241
C
APACITANCE AND
D
IELECTRICS
24.18.
I
DENTIFY
:
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
=
.
S
ET
U
P
:
1
C
and
2
C
are in parallel and
3
C
is in series with the parallel combination of
1
C
and
2
C
.
E
XECUTE
:
(a)
1
2
and
C
C
are in parallel and so have the same potential across them:
6
2
1
2
6
2
40.0
10
C
13.33 V
3.00
10
F
Q
V
V
C
−
−
×
=
=
=
=
×
. Therefore,
6
6
1
1
1
(13.33 V)(3.00
10
F)
80.0
10
C
Q
V C
−
−
=
=
×
=
×
. Since
3
C
is
in series with the parallel combination of
1
2
and
C
C
, its charge must be equal to their combined charge:
6
6
6
3
40.0
10
C
80.0
10
C
120.0
10
C
C
−
−
−
=
×
+
×
=
×
.
(b)
The total capacitance is found from
6
6
tot
12
3
1
1
1
1
1
9.00
10
F
5.00
10
F
C
C
C
−
−
=
+
=
+
×
×
and
tot
3.21 F
C
µ
=
.
6
tot
6
tot
120.0
10
C
37.4 V
3.21 10
F
ab
Q
V
C
−
−
×
=
=
=
×
.
E
VALUATE
:
6
3
3
6
3
120.0
10
C
24.0 V
5.00
10
F
Q
V
C
−
−
×
=
=
=
×
.
1
3
ab
V
V
V
=
+
.
24.28.
I
DENTIFY
:
After the two capacitors are connected they must have equal potential difference, and their combined
charge must add up to the original charge.
S
ET
U
P
:
/
C
Q V
=
. The stored energy is
2
2
1
2
2
Q
U
CV
C
=
=
E
XECUTE
:
(a)
0
.
Q
CV
=
(b)
1
2
1
2
Q
Q
V
C
C
=
=
and also
1
2
0
Q
Q
Q
CV
+
=
=
.
1
C
C
=
and
2
2
C
C
=
so
1
2
(
2)
Q
Q
C
C
=
and
1
2
2
Q
Q
=
.
1
3
2
Q
Q
=
.
1
2
3
Q
Q
=
and
1
0
2
2
3
3
Q
Q
V
V
C
C
=
=
=
.
(c)
2
2
2
2
2
2
1
2
1
2
3
3
0
1
2
(
)
2(
)
1
1
1
1
2
2
3
3
Q
Q
Q
Q
Q
U
CV
C
C
C
C
C
=
+
=
+
=
=
(d)
The original
U
was
2
1
0
2
U
CV
=
, so
2
0
1
6
U
CV
∆
= −
.
(e)
Thermal energy of capacitor, wires, etc., and electromagnetic radiation.
E
VALUATE
:
The original charge of the charged capacitor must distribute between the two capacitors to make the
potential the same across each capacitor. The voltage
V
for each after they are connected is less than the original
voltage
0
V
of the charged capacitor.
24.37.
I
DENTIFY
:
Use the rules for series and for parallel capacitors to express the voltage for each capacitor in terms of
the applied voltage. Express
U, Q,
and
E
in terms of the capacitor voltage.
S
ET
U
P
:
Le the applied voltage be
V.
Let each capacitor have capacitance
C.
2
1
2
U
CV
=
for a single capacitor with
voltage
V.
E
XECUTE
:
(a) series
Voltage across each capacitor is
/2.
V
The total energy stored is
(
)
2
2
1
1
s
2
4
2
[ /2]
U
C V
CV
=
=
parallel
Voltage across each capacitor is
V
. The total energy stored is
(
)
2
2
1
p
2
2
U
CV
CV
=
=
p
s
4
U
U
=
24
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242
Chapter 24
(b)
Q
CV
=
for a single capacitor with voltage
V
.
(
)
s
p
p
s
2
[ /2]
;
2(
)
2
; Q
2
Q
C V
CV
Q
CV
CV
Q
=
=
=
=
=
(c)
/
E
V d
=
for a capacitor with voltage
V.
s
p
p
s
/2 ;
/ ;
2
E
V
d
E
V d
E
E
=
=
=
E
VALUATE
:
The parallel combination stores more energy and more charge since the voltage for each capacitor is
larger for parallel. More energy stored and larger voltage for parallel means larger electric field in the parallel case.
24.39.
I
DENTIFY
and
S
ET
U
P
:
Q
is constant so we can apply Eq.(24.14). The charge density on each surface of the
dielectric is given by Eq.(24.16).
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 Spring '11
 Zhu
 Energy, Capacitors, Ceq

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