homework 04 – LIN, KEVIN – Due: Feb 10 2008, 4:00 am
1
Question 1, chap 26, sect 1.
part 1 of 1
10 points
An airfilled parallelplate capacitor is to
have a capacitance of 1
.
2 F.
The permittivity of free space is 8
.
85419
×
10
−
12
C
2
/
N
·
m
2
.
If the distance between the plates is 1
.
1 mm,
calculate the required surface area of each
plate.
Correct answer: 149
.
082 km
2
(tolerance
±
1
%).
Explanation:
Let :
C
= 1
.
2 F
,
d
= 1
.
1 mm = 0
.
0011 m
,
and
ǫ
0
= 8
.
85419
×
10
−
12
C
2
/
N
·
m
2
.
The capacitance is
C
=
ǫ
0
A
d
A
=
C d
ǫ
0
=
(1
.
2 F) (0
.
0011 m)
8
.
85419
×
10
−
12
C
2
/
N
·
m
2
×
parenleftbigg
1 km
2
10
6
m
2
parenrightbigg
=
149
.
082 km
2
.
Question 2, chap 26, sect 1.
part 1 of 1
10 points
A variable air capacitor used in tuning cir
cuits is made of
N
very thin semicircular con
ducting plates each of radius
R
and positioned
d
from each other. A second identical set of
plates that is free to rotate is enmeshed mid
way between the gap of the first set of plates.
Note:
One of the outside moving plates lies
outside the first set of plates spaced
d
2
.
R
θ
d
Determine the capacitance as a function of
the angle of rotation
θ
(in rad) , where
θ
= 0
corresponds to the maximum capacitance.
1.
C
=
ǫ
0
N d
2
θ
R
2.
C
=
2
ǫ
0
R
2
(2
π
−
θ
)
d
3.
C
=
ǫ
0
N R
2
π
d
4.
C
=
ǫ
0
R
2
(
π
−
θ
)
d
5.
C
=
N R
2
(
π
−
θ
)
d
6.
C
=
ǫ
0
N R
2
θ
d
7.
C
=
ǫ
0
(2
N
)
R
2
θ
d
8.
C
=
ǫ
0
(2
N
)
d θ
R
2
9.
C
=
ǫ
0
(2
N
−
1)
R
2
(
π
−
θ
)
d
correct
10.
C
=
ǫ
0
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
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homework 04 – LIN, KEVIN – Due: Feb 10 2008, 4:00 am
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
.
Question 3, chap 26, sect 1.
part 1 of 3
10 points
A
capacitor
consists
of
two
thin
coax
ial cylindrical conducting shells of length
L
.
The inner shell has radius
a
and the outer
shell has radius
b
.
Assume the length is
much greater than the radii of the cylin
ders,
L
≫
b
.
There is charge
−
Q
on the
inner shell and charge +
Q
on the outer shell.

Q
+
Q
r
a
b
The magnitude of the electric field at radius
r
, between the two shells, is given by
1.
bardbl
vector
E
bardbl
=
Q
2
π rL
2.
bardbl
vector
E
bardbl
=
Q
2
π ǫ
0
r L
correct
3.
bardbl
vector
E
bardbl
=
Q
2
2
π ǫ
0
r L
4.
bardbl
vector
E
bardbl
=
Q
4
π r
2
5.
bardbl
vector
E
bardbl
=
Q
2
π ǫ
0
r
2
6.
None of these
7.
bardbl
vector
E
bardbl
=
Q
2
4
π ǫ
0
r
2
8.
bardbl
vector
E
bardbl
=
Q
2
2
π rL
9.
bardbl
vector
E
bardbl
=
Q
2
4
π r
2
Explanation:
Construct a Gaussian cylinder at radius
r
(for
a < r < b
).
The charge enclosed is the
charge on the inner shell,
−
Q
.
Thus,
Φ =
contintegraldisplay
vector
E
·
d
vector
A
=
−
Q
ǫ
0
E
2
π r L
=
−
Q
ǫ
0
bardbl
vector
E
bardbl
=
Q
2
π ǫ
0
r L
.
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 Spring '08
 Turner
 Physics, Capacitance, Work, Correct Answer, Kevin –, Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub

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