howard (cah3459) – homework 11 – turner – (56705)
1
This printout should have 17 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
A parallelplate capacitor oF dimensions
2
.
39 cm
×
5
.
77 cm is separated by a 1
.
18 mm
thickness oF paper.
±ind the capacitance oF this device. The
dielectric constant
κ
For paper is 3.7.
Correct answer: 38
.
2862 p±.
Explanation:
Let :
κ
=3
.
7
,
d
=1
.
18 mm = 0
.
00118 m
,
and
A
=2
.
39 cm
×
5
.
77 cm = 0
.
00137903 m
2
.
We apply the equation For the capacitance oF
a parallelplate capacitor and fnd
C
=
κ!
0
A
d
= (3
.
7) (8
.
85419
×
10

12
C
2
/
N
·
m
2
)
×
±
0
.
00137903 m
2
0
.
00118 m
²
1
×
10
12
p±
1±
=
38
.
2862 p±
.
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
.
722843
μ
c.
Explanation:
Let :
E
max
.
6
×
10
7
V
/
m
.
Since the thickness oF the paper is 0
.
00118 m,
the maximum voltage that can be applied
beFore breakdown is
V
max
=
E
max
d.
Hence, the maximum charge is
Q
max
=
CV
max
=
CE
max
d
= (38
.
2862 p±)(18880 V)
·
1
×
10

12
±
1 p±
·
1
×
10
6
μ
C
1C
=
0
.
722843
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
infnite radius.
1.
U
=
Q
2
a
4
π!
0
2.
U
=
Q
2
π
8
!
0
a
3.
U
=
Q
8
0
a
4.
U
=
Q
2
16
0
a
5.
U
=
Q
2
a
6.
U
=
Q
2
8
0
a
2
7.
U
=
Q
2
8
0
a
correct
8.
U
=
Q
2
4
0
a
Explanation:
The capacitance Formula For a spherical ca
pacitor oF inner radius
a
and outer radius
b
is
C
=
ab
k
e
(
b

a
)
.
IF we let
b
→∞
, we fnd we can neglect
a
in
the denominator compared to
b
, so
C
→
a
k
e
=4
0
a.
The total energy stored is
U
=
Q
2
2
C
=
Q
2
8
0
a
.
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View Full Documenthoward (cah3459) – homework 11 – turner – (56705)
2
004
(part 2 of 4) 10.0 points
Find the energy stored in a capacitor of charge
Q
±lled with dielectric, use
C
κ
=
κ C
.
1.
U
=
Q
2
3(
κ

1)
C
2.
U
=
Q
4
3.
U
=
Q
2
3
C
4.
U
=
Q
2
5.
U
=
Q
2
2
correct
6.
U
=
Q
2
7.
U
=
Q
2
3
8.
U
=
Q
2
2(
κ

1)
C
9.
U
=
Q
2
2
C
Explanation:
The energy stored in a capacitor of capaci
tance
C
κ
is
U
=
Q
2
2
C
κ
=
Q
2
2
.
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
±nal 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
2.
W
if
=
Q
2
2
C
±
κ
+
1
κ
²
3.
W
if
=
Q
2
2
4.
W
if
=
Q
2
2
C
±
1

1
κ
²
correct
5.
W
if
=
Q
2
2
C
±
1
κ

1
²
6.
W
if
=
Q
2
2
C
±
κ

1
κ
²
7.
W
if
=
Q
2
5
8.
W
if
=
Q
2
2
(1

κ
)
9.
W
if
=
Q
2
2
C
(1

κ
)
Explanation:
The energy stored in the capacitor with the
dielectric is
U
i
=
Q
2
2
.
After pulling the slab out, the charge on the
capacitor remains the same (charge conserva
tion). The energy stored is
U
f
=
Q
2
2
C
,
so the work that must be done to pull the slab
out is
W
=
U
f

U
i
=
Q
2
2
C

Q
2
2
,
or
W
=
Q
2
2
C
±
1

1
κ
²
.
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 Spring '10
 Turner
 Correct Answer, Electric charge, Howard

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