26
Capacitance and Dielectrics
CHAPTER OUTLINE
26.1
Def nition oF Capacitance
26.2
Calculating Capacitance
26.3
Combinations oF Capacitors
26.4
Energy Stored in a Charged
Capacitor
26.5
Capacitors with Dielectrics
26.6
Electric Dipole in an Electric ±ield
26.7
An Atomic Description oF
Dielectrics
ANSWERS TO QUESTIONS
*Q26.1
(a) False.
(b) True. In
Q = C
∆
V
, the capacitance is the
proportionality constant relating the variables
Q
and
∆
V
.
Q26.2
Seventeen combinations:
Individual
CCC
123
,,
Parallel
C C CC CC CC C
12
13
23
++
+
+
+
,,,
SeriesParallel
11
1
3
CC
C
+
⎛
⎝
⎜
⎞
⎠
⎟
+
−
,
1
2
C
+
⎛
⎝
⎜
⎞
⎠
⎟
+
−
,
1
1
C
+
⎛
⎝
⎜
⎞
⎠
⎟
+
−
3
1
CC C
+
+
⎛
⎝
⎜
⎞
⎠
⎟
−
,
13 2
1
+
+
⎛
⎝
⎜
⎞
⎠
⎟
−
,
231
1
+
+
⎛
⎝
⎜
⎞
⎠
⎟
−
Series
111
1
⎛
⎝
⎜
⎞
⎠
⎟
−
,
1
+
⎛
⎝
⎜
⎞
⎠
⎟
−
,
1
+
⎛
⎝
⎜
⎞
⎠
⎟
−
,
1
+
⎛
⎝
⎜
⎞
⎠
⎟
−
*Q26.3
Volume is proportional to radius cubed. Increasing the radius by a factor of 3
1/3
will triple the
volume. Capacitance is proportional to radius, so it increases by a factor of 3
1/3
. Answer (d).
*Q26.4
Let
C
2
=
NC
1
be the capacitance of the large capacitor and
C
1
that of the small one.
The equivalent capacitance is
C
eq
=
C
CN
C
NN
C
N
N
C
eq
=
+
=
+
=
+
1
1
1
1
//
(
)
/
This is slightly less than
C
1
, answer (d).
*Q26.5
We ±
nd the capacitance, voltage, charge, and energy for each capacitor.
(a)
C
=
20
µ
F
∆
V
= 4 V
Q = C
∆
V
= 80
C
U
= (1/2)
Q
∆
V
= 160
J
(b)
C
=
30
F
∆
V
=
Q
/
C
= 3 V
Q =
90
C
U
= 135
J
(c)
C
=
Q
/
∆
V
=
40
F
∆
V
= 2 V
Q =
80
C
U
= 80
J
(d)
C
=
10
F
∆
V
= (2
U
/
C
)
1/2
= 5 V
Q =
50
C
U
= 125
J
(e)
C
=
2
U
/
∆
V
2
= 5
F
∆
V
= 10 V
Q
= 50
C
U
= 250
J
(f)
C
=
Q
2
/2
U
= 20
F
∆
V
= 6 V
Q
= 120
C
U
= 360
J
Then (i) the ranking by capacitance is c > b > a = f > d > e.
(ii) The ranking by voltage
∆
V
is e > f > d > a > b > c.
(iii) The ranking by charge
Q
is f > b > a = c > d = e.
(iv) The ranking by energy
U
is f > e > a > b > d > c.
75
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Chapter 26
Q26.6
A capacitor stores energy in the electric f
eld between the plates. This is most easily seen when
using a “dissectible” capacitor. IF the capacitor is charged, careFully pull it apart into its compo
nent pieces. One will f
nd that very little residual charge remains on each plate. When reassem
bled, the capacitor is suddenly “recharged”—by induction—due to the electric f
eld set up and
“stored” in the dielectric. This proves to be an instructive classroom demonstration, especially
when you ask a student to reconstruct the capacitor without supplying him/her with any rubber
gloves or other insulating material. (OF course, this is
after
they sign a liability waiver.)
*Q26.7
(i) According to
Q = C
∆
V
, the answer is (b).
(ii) ±rom
U =
(1/2)
C
∆
V
2
, the answer is
(a).
*Q26.8
The charge stays constant as
C
is cut in halF, so
U = Q
2
/2
C
doubles: answer (b).
*Q26.9
(i) Answer (b).
(ii) Answer (c).
(iii) Answer (c).
(iv) Answer (a).
(v) Answer (a).
*Q26.10
(i) Answer (b).
(ii) Answer (b).
(iii) Answer (b).
(iv) Answer
(c).
(v) Answer (b).
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 Spring '11
 williams,frank
 Capacitance, Charge, Energy, Dielectrics, ∆V

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