PHY2049 Chapter 25 Solutions - Chapter 25 1. (a) The...

1011
Chapter 25
1. (a) The capacitance of the system is
C
q
V
= = =
Δ
70
20
35
pC
V
pF. .
(b) The capacitance is independent of q ; it is still 3.5 pF.
(c) The potential difference becomes
Δ V
q
C
= = =
200
35
57
pC
pF
V.
.
2. 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.
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,
( ) ( )
2
12 2
2
10 0
3
8.85 10 F m 8.2 10 m
1.44 10 F 144 pF.
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. (a) We use Eq. 25-17:
( )( )
( )
( )
2
2
0
9
N m
C
40.0mm 38.0mm
4 84.5 pF.
8.99 10 40.0mm 38.0mm
ab
C
b a
πε
⋅
= = =
−
× −
(b) Let the area required be A . Then C = ε
0
A /( b – a ), or

CHAPTER 25 1012
( ) ( )( )
( )
2
12 2 2
0
84.5pF 40.0mm 38.0mm
191cm .
8.85 10 C /N m
C b a
A
ε −
− −
= = =
× ⋅
5. Assuming conservation of volume, we find the radius of the combined spheres, then
use C = 4 πε
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
( ) 3
3
4 4
2
3 3
R R
′
= ⇒
p p
′ = 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
′ ′
= = = p p p
With R = 2.00 mm, we obtain
( )( )
12 3 13 5.04 8.85 10 F m 2.00 10 m 2.80 10 F C π − − − = × × = × .
6. We use C = A ε
0
/ d .
(a) The distance between the plates is
( )( )
2 12 2 2
12
0
1.00m 8.85 10 C /N m
8.85 10 m.
1.00F
A
d
C
ε
−
−
× ⋅
= = = ×
(b) Since d is much less than the size of an atom ( ∼ 10
–10
m), this capacitor cannot be
constructed.
7. For a given potential difference V , the charge on the surface of the plate is
( ) q Ne nAd e = =
where d is the depth from which the electrons come in the plate, and n is the density of
conduction electrons. The charge collected on the plate is related to the capacitance and
the potential difference by q CV = (Eq. 25-1). Combining the two expressions leads to
C d
ne
A V
= .
With 14 / / 5.0 10 m/V
s s
d V d V − = = × and 28 3 8.49 10 / m n = × (see, for example, Sample
Problem — “Charging the plates in a parallel-plate capacitor”), we obtain
28 3 19 4 2 (8.49 10 / m )(1.6 10 C)(5.0 10 14 m/V) 6.79 10 F/m
C
A
− − = × × × − = × .

1013
8. 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
( )
( )
3
6
1 00C
9 09 10
110V 1 00 10 F
q .
N . .
VC
.
−
= = = ×
×
9. 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
10. 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 .