Physics 2213
HW #6 – Solutions
Spring 2010
24.14.
Let
C
1
15 pF
,
C
2
9.0 pF
and
C
3
11 pF
.
The capacitors between
b
and
c
, that is C
2
and C
3
, are in
parallel. The equivalent capacitance of this combination is
C
23
C
2
±
C
3
20 pF
.
The combination
C
23
is in series with the C
1
(15 pF) capacitor, so the total effective capacitance is given
by
1
C
123
1
C
1
±
1
C
23
. Solving for C
123
,
C
123
C
1
C
23
C
1
±
C
23
(15 pF)(20 pF)
15 pF
±
20 pF
8.6 pF
Notice that for capacitors in parallel the equivalent capacitance is larger than any of the individual
capacitors. For capacitors in series the equivalent capacitance is smaller than any of the individual
capacitors.
24.66.
If there is charge +Q on the top plate and –Q on the bottom plate, then the metal slab, being a
conductor, will acquire an induced charge of –Q on its top and +Q on its bottom. This situation is
analogous to having two capacitors
C
1
in series, each with separation
1
2
(
d
²
a
).
Recall that for capacitors in series
1
C
eq
1
C
1
±
1
C
2
.
In our case, C
1
=C
2
.
The capacitance of a parallel
plate capacitor is
0
A
C
s
H
, where
s
is the plate separation.
For C
1
,
s=
(
da
)/2.
For C
0
(the capacitor
without the metal slab),
s=d
.
(a)
C
1
C
1
±
1
C
1
§
©
¨
·
¹
¸
²
1
1
2
C
1
1
2
0
A
(
d
²
a
) 2
0
A
d
²
a
(b)
C
0
A
d
²
a
0
A
d
d
d
²
a
C
0
d
d
²
a
(c)
As
a
o
0
,
C
o
C
0
. The metal slab has no effect if it is very thin. And as
a
o
d
,
C
o f
.
V
Q
/
C
.
V
Ey
is the potential difference between two points separated by a distance
y
parallel to a uniform
electric field. When the distance is very small, it takes a very large field and hence a large
Q
on the
plates for a given potential difference. Since
Q
CV
this corresponds to a very large
C
.
#1 [Smile!]
(a)
To find the capacitance, we will first find time constant of the circuit,
W
= RC.
The charge on the capacitor at time t is q(t)
=
Q
o
e
t/
W
³
v(t)
=
V
o
e
t/
W
,
with V
o
=
Q
o
C
.
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 Spring '10
 Capacitance, Electric Potential, Energy, Potential Energy, Potential difference

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