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ID: B
1
UCSD Physics 2B
Unit Exam 2B
Potential & Capacitors
Answer Section
MULTIPLE CHOICE
1. ANS: A
If you remember the formula, just plug & play, otherwise derive it from first principles. First find the
potential in cylindrical symmetry, and then apply the capacitor definition formula
V
=
E
⋅
d
r
a
b
∫
=
2
k
λ
r
⋅
dr
a
b
∫
=
2
k
ln
b
a
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
C
=
Q
V
=
L
2
k
b
a
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
=
L
2
k
b
a
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
Now plug in the numbers:
C
=
L
2
k
b
a
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
=
18
m
29
.0
×
10
9
Nm
2
/
C
2
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
1.0
×
−
3
m
0.90
×
−
3
m
Ê
Ë
Á
Á
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
˜
˜
=
9.49
×
−
9
F
=
9.49
nF
On the other hand
you could get a rough answer by treating the coaxial cylinders like a parallel plates (the
easiest case). Not exact, but the plates are fairly close together, so the field shouldn’t diverge too much. The
surface area
A
=
2
π
rL
(you could use
r
=
a
or
r
=
b
or something in between) and the separation distance is
d
=
b
−
a
. Plug in the formula:
C
=
ε
0
A
d
=
0
2
a
+
b
2
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
L
b
−
a
=
0
2
0.95
()
0.10
=
9.50
Very close to the same answer.
.. and a great way to check your work, even if you know the correct formula.
Note that I used the mean distance, and since the units of millimeter cancel in numerator & denominator, I
didn’t bother.
TOP: CAP cylinder
2. ANS: D
E
=−
∂
V
∂
x
,
∂
V
∂
y
,
∂
V
∂
z
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
33
V
m
,
−
22
V
m
2
y
,66
V
m
3
z
2
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
V
m
i
8
+
V
m
2
y
j
8
−
66
V
m
3
z
2
k
8
TOP:
E from V
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2
3. ANS: B
Since
V
=
Ed
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 Summer '09
 Lever
 Physics

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