ID: A
UCSD Physics 2B
Answer Section
Unit Exam 2A
Potential & Capacitors
MULTIPLE CHOICE
1. ANS: B
The field is uniform between parallel plates, so
|
|V | = | −
||
∫
d
0
|
E ⋅ dL | = Ed =
||
ˆÊ
ˆ
ÊÁ
Á 150 × 10 3 V / m˜˜ ÁÁ 3.77 × 10−2 m˜˜ = 5.66 × 103 V = 5.66kV
¯Ë
¯
Ë
TOP: V from E
2. ANS: D
Just apply the formula & plug & play
1
1Ê
2
ˆ
W = CV 2 = ÁÁ 55 × 10 −3 F ˜˜ (790V ) = 1.72 × 10 4 J
2
2Ë
¯
TOP: CAP Energy
3. ANS: A
We first have to reduce the circuit before we can calculate anything at all.
C1 and C2 are in parallel so the capacitances ADD: C 12 = C 1 + C 2 = 9.0μF .
The network C12 is in series with C3, so
1
1
1
1
1
4
=
+
=
+
=
C series C 12 C 3 9.0μF 3.0μF 9.0μF
Hence the equivalent capacitance C series = 2.25μF .
One way to proceed is to use U 3 =
Q 23
2C 3
and the fact that Q3 must be the same as QTOT for the whole network.
Ê
ˆ
Thus Q TOT = C TOT V TOT = ÁÁ 2.25 × 10−6 F ˜˜ (12V ) = 2.7 × 10 −5 C so that
Ë
¯
2
ÊÁ
ˆ
Á 2.7 × 10 −5 C ˜˜
Q 23
Ë
¯
U3 =
=
= 1.22 × 10 −4 J
2C 3 2 ÊÁÁ 3.0 × 10−6 F ˆ˜˜
Ë
¯
But the fastest way is to use Q = C 3 V 3 = C 12 V 12 which splits the voltages inversely to the ratio of the
C 12 9μF 3
V3
=
=
= . Since the two voltages must add to 12V total, we V 12 = 3V and
capacitances.
V 12
C3
3μF 1
V 3 = 9V and we can immediately use the equivalent energy formula:
1
1
2
U 3 = C 3 V 23 = ÁÊË 3.0μF ˜ˆ¯ (9.0V ) = 1.22 × 10 −4 J = 122 μ J
2
2
TOP: CAP network Energy
1
ID: A
4. ANS: B
We solve this by equating the Energy gain through Work done on the electron by the field W = qΔV with the
1
final kinetic energy of the moving electron KE = mv 2
2
W = KE ⇒ v =
2qΔV
=
m
Ê
ˆÊ
ˆ
2 ÁÁ 1.60 × 10 −19 C ˜˜ ÁÁ 24.0 × 10 3 V ˜˜
m
Ë
¯Ë
¯
= 9.18 × 10 7
ÊÁ
ˆ˜
−31
s
Á 9.11 × 10 kg ˜
Ë
¯
TOP: PE = KE
5. ANS: C
Apply the formula for concentric spheres:
1
1
C= Ê
= 11 × 10−12 F
ˆ
ˆ˜ = Ê
Ê
˜
Á
Á
1
1
1
1
ˆÁ
9
2
2˜
˜˜
˜ Á
−
k ÁÁ
−
˜˜
b ˜¯ ÁË 9.0 × 10 Nm / C ˜¯ ÁÁÁ
−3
−3
Ë a
m
m
1.0
×
10
0.99
×
10
¯
Ë
If you can’t find the formula and your mind is frazzled, you can make a pretty good guess.
Note in this case that the spheres are almost the same radius, so they’re VERY close together. If you draw a
picture, the field lines hardly diverge - that means the field is ALMOST uniform. Why not treat it like a
parallel plate capacitor? If you can remember that formula, just substitute the appropriate area:
A
4π b 2
b
1 10 −3
1
C = ε0 ≈ ε0
= 4πε 0
=
=
= 11.1pF
d
a
b−a
k 1 − 0.99 10k
1−
b
TOP: CAP sphere
6. ANS: E
V
or Volts per meter
d
È ˘ È
˘ È
˘ È
˘ È ˘
F ÍÍ N ˙˙ ÍÍ N m ˙˙ ÍÍ N ⋅ m 1 ˙˙˙˙ ÍÍÍÍ J / C ˙˙˙˙ ÍÍÍÍ V ˙˙˙˙
=
× ˙˙ = ÍÍ
Or, by definition, E = = ÍÍÍÍ ˙˙˙˙ = ÍÍÍÍ × ˙˙˙˙ = ÍÍÍÍ
q ÍÎ C ˙˚ ÍÎ C m ˙˚ ÍÎ C
m ˙˚ ÍÎ m ˙˙˙˚ ÍÍÍÎ m ˙˙˙˚
Since V = Ed for parallel plates, invert the formula to get E =
TOP: UNITS
2
ID: A
7. ANS: C
By superposition, the potential at the point charge Q is given by
V = V1 + V2 =
kq 1 kq 2
−
=
r1
r2
kq
a2 + a2
+
k(−q)
a2 + a2
=0
The work required is to get Q there from ‘far away’ is the potential times its charge
Ê
ˆ
W = QV = ÁÁ 5.8 × 10−6 C ˜˜ × 0 = 0J
Ë
¯
Another way the get this with no calculation whatsoever, is to draw a force diagram at the point Q
- just draw the correct field arrow at Q from each of the other charges. Adding the vectors, you can
see that the resulting net field arrow points to the right, perpendicular to the y-axis. If charge Q is
brought in from infinity along that axis, the work W = ∫ F • d y = 0 since there is never any force in
the direction of motion!
