oliver (jmo673) – HW04 – Tsoi – (59090)
1
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printout
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have
30
questions.
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before answering.
001
(part 1 of 4) 10.0 points
An airfilled capacitor consists of two parallel
plates, each with an area of 9
.
6 cm
2
,
sepa
rated by a distance 3
.
5 mm
.
A 21 V potential
difference is applied to these plates.
The permittivity of a vacuum is 8
.
85419
×
10
−
12
C
2
/
N
·
m
2
. 1 pF is equal to 10
−
12
F
.
The magnitude of the electric field between
the plates is
1.
E
= (
V d
)
2
.
2.
E
=
parenleftbigg
V
d
parenrightbigg
2
.
3.
None of these
4.
E
=
V
d
.
correct
5.
E
=
V d .
6.
E
=
parenleftbigg
d
V
parenrightbigg
2
.
7.
E
=
1
(
V d
)
2
.
8.
E
=
1
V d
.
9.
E
=
d
V
.
Explanation:
Since
E
is constant between the plates,
V
=
integraldisplay
vector
E
·
d
vector
l
=
E d
E
=
V
d
.
002
(part 2 of 4) 10.0 points
The magnitude of the surface charge density
on each plate is
1.
None of these
2.
σ
=
ǫ
0
parenleftbigg
V
d
parenrightbigg
2
.
3.
σ
=
ǫ
0
d
V
.
4.
σ
=
ǫ
0
V d
.
5.
σ
=
ǫ
0
(
V d
)
2
.
6.
σ
=
ǫ
0
(
V d
)
2
.
7.
σ
=
ǫ
0
V d
8.
σ
=
ǫ
0
V
d
.
correct
9.
σ
=
ǫ
0
parenleftbigg
d
V
parenrightbigg
2
.
Explanation:
Use Gauss’s Law. We find that a pillbox of
cross section
S
which sticks through the sur
face on one of the plates encloses charge
σ S.
The flux through the pillbox is only through
the top, so the total flux is
E S.
Gauss’ Law
gives
σ
=
ǫ
0
E
=
ǫ
0
V
d
Alternatively, we could just recall this result
for an infinite conducting plate (meaning we
neglect edge effects) and apply it.
003
(part 3 of 4) 10.0 points
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oliver (jmo673) – HW04 – Tsoi – (59090)
2
Calculate the capacitance.
Correct answer: 2
.
42858 pF.
Explanation:
Let :
A
= 0
.
00096 m
2
,
d
= 0
.
0035 m
,
V
= 21 V
,
and
ǫ
0
= 8
.
85419
×
10
−
12
C
2
/
N
·
m
2
.
The capacitance is given by
C
=
ǫ
0
A
d
= 8
.
85419
×
10
−
12
C
2
/
N
·
m
2
×
0
.
00096 m
2
0
.
0035 m
= 2
.
42858
×
10
−
12
F
=
2
.
42858 pF
.
004
(part 4 of 4) 10.0 points
Calculate plate charge;
i.e.
, the magnitude of
the charge on each plate.
Correct answer: 51
.
0001 pC.
Explanation:
The charge
Q
on one of the plates is simply
Q
=
C V
= (2
.
42858
×
10
−
12
F) (21 V)
= 5
.
10001
×
10
−
11
C
=
51
.
0001 pC
.
005
10.0 points
A spherical capacitor consists of a conducting
ball of radius 16 cm that is centered inside a
grounded conducting spherical shell of inner
radius 19 cm.
What
charge
is
required
to
achieve
a
potential
of
1224
V
on
the
ball
of
ra
dius
16
cm?
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 1
.
38004
×
10
−
7
C.
Explanation:
Let :
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
,
a
= 16 cm = 0
.
16 m
,
b
= 19 cm = 0
.
19 m
and
V
= 1224 V
.
The capacitance of a spherical capacitor is
given by
C
=
a b
k
e
(
b

a
)
. The charge required
to achieve a potential difference
V
between
the shell and the ball is given by
Q
=
C V
=
1
k
e
a b V
(
b

a
)
=
1
(8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(0
.
16 m) (0
.
19 m) (1224 V)
(0
.
19 m)

(0
.
16 m)
=
1
.
38004
×
10
−
7
C
.
keywords:
006
(part 1 of 2) 10.0 points
A capacitor network is shown below.
112 V
16
μ
F
16
μ
F
16
μ
F
16
μ
F
14
μ
F
16
μ
F
16
μ
F
y
z
What is the equivalent capacitance between
points
y
and
z
of the entire capacitor net
work?
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 Fall '09
 TSOI
 Physics, Energy, Correct Answer, Electric charge, Oliver, C12 C12 C12

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