Physics 2B
Quiz 2
Summer Session I 2009
Name:
PID:
Version 0
Quiz ID:
READ FIRST: Make sure to write/fill in your scantron form your name,
quiz ID and the test version.
−19
e ≈ 1.6 × 10
9
2
−12
2
C , k ≈ 9.0 × 10 N m /C , 0 ≈ 8.9 × 10
2
2
Z
C /N m ,
dx
= ln |x − a|
x−a
Unless otherwise specified, assume potential is zero at infinity.
~ = 2.5x̂ N/C.
1. An electron move in the −x̂ direction 0.5m through an electric field E
What is the change in potential energy?
~ · ~l = 1.25V. An electron carries
Solution: The potential difference is: ∆V = −E
charge −e, So the change in potential energy is ∆U = q∆V = −1.25 eV.
A. -1.25 eV
B. 1.25 eV
C. 1.25 J
D. -1.25 J
2. At each of the four corners of a square is a 2 nC charge. The side of the square is 1.41
m. What is electric potential at the center of the square?
√
Solution: The center is the same distance away from each charges: 1.41/ 2 ≈ 1
(m). Calculate the potential using principle of superposition:
J·m
9 × 109 2 · 2 × 10−9 C
kq
C
= 72V
φ=4 ≈4×
d
1m
A. 72 V
B. 36 V
C. 18 V
D. 144 V
3. How much energy is needed to assemble four 2 nC charges to form a square of side 1.41
cm?
Solution: Each charge see three other charges: two at distance 1.41 cm, one at
distance 2 cm. So the total energy is just four times the energy to move the last
charge in placewith the factor of 1/2 to take into account double-counting.
9 × 109 N m2
(2 × 10−9 C)2 (2 × 10−9 C)2
U =4×
2×
+
≈ 13.8 × 10−6 J
2
2
0.0141m
0.02m
C
Physics 2B
Quiz 2
Summer Session I 2009
Version 0
Alternative: Assemble the charges one by one. With q being the charge of each
charges and a being the length of a side of the square. The first one does not cost any
2
energy. Moving the one across the diagonal costs √kq2 a . Moving in the third corner
2
2
costs 2 kqa . And moving in the final corner costs 2 kqa +
same answer.
A. 14 µJ
B. 3.5 µJ
C. 904 µJ
kq 2
√
.
2a
They add up to the
D. 251 µJ
4. A thin, uniformly charged rod has charge 3 nC. One end is located at the origin, and
the other end at x = −2 m. What is the electric potential at x = 1 m?
Solution: Use the principle of superposition:
Z 0
3 × 10−9 C dx
9J m
V = 9 × 10 2
= (27 ln 3)/2 ≈ 14.8V
C −2 1 − x 2m
A. 14.8 V
B. 29.7 V
C. 0 V
D. 18.7 V
5. A capacitor is form with two long concentric cylinders with radii ra = 10.0 cm and
rb = 30.0 cm. They are charged ±4.0 µC/m. How much energy is stored per unit
length?
λ
. Pick a section
2π0 r
of length L and integrating the energy density over the volume between the cylinder
gives the energy stored:
Z rb
Z
0 rb λ2
0
2
E (r)2πrdr =
2πrdr
U= L
2
2 ra 4π 2 20 r2
ra
Z rb
U
1
λ2
λ2
rb
=
dr =
ln
= 9 × 109 (4 × 10−6 )2 ln 3 = 0.16J/m
L
4π0 ra r
4π0 ra
Solution: The electric field between the cylinders is E(r) =
Alternative: The potential difference between the inner and outter cylinders is:
Z
λ
rb
V = − E(r)dr =
ln
2π0 ra
Pick a segment of the capacitor of length L. The energy stored in the segment is is
U = 12 QV , where Q = λL is the charge stored in the segment. Divide both sides by
length gives energy per length:
U
1Q
1
λ2
rb
=
V = λV =
ln ,
L
2 L
2
4π0 ra
Page 2
Physics 2B
Quiz 2
Summer Session I 2009
Version 0
which is same as above.
A. 0.16 J/m
B. 1.6 J/m
C. 0.96 J/m
D. 0.025 J/m
6. Three capacitors of equal capacitance are arranged as shown. The voltage across C1 is
60.0 V. What is the voltage across C3 ?
Solution: C1 and C2 in series, so they have the same charge. Since they also have
the same capacitance, the voltage of C2 is also 60.0V. And the total voltage across
both is 120 V. C3 is in parallel with them. So its voltage is also 120 V.
A. 120 V
B. 60 V
C. 240 V
D. 180 V
7. Two capacitors, one a 5.0 µF capacitor, C1 , and the other a 6.0 µF capacitor, C2 , are
connected in series. If a 150.0 V voltage source is applied to the capacitors, find the
voltage drop across the 5.0 µF capacitor.
Solution: The equivalent total capacitance is:
CT−1 = C1−1 + C2−1
⇒
CT =
C1 C2
= 30/11µF
C1 + C2
The charge on the capacitors is:
Q = CT VT = 150 × 30/11µC ≈ 409µC
The voltage on the 5.0 µF capacitor is thus:
V1 = Q/C1 ≈ 82V
A. 82 V
B. 68 V
C. 130 V
D. 9.0 V
Page 3
Physics 2B
Quiz 2
Summer Session I 2009
Version 0
8. Two capacitors, C1 =5.0 µF and C2 = 6.0 µF, are charged separately to 400 µC. After
the batteries are removed, the positive plates of both capacitors are connected together
and the negative plates are connected together. What is the final charge on C1 ?
Solution: Suppose the final charges are Q1 and Q2 respectively. Conservation of
charge gives: Q1 + Q2 = 2Q = 800µC. When the capacitors are connected, they
have the same voltage: Q1 /C1 = V1 = V2 = Q2 /C2 . Solving these two equations
C1
gives Q1 =
2Q ≈ 363µC.
C1 + C2
A. 363 µC
B. 437 µC
C. 400 µC
D. 667 µC
9. Two identical sphere carry equal-magnitude, 1.0 nC charges that differ in sign. If their
radii are 3.0 mm, and they are 5.0 m apart, What is the potential difference between
them?
Solution: The two spheres are so far apart (5 m 3 mm) that we assume the
over all electric field is essentially that of two isolated point charges. The potential
difference is then:
2 1.0 × 10−9 C −1.0 × 10−9 C
9N m
∆V = 9 × 10
−
= 6000 V
0.003 m
0.003 m
C2
A. 6 kV
B. 0.0 V
C. 2 kV
D. 3 kV
10. What is the electric energy density inside a 1 m2 parallel plates capacitor charged to
±55 µC? Assume the separation of the plates is much smaller than 1 m.
Solution: We can compute the electric field from infinite plate approximation. The
charge density is σ = Q/A = 55 × 10−6 C/m2 , given electric field:
E=
σ
≈ 6.18 × 106 N/C
0
The energy density is:
u=
A. 171 J/m3
B. 342 J/m3
0 2
E ≈ 170 J/m3
2
C. 0.17 J/m3
Page 4
D. 86 J/m3