F)(2.25 V) = 22.5
C.
57. The pair
C
3
and
C
4
are in parallel and consequently equivalent to 30
F.
Since this
numerical value is identical to that of the others (with which it is in series, with the
battery), we observe that each has one-third the battery voltage across it.
Hence, 3.0 V is
across
C
4
, producing a charge
q
4
=
C
4
V
4
= (15
F)(3.0 V) = 45
C.
58. (a) Here
D
is not attached to anything, so that the 6
C
and 4
C
capacitors are in series
(equivalent to 2.4
C
). This is then in parallel with the 2
C
capacitor, which produces an
equivalence of 4.4
C
. Finally the 4.4
C
is in series with
C
and we obtain
eq
4.4
0.82
0.82(40
F)
33
F
4.4
C
C
C
C
C
C
where we have used the fact that
C
= 40
F.
(b) Now,
B
is the point that is not attached to anything, so that the 6
C
and 2
C
capacitors
are now in series (equivalent to 1.5
C
), which is then in parallel with the 4
C
capacitor (and
thus equivalent to 5.5
C
). The 5.5
C
is then in series with the
C
capacitor; consequently,
eq
5.5
0.85
34
F .
5.5
C
C
C
C
C
C
59. The pair
C
1
and
C
2
are in parallel, as are the pair
C
3
and
C
4
; they reduce to equivalent
values 6.0
F and 3.0
F, respectively.
These are now in series and reduce to 2.0
F,
across which we have the battery voltage. Consequently, the charge on the 2.0
F
equivalence is (2.0
F)(12 V) = 24
C.
This charge on the 3.0
F equivalence (of
C
3
and
C
4
) has a voltage of
V
=
q
C
=
24
C
3
F
= 8.0 V.
CHAPTER 25
1032
Finally, this voltage on capacitor
C
4
produces a charge (2.0
F)(8.0 V) = 16
C.
60. (a) Equation 25-22 yields
U
CV
1
2
1
2
200
10
7 0
10
4 9
10
2
12
3
2
3
F
V
J
c
hc
h
.
.
.
(b) Our result from part (a) is much less than the required 150 mJ, so such a spark should
not have set off an explosion.
61. Initially the capacitors
C
1
,
C
2
, and
C
3
form a series combination equivalent to a single
capacitor, which we denote
C
123
. Solving the equation
1
2
2
3
1
3
123
1
2
3
1
2
3
1
1
1
1
C C
C C
C C
C
C
C
C
C C C
,
we obtain
C
123
= 2.40
F.
With
V =
12.0 V, we then obtain
q
=
C
123
V
= 28.8
C.
In the
final situation,
C
2
and
C
4
are in parallel and are thus effectively equivalent to
24
12.0
F
C
.
Similar to the previous computation, we use
1
24
24
3
1
3
1234
1
24
3
1
24
3
1
1
1
1
C C
C C
C C
C
C
C
C
C C C
and find
C
1234
= 3.00
F.
Therefore, the final charge is
q
=
C
1234
V
= 36.0
C.
(a) This represents a change (relative to the initial charge) of
q
= 7.20
C.
(b) The capacitor
C
24
which we imagined to replace the parallel pair
C
2
and
C
4
,
is in series
with
C
1
and
C
3
and thus also has the final charge
q
=36.0
C found above.
The voltage
across
C
24
would be
24
24
36.0
C
3.00 V
12.0
F
q
V
C
.
This is the same voltage across each of the parallel pairs. In particular,
V
4
= 3.00 V
implies that
q
4
=
C
4
V
4
= 18.0
C.
