Chapter 24
Young and Friedman, 13 th ed.
24.1.
IDENTIFY: The capacitance depends on the geometry (area and plate separation) of the plates.
Q
Q
SET UP: For a parallel-plate capacitor, Vab  Ed , E 
, and C 
.
Vab
e0 A
EXECUTE: (a) Vab  Ed  (400 106 V/m)(250 103 m)  100 104 V.
(b) Solving for the area gives
Q
800  109 C
A

 226  103 m 2  226 cm 2 .
Ee0 (400  106 V/m)[8854  1012 C 2 /(N  m 2 )]
Q 800  109 C

 800  1012 F  800 pF.
Vab 100  104 V
EVALUATE: The capacitance is reasonable for laboratory capacitors, but the area is rather large.
(c) C 
24.4.
A
d
IDENTIFY:
when there is air between the plates.
A  (30 102 m)2
SET UP:
is the area of each plate.
(8854 1012 F/m)(30 102 m) 2
C
 159 1012 F  159 pF
50 103 m
EXECUTE:
EVALUATE: C increases when A increases and C increases when d decreases.
C  e0
C
24.6.
24.9.
Q
e0 A
.
Vab C  d .
IDENTIFY:
SET UP: When the capacitor is connected to the battery, enough charge flows onto the plates to make
Vab  120 V.
EXECUTE: (a) 12.0 V
Q
(b) (i) When d is doubled, C is halved. Vab 
and Q is constant, so V doubles. V  240 V.
C
(ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and V  30 V.
EVALUATE: The electric field between the plates is E  Q/e0 A. Vab  Ed . When d is doubled E is unchanged
and V doubles. When A is increased by a factor of 4, E decreases by a factor of 4 so V decreases by a factor of 4.
IDENTIFY: The energy stored in a capacitor depends on its capacitance, which in turn depends on its geometry.
C  Q/V for any capacitor, and C  e0 A for a parallel-plate capacitor.
SET UP:
d
10
Q 240  10
C
C 
 5714  1012 F.
eA
V
42

0
V
EXECUTE: (a)
Using C  0 gives
d
d
e0 A (8854 1012 C2/(N  m2 ))(680 104 m2 )

 105 mm.
C
5714 1012 F
Q
e0 A 5714  1012 F

 2857  1012 F. V  , so V  2(420 V)  840 V.
d
2
C
EVALUATE: Doubling the plate separation halves the capacitance, so twice the potential difference is required
to keep the same charge on the plates.
EVALUATE: It requires many small cells to produce a large voltage surge.
IDENTIFY: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor.
SET UP: Let C1  15 pF, C2  90 pF and C3  11 pF.
(b) d  210 103 m. C 
24.16.
EXECUTE: (a) For capacitors in parallel, Ceq  C1  C2 
so C23  C2  C3  20 pF.
(b)
C1  15 pF
and C123 
is in series with C23  20 pF. For capacitors in series,
C1C23
(15 pF)(20 pF)

 86 pF.
C1  C23 15 pF  20 pF
1
1
1



Ceq C1 C2
so
1
1
1
 
C123 C1 C23
24.20.
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.
IDENTIFY: For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the
charges are the same and the voltages add. C  Q/V .
SET UP:
C1
and C2 are in parallel and C3 is in series with the parallel combination of C1 and C2 .
EXECUTE: (a)
C1 and C2
are in parallel and so have the same potential across them:
6
Q2 400  10 C

 1333 V. Therefore, Q1  V1C1  (1333 V)(600 106 F)  800 106 C. Since
C2 300  106 F
C3 is in series with the parallel combination of C1 and C2 , its charge must be equal to their combined charge:
V1  V2 
Q3  400 106 C  800 106 C  1200 106 C.
(b) The total capacitance is found from
Vab 
Qtot 1200  106 C