8. ANS: C
ÁÊÁ kq ˜ˆ˜ ke 2
PE = qV = q ÁÁÁ ˜˜˜ =
=
Á r ˜
r
Ë ¯
2
ÁÊÁ 9.0 × 10 9 Nm2 / C 2 ˜ˆ˜ ÊÁÁ 1.6 × 10 −19 C ˆ˜˜
Ë
¯Ë
¯
= 8.86 × 10 −14 J
ÁÊÁ 2.6 × 10 −15 m˜ˆ˜
Ë
¯
TOP: Energy of Assembly
9. ANS: B
ÊÁ
ˆ
Á 4200m2 ˜˜
¯
Ë
A
A ÊÁ
ˆ
C = ε 0 ⇒ d = ε 0 = Á 8.85 × 10 −12 C 2 / Nm2 ˜˜ Ê
= 5.24 × 10 −4 m = 0.524mm
−6
d
C Ë
¯ ÁÁ 71 × 10 F ˆ˜˜
Ë
¯
TOP: CAP plate
10. ANS: D
remember to take the partial derivatives with respect to x, y and z:
ÊÁ ∂V ∂V ∂V ˆ˜
Ê
ˆ
˜˜ = − ÁÁÁ 33 V ,− 22 V y,66 V z 2 ˜˜˜ = −33 V 8i + 22 V y 8j − 66 V z 2 k8
E = − ÁÁÁÁ
,
,
˜
Á
˜
˜
Á m
m
m2
m3 ˜¯
m2
m3
Ë ∂x ∂y ∂z ¯
Ë
TOP: E from V
3
ID: A
11. ANS: B
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=
∫
b
a
C=
E ⋅ dr =
∫
b
a
Ê
ˆ
2k λ
⋅ dr = 2k λ ln ÁÁÁ b ˜˜˜
r
Ë a¯
Q
λL
L
=
=
Ê
ˆ
V 2k λ ln ÁÁ b ˜˜ 2k ln ÊÁÁ b ˆ˜˜
Á a˜
Á a˜
Ë
Ë
¯
¯
Now plug in the numbers:
L
C=
ÊÁ b ˆ˜ =
˜˜
2k ln ÁÁ
Ë a¯
18m
ˆ
Ê
−3
ˆ ÁÁ 1.0 × 10 m ˜˜˜˜
Ê
2 ÁÁ 9.0 × 10 9 Nm2 / C 2 ˜˜ ln ÁÁÁÁ
¯ Á 0.90 × 10 −3 m ˜˜˜
Ë
¯
Ë
= 9.49 × 10 −9 F = 9.49nF
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 = 2π 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:
ÁÊ a + b ˜ˆ˜
˜˜ L
2π ÁÁÁÁ
˜
2π (0.95) (18)
Ë 2 ¯
A
C = ε0 = ε0
= ε0
= 9.50nF
b−a
0.10
d
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
TOP: CAP cylinder
12. ANS: D
If the field is uniform and the line is straight then the differential element of potential is
dV = E ⋅ dL = E (dL) cos θ
Now the work to move the charge along that line is the integral (the constants move outside)
r
r
r
0
0
0
W = ∫ qdV = ∫ qE ⋅ dL = qE cos θ ∫ dL = qEr cos θ
Ê
ˆÊ
ˆÊ
ˆ
= ÁÁ 5.00 × 10 −6 C ˜˜ ÁÁ 3.50 × 105 N / C ˜˜ ÁÁ 50.0 × 10 −2 m˜˜ cos (57.0°)
Ë
¯Ë
¯Ë
¯
= 4.77 × 10 −1 J = 477mJ
13. ANS: B
N
For a parallel network, the capacitors add: C parallel =
∑C
i
i=1
If all the values are the same, then C parallel = NC = 8 ÊÁË 120μF ˆ˜¯ = 960μF
4
ID: A
14. ANS: D
Here we use the Capacitor Formula but remember to express the charge in Coulombs:
Q = CV ⇒
Ê
ˆ ÁÊ
12 ˜
−19
˜ˆ
− Á
Q n e e ÁË 3.71 × 10 e ˜¯ ÁË 1.60 × 10 C / e ˜¯
=
= 6.45 × 10 −8 F = 64.5 nF
C=
V
V
9.20 V
TOP: Q = CV
15. ANS: C
The Work on a charge by an Electric Field is the Charge times Potential Difference:
W = q (ΔV ) = (5.0C ) (12.0V ) = 60J
Joule
You might recall that in units, 1Volt = 1
Coulomb
TOP: WORK
5