malsam (wgm329) – homework 11 – turner – (56705)
1
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printout
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have
17
questions.
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before answering.
001
(part 1 of 2) 10.0 points
A
parallelplate
capacitor
of
dimensions
1
.
75 cm
×
5
.
97 cm is separated by a 0
.
54 mm
thickness of paper.
Find the capacitance of this device.
The
dielectric constant
κ
for paper is 3.7.
Correct answer: 63
.
3825 pF.
Explanation:
Let :
κ
= 3
.
7
,
d
= 0
.
54 mm = 0
.
00054 m
,
and
A
= 1
.
75 cm
×
5
.
97 cm = 0
.
00104475 m
2
.
We apply the equation for the capacitance of
a parallelplate capacitor and find
C
=
κ ǫ
0
A
d
= (3
.
7) (8
.
85419
×
10
−
12
C
2
/
N
·
m
2
)
×
parenleftbigg
0
.
00104475 m
2
0
.
00054 m
parenrightbigg
1
×
10
12
pF
1 F
=
63
.
3825 pF
.
002
(part 2 of 2) 10.0 points
What is the maximum charge that can be
placed on the capacitor? The electric strength
of paper is 1
.
6
×
10
7
V
/
m.
Correct answer: 0
.
547624
μ
c.
Explanation:
Let :
E
max
= 1
.
6
×
10
7
V
/
m
.
Since the thickness of the paper is 0
.
00054 m,
the maximum voltage that can be applied
before breakdown is
V
max
=
E
max
d .
Hence, the maximum charge is
Q
max
=
C V
max
=
C E
max
d
= (63
.
3825 pF)(8640 V)
·
1
×
10
−
12
F
1 pF
·
1
×
10
6
μ
C
1 C
=
0
.
547624
μ
c
.
003
(part 1 of 4) 10.0 points
Determine the total energy stored in a con
ducting sphere with charge
Q
.
Hint:
Use the capacitance formula for
a spherical capacitor which consists of two
spherical shells.
Take the inner sphere to
have a radius
a
and the outer shell to have an
infinite radius.
1.
U
=
Q
2
16
π ǫ
0
a
2.
U
=
Q
2
4
π ǫ
0
a
3.
U
=
Q
2
8
π ǫ
0
a
correct
4.
U
=
Q
8
π ǫ
0
a
5.
U
=
Q
2
a
4
π ǫ
0
6.
U
=
Q
2
π
8
ǫ
0
a
7.
U
=
Q
2
8
π ǫ
0
a
2
8.
U
=
Q
2
a
Explanation:
The capacitance formula for a spherical ca
pacitor of inner radius
a
and outer radius
b
is
C
=
a b
k
e
(
b

a
)
.
If we let
b
→ ∞
, we find we can neglect
a
in
the denominator compared to
b
, so
C
→
a
k
e
= 4
π ǫ
0
a .
The total energy stored is
U
=
Q
2
2
C
=
Q
2
8
π ǫ
0
a
.
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malsam (wgm329) – homework 11 – turner – (56705)
2
004
(part 2 of 4) 10.0 points
Find the energy stored in a capacitor of charge
Q
filled with dielectric, use
C
κ
=
κ C
.
1.
U
=
Q
2
3
κ C
2.
U
=
Q
2
3
C
3.
U
=
Q
2
2
κ C
correct
4.
U
=
Q
4
κ C
5.
U
=
Q
2
κ C
6.
U
=
Q
2
3 (
κ

1)
C
7.
U
=
Q
2
2 (
κ

1)
C
8.
U
=
Q
2
2
C
9.
U
=
Q
2
κ C
Explanation:
The energy stored in a capacitor of capaci
tance
C
κ
is
U
=
Q
2
2
C
κ
=
Q
2
2
κ C
.
005
(part 3 of 4) 10.0 points
Work =
U
f

U
i
, where “i” is the initial state
where there is a slab in the gap and “f” is the
final state where there is no slab in the gap.
Find the work done in pulling a dielectric
slab of dielectric constant
κ
from the gap of a
parallel plate capacitor of plate charge
Q
and
capacitance
C
.
1.
W
if
=
Q
2
2
C
parenleftbigg
κ
+
1
κ
parenrightbigg
2.
W
if
=
Q
2
2
C
3.
W
if
=
Q
2
2
C
(1

κ
)
4.
W
if
=
Q
2
2
κ C
(1

κ
)
5.
W
if
=
Q
2
2
C
parenleftbigg
κ

1
κ
parenrightbigg
6.
W
if
=
Q
2
2
C
parenleftbigg
1

1
κ
parenrightbigg
correct
7.
W
if
=
Q
2
2
C
parenleftbigg
1
κ

1
parenrightbigg
8.
W
if
=
Q
2
5
κ C
9.
W
if
=
Q
2
2
κ C
Explanation:
The energy stored in the capacitor with the
dielectric is
U
i
=
Q
2
2
κ C
.
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 Spring '08
 Turner
 Electric charge, Coaxial cable, Ucap, 0.001808 m, 1 3 164 mg, 2 0.001808 m

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