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ID: A
1
UCSD Physics 2B
Unit Exam 2A
Potential & Capacitors
Answer Section
MULTIPLE CHOICE
1. ANS: B
The field is uniform between parallel plates, so
V
=−
E
⋅
dL
0
d
∫








=
Ed
=
150
×
10
3
V
/
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
3.77
×
−
2
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
=
5.66
×
3
V
=
5.66
kV
TOP:
V from E
2. ANS: D
Just apply the formula & plug & play
W
=
1
2
CV
2
=
1
2
55
×
−
3
F
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
790
V
()
2
=
1.72
×
4
J
TOP: CAP Energy
3. ANS: A
We first have to
reduce the circuit
before we can calculate anything at all.
C
1
and C
2
are in
parallel
so the capacitances ADD:
C
12
=
C
1
+
C
2
=
9.0
μ
F
.
The network C12 is in series with C3, so
1
C
series
=
1
C
+
1
C
3
=
1
9.0
F
+
1
3.0
F
=
4
9.0
F
Hence the equivalent capacitance
C
=
2.25
F
.
One way to proceed is to use
U
3
=
Q
3
2
2
C
3
and the fact that
Q
3
must be the same as
Q
TOT
for the whole network.
Thus
Q
TOT
=
C
TOT
V
TOT
=
2.25
×
−
6
F
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
V
=
2.7
×
−
5
C
so that
U
3
=
Q
3
2
2
C
3
=
2.7
×
−
5
C
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
2
23
.0
×
−
6
F
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
=
1.22
×
−
4
J
But the
fastest
way is to use
Q
=
C
3
V
3
=
C
V
which splits the voltages inversely to the ratio of the
capacitances.
V
3
V
=
C
C
3
=
9
F
3
F
=
3
1
. Since the two voltages must add to 12V total, we
V
=
3
V
and
V
3
=
9
V
and we can immediately use the equivalent energy formula:
U
3
=
1
2
C
3
V
3
2
=
1
2
3.0
F
Ê
Ë
Á
ˆ
¯
˜
9.0
V
2
=
1.22
×
−
4
J
=
122
J
TOP:
CAP network Energy
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View Full DocumentID: A
2
4. ANS: B
We solve this by equating the Energy gain
through
Work done
on
the electron by the field
W
=
q
Δ
V
with the
final kinetic energy of the moving electron
KE
=
1
2
mv
2
W
=
KE
⇒
v
=
2
q
Δ
V
m
=
21
.6
0
×
10
−
19
C
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
24.0
×
3
V
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
9.11
×
−
31
kg
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
=
9.18
×
7
m
s
TOP:
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 Summer '09
 Lever
 Physics

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