Physics 2B
Quiz 1
Summer Session I 2009
Name:
PID:
Version 0
Quiz ID:
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and the test version.
1. A proton is located at x = 3.0 nm, y = 0.0 nm and an electron is located at x = 0.0 nm,
y = 1.0 nm. Find the attractive Coulomb force between them.
Solution:
F =
A. 2.3 × 10−11 N
ke2
9 × 109 (1.6 × 10−19 )2
=
= 2.3 × 10−11 (N)
r2
(3 × 10−9 )2 + (1 × 10−9 )2
B. 9.0 × 10−18 N
C. 9.0 × 108 N
D. 3.7 × 10−15 N
1.
2. A 10.0 C charge experience a 45.0x̂ N force. What is the electric field at the location of
the charge?
Solution:
~
~ = F = 4.50x̂ N/C
E
q
A. 4.5 x̂ N/C
B. −4.5 x̂ N/C
C. 0.45 x̂ N/C
D. 4.5 ŷ N/C
2.
3. A thin wire 9.0 m long has a charge of 1.0 nC. Find the electric field strength 2.0 mm
from the end of the wire.
Solution: Use the principle of superposition:
Z
E=
k dq
=
r2
Z
0
9
9
9 × 109 1 × 10−9 dx
1
≈ 500N/C
=
(x + 0.002)2
9
(x + 0.002)2 0
Physics 2B
Quiz 1
A. 500 N/C
Summer Session I 2009
B. 20 N/C
C. 10 N/C
Version 0
D. 2.3 × 106 N/C
3.
4. A dipole centered at the origin with moment 3.0 µC · m is oriented at an angle of π/6
with respect to the x-axis. The positive charge is in the first quadrant. Find the torque
an electric field of 9.0 × 106 x̂ N/C exerts on the dipole.
π
π
Solution: The dipole moment is ~p = 3.0(cos x̂ + sin ŷ) µC · m. The torque is
6
6
therefore:
!
√
3
1
~ = 27N · m
x̂ + ŷ × x̂ = −13.5ẑ
~τ = ~p × E
2
2
A. −14 ẑ N·m
C. −23 ẑ N·m
B. 14 ẑ N·m
D. 23 ẑ N·m
4.
~ = 2.0 x̂ − 4.0 ŷ N/C. A verticle (one edge along the z
5. In a constant electric field E
axis) 3.0m × 5.0m flat surface is oriented in such a way that gives minimum electric
flux. What is the area vector of the surface? Give your answer with a non-negative x
component.
~ = Ax x̂ + Ay ŷ. Its magnitude
Solution: Suppose the area vector has the form A
equal to the area:
q
A = A2x + A2y = 3 × 5 = 15
The smallest possible magnitude for electric flux is zero:
~ ·A
~ = 2Ax − 4Ay = 0
E
Solve the two above equations gives Ax = 13.4 and Ay = 6.7.
We did not catch the error in time, so everyone gets the point for this problem.
A. 5.5 x̂ + 2.7 ŷ
B. 0 x̂ + 0 ŷ
C. 3.0 x̂ + 5.0 ŷ
D. 7.7 x̂ + 1.9 ŷ
5.
6. Ten dipoles, each consisting of two charges ±17µC, are located within a spherical surface,
the radius of which is 9mm. What is the electric flux through the sphere?
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Physics 2B
Quiz 1
Summer Session I 2009
Version 0
Solution: There are just as much positive charge as there are negative charge inside
the sphere. Total charge inside is zero. The electric flux is zero.
A. 0 N·m2 /C
B. 1.9 × 106 N·m2 /C
C. 1.8 × 109 N·m2 /C
D. 170 N·m2 /C
6.
7. A hollow spherical conductor of inner radius 2.0 cm has a 4.0 C point charge at its center.
Find the surface charge density at the inner surface of the sphere.
Solution: The electric field in the conductor is zero. From Gauss’s law, the total
charge enclosed by any surface in the conductor is zero. This means the inner surface
has exactly the same amount of charge in opposite sign:
σ = −4/4π0.022 ≈ −795C/m2
A. −800C/m2
B. 800C/m2
C. −1200C/m2
D. 1200C/m2
7.
8. At each corner of a square is a particle of charge q. Fixed at the center of the square is a
particle of charge Q. For what value of Q is the force on each of the four corner charges
zero?
Solution: Pick any corner—e.g. upper right—and use the principle of superposition
to compute the total force. It will suffice to compute just the x-component:
√
kq 2
π
kqQ
π
kq
kq 2
q
cos + √
cos = 2 q + √ + 2 Q
Fx = 2 + √
a
4 ( 2 a/2)2
4
a
( 2 a)2
2 2
Setting Fx = 0 yields Q = − 14 + √12 q ≈ −0.96q.
A. −0.96q
B. −4.0q
C. −1.25q
D. +1.25q
8.
9. A long solid rod 5 cm in radius carries a uniform volume charge density. If the electric
field strength at the furface of the rod (not near either end) is 20 kN/C. What is the
volume charge density?
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Physics 2B
Quiz 1
Summer Session I 2009
Version 0
Solution: For a uniformly charged rod of radius a, the relationship between volume
charge density and linear charge density is:
λ = ρπa2
The electric field of on the surface of the long rod is, as shown in class:
E=
ρa
λ
=
.
2π0 a
20
From this one can solve for volume charge density ρ:
ρ = 2ε0 E/a ≈ 7.07 × 10−6 C/m3
A. 7.1×10−6 C/m3
B. 14×10−6 C/m3
C. 0.35×10−6 C/m3
D. 3.0×10−6 C/m3
9.
10. The electric field 22 cm from a very long wire carrying a uniform line charge density is
1.9 kN/C. What will be the field strength 38 cm from the wire?
Solution: The electric field of a very long wire decrease with distance like ∼ 1/r,
so:
22
E = 1.9 ×
= 1.1kN/C
38
A. 1.1 kN/C
B. 3.3 kN/C
C. 0.64 kN/C
D. 1.9 kN/C
10.
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