Version 063 – Midterm 1 – Shih – (58250)
1
This
printout
should
have
26
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
A 60 m length of coaxial cable has a solid
cylindrical wire inner conductor with a di
ameter of 2
.
939 mm and carries a charge of
11
.
04
μ
C.
The surrounding conductor is a
cylindrical shell and has an inner diameter of
7
.
048 mm and a charge of

11
.
04
μ
C.
Assume the region between the conductors
is air.
The Coulomb constant is 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
What is the capacitance of this cable?
1. 2.21279
2. 1.71512
3. 3.81622
4. 2.41954
5. 2.49337
6. 2.78068
7. 1.09339
8. 1.1855
9. 0.910748
10. 4.34197
Correct answer: 3
.
81622 nF.
Explanation:
Let :
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
,
Q
= 11
.
04
μ
C
,
ℓ
= 60 m
,
a
= 2
.
939 mm
,
and
b
= 7
.
048 mm
.
The charge per unit length is
λ
≡
Q
ℓ
.
V
=

integraldisplay
b
a
vector
E
·
dvectors
=

2
k
e
λ
integraldisplay
b
a
dr
r
=

2
k
e
Q
ℓ
ln
parenleftbigg
b
a
parenrightbigg
.
The capacitance of a cylindrical capacitor
is given by
C
≡
Q
V
=
ℓ
2
k
e
1
ln
parenleftbigg
b
a
parenrightbigg
=
60 m
2 (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
1
ln
parenleftbigg
7
.
048 mm
2
.
939 mm
parenrightbigg
·
parenleftbigg
1
×
10
9
nF
1 F
parenrightbigg
=
3
.
81622 nF
002
(part 1 of 3) 10.0 points
Consider the setup shown in the figure be
low, where the arc is a semicircle with radius
r
.
The total charge
Q
is negative, and dis
tributed uniformly on the semicircle.
The
charge on a small segment with angle Δ
θ
is
labeled Δ
q
.
x
y


















Δ
θ
θ
r
x
y
I
II
III
IV
B
A
O
Δ
q
is given by
1.
Δ
q
=
Q
2
π
2.
Δ
q
=
Q
Δ
θ
π
correct
3.
None of these
4.
Δ
q
=
π Q
5.
Δ
q
=
Q
6.
Δ
q
=
2
Q
π
7.
Δ
q
= 2
π Q
8.
Δ
q
=
Q
π
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Version 063 – Midterm 1 – Shih – (58250)
2
9.
Δ
q
=
Q
Δ
θ
2
π
10.
Δ
q
=
2
Q
Δ
θ
π
Explanation:
The angle of a semicircle is
π
, thus the
charge on a small segment with angle Δ
θ
is
Δ
q
=
Q
Δ
θ
π
.
003
(part 2 of 3) 10.0 points
The magnitude of the
x
component of the
electric field at the center, due to Δ
q
, is given
by
1.
Δ
E
x
=
k

Δ
q

(cos
θ
)
r
2.
Δ
E
x
=
k

Δ
q

sin
θ
r
2
3.
Δ
E
x
=
k

Δ
q

(cos
θ
)
r
2
4.
Δ
E
x
=
k

Δ
q

(sin
θ
)
r
5.
Δ
E
x
=
k

Δ
q

cos
θ
r
2
correct
6.
Δ
E
x
=
k

Δ
q

r
2
7.
Δ
E
x
=
k

Δ
q

(sin
θ
)
r
2
8.
Δ
E
x
=
k

Δ
q

sin
θ
r
9.
Δ
E
x
=
k

Δ
q

cos
θ
r
10.
Δ
E
x
=
k

Δ
q

r
2
Explanation:
Negative charge attracts a positive test
charge. At
O
, Δ
E
points toward Δ
q .
Accord
ing to the sketch, the vector Δ
E
x
is pointing
along the negative
x
axis. The magnitude of
the Δ
E
x
is given by
Δ
E
x
= Δ
E
cos
θ
=
k

Δ
q

r
2
cos
θ .
004
(part 3 of 3) 10.0 points
Determine the magnitude of the electric field
at
O .
The total charge is

26
.
1
μ
C, the radius
of the semicircle is 18 cm, and the Coulomb
constant is 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
1. 717473.0
2. 52476.7
3. 62049.0
4. 939376.0
5. 1151900.0
6. 1217810.0
7. 102868.0
8. 39658.6
9. 1064060.0
10. 4609110.0
Correct answer: 4
.
60911
×
10
6
N
/
C.
Explanation:
Let :
Q
=

26
.
1
μ
C
,
r
= 18 cm
,
and
k
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
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 Spring '08
 Turner
 Physics, Electrostatics, Correct Answer, Electric charge

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