Version PREVIEW – HW4 – feng – (58230)
1
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printout
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have
20
questions.
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before answering.
Capacitance
001
10.0 points
A capacitor that is connected to a 49 V source
contains 99
μ
C of charge.
What is its capacitance?
Correct answer: 2
.
02041
μ
F.
Explanation:
Let :
V
= 49 V
and
q
= 99
μ
C
.
The capacitance is
C
=
q
V
=
99
μ
C
49 V
=
2
.
02041
μ
F
Potential of a Capacitor
002
10.0 points
A 5
.
1
μ
F capacitor is charged with 3
.
3
×
10
5
C.
What potential difference exists across it?
Correct answer: 6
.
47059
×
10
10
V.
Explanation:
Let :
C
= 5
.
1
μ
F = 5
.
1
×
10
−
6
F
and
q
= 3
.
3
×
10
5
C
.
The capacitance is
C
=
q
V
,
so
V
=
q
C
=
3
.
3
×
10
5
C
5
.
1
×
10
−
6
F
=
6
.
47059
×
10
10
V
Capacitance of a Sphere
003
(part 1 of 2) 10.0 points
An isolated conducting sphere can be consid
ered as one element of a capacitor (the other
element being a concentric sphere of infinite
radius).
If
k
=
1
4
π ǫ
0
, the capacitance of the system
is
C
, and the charge on the sphere is
Q
, what
is the radius
r
of the sphere?
1.
r
=
Q
2
C
k
2.
r
=
k C
Q
3.
r
=
k Q C
4.
r
=
C
k
5.
r
=
k C
correct
6.
r
=
k Q
Explanation:
The definition of capacitance is
C
=
Q
V
Q
=
C V .
Let
V
(
∞
) = 0.
Then the potential of the
conducting sphere is
V
=
k
Q
r
=
k
C V
r
,
so
r
=
k C .
004
(part 2 of 2) 10.0 points
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
If
the
potential
on
the
surface
of
the
sphere
is
6000
V
and
the
capacitance
is
8
.
94
×
10
−
11
F, what is the surface charge
density?
Correct answer: 66
.
1182 nC
/
m
2
.
Explanation:
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Version PREVIEW – HW4 – feng – (58230)
2
Let :
k
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
,
V
= 6000 V
,
and
C
= 8
.
94
×
10
−
11
F
.
The charge is
Q
=
C V .
Assuming a uniform surface charge density
we have
σ
=
Q
4
π r
2
=
C V
4
π
(
k C
)
2
=
1
4
π k
2
V
C
=
1
4
π
(8
.
98755
×
10
9
N
·
m
2
/
C
2
)
2
×
6000 V
8
.
94
×
10
−
11
F
=
66
.
1182 nC
/
m
2
.
Variable Capacitor
005
10.0 points
A variable air capacitor used in tuning circuits
is made of
N
semicircular plates each of radius
R
and positioned
d
from each other. A second
identical set of plates that is free to rotate is
enmeshed with the first set.
R
θ
d
Determine the capacitance as a function of
the angle of rotation
θ
, where
θ
= 0 corre
sponds to the maximum capacitance.
1.
C
=
ǫ
0
(2
N
)
R
2
θ
d
2.
C
=
ǫ
0
N R
2
π
d
3.
C
=
ǫ
0
N R
2
(
π

θ
)
d
4.
C
=
ǫ
0
N d
2
θ
R
5.
C
=
ǫ
0
(2
N

1)
R
2
(
π

θ
)
d
correct
6.
C
=
ǫ
0
(2
N
)
d θ
R
2
7.
C
=
2
ǫ
0
R
2
(2
π

θ
)
d
8.
C
=
ǫ
0
N R
2
θ
d
9.
C
=
ǫ
0
R
2
(
π

θ
)
d
10.
C
=
N R
2
(
π

θ
)
d
Explanation:
Considering the situation of
θ
= 0, the two
sets of semicircular plates in fact form 2
N

1
capacitors connected in parallel, with each
one having capacitance
C
=
ǫ
0
A
d
2
=
ǫ
0
π R
2
2
d
2
=
ǫ
0
π R
2
d
.
So the total capacitance would be
C
= (2
N

1)
ǫ
0
π R
2
d
.
Note:
The common area of the two sets of
plates varies linearly when one set is rotating,
so the capacitance at angle
θ
is
C
=
ǫ
0
(2
N

1)
R
2
(
π

θ
)
d
.
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 Spring '08
 Turner
 Physics, Capacitance, Vseries

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