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Capacitance and Dielectrics
2421
24.66.
IDENTIFY:
This situation is analogous to having two capacitors
1
C
in series, each with separation
1
2
()
.
da
−
SET UP:
For capacitors in series,
eq
1
2
111
CC
C
=+
.
EXECUTE:
(a)
1
00
11
1
22
2
AA
−
⎛⎞
==
=
⎜⎟
−
−
⎝⎠
PP
(b)
0
d
d
da dda
=
−−
−
(c)
As
0
a
→
,
0
→
. The metal slab has no effect if it is very thin. And as
ad
→
,
C
→∞
.
/
VQC
=
.
VE
y
=
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
QC
V
=
this corresponds to a very large
C
.
24.67.
(a)
IDENTIFY:
The conductor can be at some potential
V
, where
V
= 0 far from the conductor. This potential
depends on the charge
Q
on the conductor so we can define
C = Q/V
where
C
will not depend on
V
or
Q.
(b) SET UP:
Use the expression for the potential at the surface of the sphere in the analysis in part (a).
EXECUTE:
For any point on a solid conducting sphere
0
/4
if
0 at
.
VQ
RV
r
π
=
=→
∞
P
0
0
4
4
QR
CQ
R
=
P
P
(c)
( )( )
12
6
4
0
4
4 8
.
8
5
41
0
F
/m 6
.
3
81
0
m 7
.
1
01
F 7
1
0
F
.
CR
πμ
×
×
=
×=
P
EVALUATE:
The capacitance of the earth is about seven times larger than the largest capacitances in this range. The
capacitance of the earth is quite small, in view of its large size.
24.68.
IDENTIFY:
The electric field energy density is
2
1
0
2
E
P
. For a capacitor
2
2
Q
U
C
=
.
SET UP:
For a solid conducting sphere of radius
R
,
0
E
=
for
rR
<
and
2
0
4
Q
E
r
=
P
for
>
.
EXECUTE:
(a)
2
1
0
2
:
0.
rRu
E
<=
=
P
(b)
2
2
2
2222
4
.
43
2
QQ
E
rr
ππ
>=
=
=
(c)
2
2
4
88
RR
Qd
r Q
U
udV
r udr
∞∞
=
=
∫∫
∫
.
(d)
This energy is equal to
2
0
1
24
Q
R
P
which is just the energy required to assemble all the charge into a spherical
distribution. (Note that being aware of double counting gives the factor of 1/2 in front of the familiar potential energy
formula for a charge
Q
a distance
R
from another charge
Q
.)
EVALUATE:
(e)
From Equation (24.9),
2
2
Q
U
C
=
.
2
0
8
Q
U
R
=
P
from part (c) ,
0
4
=
P
, as in Problem (24.67).
24.69.
IDENTIFY:
We model the earth as a spherical capacitor.
SET UP:
The capacitance of the earth is
0
4
ab
ba
C
=
−
P
and, the charge on it is
Q = CV
, and its stored energy is
2
1
2
.
UC
V
=
EXECUTE:
(a)
( )( )
66
2
92
2
6
6
6.38 10 m 6.45 10 m
1
6.5 10 F
9.00 10 N m /C 6.45 10 m
6.38 10 m
C
−
××
×
×⋅
×
−×
(b)
( )
6.54 10 F (350,000 V) = 2.3 10 C
V
−
×
×
(c)
( )
2
9
6.54 10 F (350,000 V)
4.0 10 J
V
−
×
=
×
EVALUATE:
While the capacitance of the earth is larger than ordinary laboratory capacitors, capacitors much larger
than this, such as 1 F, are readily available.
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View Full Document2422
Chapter 24
24.70.
IDENTIFY:
The electric field energy density is
2
1
0
2
uE
=
P
.
2
2
Q
U
C
=
.
SET UP:
For this charge distribution,
0
E
=
for
a
rr
<
,
0
2
E
r
λ
π
=
P
for
ab
rrr
<
<
and
0
E
=
for
b
>
.
Example 24.4 shows that
0
2
ln( / )
ba
U
L
=
P
for a cylindrical capacitor.
EXECUTE:
(a)
2
2
2
11
00
22
28
λλ
ππ
⎛⎞
==
=
⎜⎟
⎝⎠
PP
(b)
2
0
2
4
b
a
r
r
L
dr
Uu
d
V
L
u
r
d
r
r
=
∫∫
∫
P
and
2
0
ln( / )
4
U
L
=
P
.
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 Spring '06
 Buchler
 Physics, Capacitance

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