CH-25 - Chapter 25 1. Charge flows until the potential...

Chapter 25
1. Charge flows until the potential difference across the capacitor is the same as the
potential difference across the battery. The charge on the capacitor is then q = CV , and
this is the same as the total charge that has passed through the battery. Thus,
q = (25 × 10 –6 F)(120 V) = 3.0 × 10 –3 C.
2. (a) The capacitance of the system is
C
q
V
= = =
∆
70
20
35
pC
V
pF. .
(b) The capacitance is independent of q ; it is still 3.5 pF.
(c) The potential difference becomes
∆ V
q
C
= = =
200
35
57
pC
pF
V.
.
3. (a) The capacitance of a parallel-plate capacitor is given by C = ε
0
A / d , where A is the
area of each plate and d is the plate separation. Since the plates are circular, the plate area
is A = π R 2 , where R is the radius of a plate. Thus,
( 29 ( 29
2
12 2
2
10
0
3
8.85 10 F m 8.2 10 m
1.44 10 F 144pF.
1.3 10 m
R
C
d
π
ε π
- -
-
-
= = = =
(b) The charge on the positive plate is given by q = CV , where V is the potential
difference across the plates. Thus,
q = (1.44 × 10
–10
F)(120 V) = 1.73 × 10
–8
C = 17.3 nC.
4. We use C = A ε
0
/ d .
(a) Thus,
d
A
C
= =
×
= ×
-
⋅ -
ε
0
12
12
100 885 10
100
885 10
2 . .
.
.
m
F
m.
2
C
N m
2
c hd i
1039

CHAPTER 25
(b) Since d is much less than the size of an atom ( ∼ 10 –10 m), this capacitor cannot be
constructed.
5. Assuming conservation of volume, we find the radius of the combined spheres, then
use C = 4 π ε
0
R to find the capacitance. When the drops combine, the volume is doubled.
It is then V = 2(4 π /3) R 3 . The new radius R' is given by
( 29 3
3
4 4
2
3 3
R R =
π π
′ = R R 2 1 3 .
The new capacitance is
1 3
0 0 0
4 4 2 5.04 . C R R R ε ε ε = = = π π π
With R = 2.00 mm, we obtain
( 29 ( 29
12 3 13 5.04 8.85 10 F m 2.00 10 m 2.80 10 F C π - - - =% =
.
6. (a) We use Eq. 25-17:
C
ab
b a
=
-
=
× -
=
⋅
4
40 0 38 0
8 99 10 40 0 38 0
84 5
0
9
π ε
. .
. . .
.
mm mm
mm mm
pF.
N m
C
2
2
b gb g
d ib
g
(b) Let the area required be A . Then C = ε
0
A /( b – a ), or
( 29 ( 29 ( 29
( 29
2
2
2
12
C
0
N m
84.5pF 40.0mm 38.0mm
191cm .
8.85 10
C b a
A
ε
-
⋅
- -
= = =
×
7. The equivalent capacitance is given by C
eq
= q / V , where q is the total charge on all the
capacitors and V is the potential difference across any one of them. For N identical
capacitors in parallel, C
eq
= NC , where C is the capacitance of one of them. Thus,
/ NC q V =
and
( 29
( 29
3
6
1 00C
9 09 10
110V 1 00 10 F
q .
N . .
VC
.
-
= = =
8. The equivalent capacitance is
C C
C C
C C
eq
F
F F
F F
F. = +
+
= +
+
=
3
1 2
1 2
4 00
10 0 500
10 0 500
7 33 .
. .
. .
. μ
μ μ
μ μ
μ
b gb g
9. The equivalent capacitance is
16

( 29 ( 29 ( 29
1 2 3
eq
1 2 3
10.0 F 5.00 F 4.00 F
3.16 F.
10.0 F 5.00 F 4.00 F
C C C
C
C C C
μ μ μ
μ
μ μ μ
+ +
= = =
+ + + +
10. The charge that passes through meter A is
q C V CV = = = =
eq
F V C. 3 3 250 4200 0 315 . . μ
b gb g
11. (a) and (b) The original potential difference V
1
across C
1
is
( 29 ( 29
eq
1
1 2
3.16 F 100.0V
21.1V.
10.0 F 5.00 F
C V
V
C C
μ
μ μ
= = =
+ +
Thus ∆ V
1
= 100.0 V – 21.1 V = 78.9 V and
∆ q
1
= C
1
∆ V
1
= (10.0 μ F)(78.9 V) = 7.89 × 10
–4
C.
12. (a) The potential difference across C
1
is V
1
= 10.0 V. Thus,
q
1
= C
1
V
1
= (10.0 μ F)(10.0 V) = 1.00 × 10
–4
C.
(b) Let C = 10.0 μ F. We first consider the three-capacitor combination consisting of C
2
and its two closest neighbors, each of capacitance C . The equivalent capacitance of this
combination is
2
eq
2
1 50
C C
C C . C.