the plates. Now what is the maximum
magnitude of charge on each plate if the electric field between the plates is not to exceed
300 104 V/m?
24.15. In Fig. 24.25, each capacitor has C 400 F and Vab 1 280 V Calculate (a) the
charge on each capacitor; (b) the potential difference across each capacitor; (c) the potential
difference between points a and d.
Figure 24.25 Exercise 24.15.
24.1. IDENTIFY: C
SET UP: 1 F 106 F
EXECUTE: Q CVab (7.28 106 F)(25.0 V) 1.82 104 C 182 C
EVALUATE: One plate has charge Q and the other has charge Q .
24.2. IDENTIFY and SET UP: C 0 , C and V Ed .
A 0.00122 m 2
(a) C P P
0 0 3.29 pF
d 0.00328 m
Q 4.35 108 C
(b) V 13.2 kV
C 3.29 1012 F
V 13.2 103 V
(c) E 4.02 106 V/m
d 0.00328 m
EVALUATE: The electric field is uniform between the plates, at points that aren't close to the edges.
24.3. IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1.
Q Q 0.148 106 C
EXECUTE: (a) C so Vab 604 V
Vab C 245 1012 F
245 1012 F 0.328 103 m
9.08 103 m2 90.8 cm2
d P0 8.854 1012 C2 / N m 2
V 604 V
(c) Vab Ed so E ab 1.84 106 V/m
d 0.328 103 m
(d) E so EP 1.84 106 V/m 8.854 1012 C2 / N m 2 1.63 105 C/m 2
Q 0.148 106 C
EVALUATE: We could also calculate directly as Q/A. 1.63 105 C/m 2 , which
A 9.08 103 m 2
24.9. IDENTIFY: Apply the results of Example 24.4. C Q / V .
SET UP: ra 0.50 mm , rb 5.00 mm
L2 P (0.180 m)2 P
EXECUTE: (a) C 0
4.35 1012 F .
ln(rb ra ) ln(5.00 0.50)
(b) V Q / C (10.0 1012 C) /(4.35 1012 F) 2.30 V
EVALUATE: 24.2 pF . This value is similar to those in Example 24.4. The capacitance is determined
entirely by the dimensions of the cylinders.
1 1 1 Q
24.32. IDENTIFY: The two capacitors are in series. C . U 1 CV 2 .
Ceq C1 C2 V
SET UP: For capacitors in series the voltages add and the charges are the same.
1 1 1 CC (150 nF)(120 nF)
EXECUTE: (a) so Ceq 1 2 66.7 nF .
Ceq C1 C2 C1 C2 150 nF 120 nF
Q CV (66.7 nF)(36 V) 2.4 106 C 2.4 C
(b) Q 2.4 C for each capacitor.
24.38. IDENTIFY: V Ed and C Q / V . With the dielectric present, C KC0 .
SET UP: V Ed holds both with and without the dielectric.
EXECUTE: (a) V Ed (3.00 104 V/m)(1.50 103 m) 45.0 V .
Q C0V (5.00 1012 F)(45.0 V) 2.25 1010 C .
(b) With the dielectric, C KC0 (2.70)(5.00 pF) 13.5 pF . V is still 45.0 V, so
Q CV (13.5 1012 F)(45.0 V) 6.08 1010 C .
EVALUATE: The presence of the dielectric increases the amount of charge that can be stored for a given
potential difference and electric field between the plates. Q increases by a factor of K.
1 1 1 1 1 1
(b) C1 15 pF is in series with C23 20 pF . For capacitors in series, so and
Ceq C1 C2 C123 C1 C23
24.14. IDENTIFY: The capacitors between b and c are in parallel. This combination is in series with
the 15 pF capacitor.
CC (15 pF)(20 pF)
C123 1 23 8.6 pF .
C1 C23 15 pF 20 pF
EVALUATE: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors.
For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors.
24.15. IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents. In each equivalent
network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work
back to the original circuit.
SET UP: Do parts (a) and (b) together. The capacitor network is drawn in Figure 24.15a.
C1 C2 C3 C4 400 F
Vab 28.0 V
EXECUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: C1 and C2 are in
series and are equivalent to C12 (Figure 24.15b).
1 1 1
C12 C1 C2
4.00 106 F 4.00 106 F 2.00 106 F
C1 C2 4.00 106 F 4.00 106 F
C12 and C3 are in parallel and are equivalent to C123 (Figure 24.15c).
C123 C12 C3
C123 2.00 106 F 4.00 106 F
C123 6.00 106 F
C123 and C4 are in series and are equivalent to C1234 (Figure 24.15d).
1 1 1
C1234 C123 C4
6.00 106 F 4.00 106 F 2.40 106 F
C123 C4 6.00 106 F 4.00 106 F
The circuit is equivalent to the circuit shown in Figure 24.15e.
V1234 V 28.0 V
Q1234 C1234V 2.40 106 F 28.0 V 67.2 C
Now build back up the original circuit, step by step. C1234 represents C123 and C4 in series (Figure 24.15f).
Q123 Q4 Q1234 67.2 C
(charge same for capacitors in series)
Q123 67.2 C
Then V123 11.2 V
C123 6.00 F
Q 67.2 C
V4 4 16.8 V
C4 4.00 F
Note that V4 V123 16.8 V 11.2 V 28.0 V, as it should.
Next consider the circuit as written in Figure 24.15g.
V3 V12 28.0 V V4
V3 11.2 V
Q3 C3V3 4.00 F11.2 V
Q3 44.8 C
Q12 C12V12 2.00 F11.2 V
Q12 22.4 C
Finally, consider the original circuit, as shown in Figure 24.15h.
Q1 Q2 Q12 22.4 C
(charge same for capacitors in series)
Q 22.4 C
V1 1 5.6 V
C1 4.00 F
Q 22.4 C
V2 2 5.6 V
C2 4.00 F
Note that V1 V2 11.2 V, which equals V3 as it should.
Summary: Q1 22.4 C, V1 5.6 V
Q2 22.4 C, V2 5.6 V
Q3 44.8 C, V3 11.2 V
Q4 67.2 C, V4 16.8 V
(c) Vad V3 11.2 V
EVALUATE: V1 V2 V4 V , or V3 V4 V . Q1 Q2 , Q1 Q3 Q4 and Q4 Q1234 .