 374 V.
Ctot
321  106 F
V3 
EVALUATE:
24.25.
1
1
1
1
1
and Ctot  321 F.




Ctot C12 C3 900 106 F 500 106 F
Q3 1200  106 C

 240 V.
C3 500  106 F
Vab  V1  V3.
IDENTIFY and SET UP: The energy density is given by Eq. (24.11): u  12 e0 E 2  Use V  Ed to solve
for E.
V
400 V

 800  104 V/m.
d 500 103 m
Then u  12 e0 E 2  12 (8854  1012 C2 /N  m2 )(800  104 V/m)2  00283 J/m3.
EXECUTE: Calculate E: E 
EVALUATE: E is smaller than the value in Example 24.8 by about a factor of 6 so u is smaller by about a factor
of 62  36.
24.30.
IDENTIFY: The two capacitors are in series.
1
1
1



Ceq C1 C2
C
Q
. U  12 CV 2 .
V
SET UP: For capacitors in series the voltages add and the charges are the same.
1
1
1


Ceq C1 C2
CC
(150 nF)(120 nF)
 667 nF.
EXECUTE: (a)
so Ceq  1 2 
C1  C2 150 nF  120 nF
Q  CV  (667 nF)(36 V)  24 106 C  24C
(b) Q  24C for each capacitor.
U  12 CeqV 2  12 (667  109 F)(36 V) 2  432 J
(c)
(d) We know C and Q for each capacitor so rewrite U in terms of these quantities.
U  12 CV 2  12 C (Q/C )2  Q 2 /2C
(24  106 C) 2
(24  106 C) 2
 240 J
2(150  10 F)
2(120  109 F)
Note that 192 J  240 J  432 J, the total stored energy calculated in part (c).
150 nF: U 
9
 192 J; 120 nF: U 
Q 24 106 C
Q 24 106 C

 16 V; 120 nF: V  
 20 V

9
C 150 10 F
C 120 109 F
Note that these two voltages sum to 36 V, the voltage applied across the network.
EVALUATE: Since Q is the same, the capacitor with smaller C stores more energy (U  Q2/2C ) and has a
larger voltage (V  Q/C).
(e) 150 nF: V 
24.31.
Q
. U  12 CV 2 .
V
SET UP: For capacitors in parallel, the voltages are the same and the charges add.
IDENTIFY: The two capacitors are in parallel. Ceq  C1  C2 . C 
EXECUTE: (a)
Ceq  C1  C2  35 nF  75 nF  110 nF.
Qtot  CeqV  (110 109 F)(220 V)  242C
(b) V  220 V for each capacitor.
35 nF: Q35  C35V  (35 109 F)(220 V)  77C; 75 nF: Q75  C75V  (75 109 F)(220 V)  165C. Note
that Q35  Q75  Qtot .
U  1 C V 2  12 (110  109 F)(220 V)2  266 mJ
(c) tot 2 eq
(d) 35 nF: U 35  12 C35V 2  12 (35  109 F)(220 V) 2  085 mJ;
75 nF: U 75  12 C75V 2  12 (75  109 F)(220 V)2  181 mJ. Since V is the same the capacitor with larger C stores
24.66.
more energy.
(e) 220 V for each capacitor.
EVALUATE: The capacitor with the larger C has the larger Q.
IDENTIFY: This situation is analogous to having two capacitors C1 in series, each with separation
SET UP: For capacitors in series,
1 (d
2
 a).
1
1
1


.
Ceq C1 C2
1
 1
1 
e0 A
e A
 0
EXECUTE: (a) C      12 C1  12
(d  a )/2 d  a
 C1 C1 
eA eA d
d
(b) C  0  0
 C0
d a
d d a
d a
(c) As a  0, C  C0 . The metal slab has no effect if it is very thin. And as a  d , C  . V  Q/C. V  Ey
is the potential difference between two points separated by a distance y parallel to a uniform electric
field. When the distance is very small, it takes a very large field and hence a large Q on the plates for a
given potential difference.