Chapter 24
p. 1
CHAPTER 24 – Capacitance, Dielectrics, Electric Energy Storage
1. From
Q
=
CV
, we have
2500
µ
C =
C
(950 V), which gives
C
=
2.6
F
.
2. From
Q
=
CV
, we have
28.0
×
10
–8
C = (12,000
×
10
–12
F)
V
, which gives
V
=
23.3 V
.
3. From
Q
=
CV
, we have
75 pC =
C
(12.0 V), which gives
C
=
6.3 pF
.
4.
The final potential on the capacitor will be the voltage of the battery.
Positive charge will move from one
plate to the other, so the charge that flows through the battery is
Q
=
CV
= (15.6
×
10
–6
F)(12 V) =
1.9
×
10
–4
C
.
5. From
Q
=
CV
, we see that
?
Q
=
C
?
V
;
16
C =
C
(48 V – 28 V), which gives
C
=
0.80
F
.
6.
When the capacitors are connected, some charge will flow from
C
1
to
C
2
until the potential
difference across the two capacitors is the same:
V
1
=
V
2
=
V
.
Because charge is conserved, we have
Q
0
=
Q
1
+
Q
2
.
For the two capacitors we have
Q
1
=
C
1
V
,
and
Q
2
=
C
2
V
.
When we form the ratio, we get
Q
2
/
Q
1
= (
Q
0
–
Q
1
)/
Q
1
=
C
2
/
C
1
, which gives
Q
1
=
Q
0
C
1
/(
C
1
+
C
2
).
For
Q
2
we have
Q
2
=
Q
0
–
Q
1
=
Q
0
{1 – [
C
1
/(
C
1
+
C
2
)]}, thus
Q
2
=
Q
0
C
2
/(
C
1
+
C
2
).
We find the potential difference from
Q
1
=
C
1
V
;
Q
0
C
1
/(
C
1
+
C
2
) =
C
1
V
, which gives
V
=
Q
0
/(
C
1
+
C
2
).
7.
We assume that the charge transferred is small compared to the initial charge on the plates so the
potential difference between the plates is constant.
The energy required to move the charge is
W
=
qV
.
Thus the charge on each plate is
Q
=
CV
=
C
(
W
/
q
) = (16
×
10
–6
F)(25 J)/(0.20
×
10
–3
C) =
2.0 C
.
Because this is much greater than the charge moved, our assumption is justified.
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View Full DocumentChapter 24
p. 2
8.
We find the initial charges on the capacitors:
Q
1
=
C
1
V
1
= (2.40
µ
F)(880 V) = 2112
C;
Q
2
=
C
2
V
2
= (4.00
F)(560 V) = 2240
C.
(
a
) When the capacitors are connected with positive plates together, some charge will flow from
C
2
to
C
1
until the potential difference across the two capacitors is the same:
V
1
′
=
V
2
′
=
V
.
Because charge is conserved, we have
Q
=
Q
1
′
+
Q
2
′
=
Q
1
+
Q
2
= 2112
C + 2240
C = 4352
C.
For the two capacitors we have
Q
1
′
=
C
1
V
,
and
Q
2
′
=
C
2
V
.
When we add these, we get
Q
1
′
+
Q
2
′
=
Q
= (
C
1
+
C
2
)
V
;
4352
C = (2.40
F + 4.00
F)
V
, which gives
V
=
680 V
.
The charge on
C
1
is
Q
1
′
=
C
1
V
= (2.40
F)(680 V) = 1.63
×
10
3
C =
1.63
×
10
–3
C
.
The charge on
C
2
is
Q
2
′
=
C
2
V
= (4.00
F)(680 V) = 2.72
×
10
3
C =
2.72
×
10
–3
C
.
(
b
) When the capacitors are connected with opposite plates together and charge flows from
C
2
to
C
1
, the combination of positive and negative charges will result in the cancellation of some
charge until the potential difference across the two capacitors is the same:
V
1
′
=
V
2
′
=
V
.
Because charge is conserved, we have
Q
=
Q
1
′
+
Q
2
′
=
Q
2
–
Q
1
= 2240
C – 2112
C = 128
C.
For the two capacitors we have
Q
1
′
=
C
1
V
,
and
Q
2
′
=
C
2
V
.
When we add these, we get
Q
1
′
+
Q
2
′
=
Q
= (
C
1
+
C
2
)
V
;
128
C = (2.40
F + 4.00
F)
V
, which gives
V
=
20 V
.
The charge on
C
1
is
Q
1
′
=
C
1
V
= (2.40
F)(20 V) =
48
C
.
The charge on
C
2
is
Q
2
′
=
C
2
V
= (4.00
F)(20 V) =
80
C
.
9.
For a parallelplate capacitor, we find the area from
C
=
Å
0
A
/
d
;
0.40
×
10
–6
F = (8.85
×
10
–12
C
2
/N
·
m
2
)
A
/(4.0
×
10
–3
m), which gives
A
=
1.8
×
10
2
m
2
.
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 Spring '09
 Fielding
 Physics, Capacitance, Charge, Energy, Electric charge, Vab, μF

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