1. The flux of the electric field
 24 i  30j  16k  N / C
through a 2.0 m2 portion
of the yz plane is:
A)
B)
C)
D)
E)
32 N  m2/C
34 N  m2/C
42 N  m2/C
48 N  m2/C
60 N  m2/C
Solution:
  E  S  24 N / C  2 m2  48 N  m2 / C
Ans: D
2. A point particle with charge q is at the center of a Gaussian surface in the form of
a cube. The electric flux through any one face of the cube is:
A)
B)
C)
D)
E)
q/0
q/40
q/60
q/80
q/160
Solution:
Ans:
6 faces in a cube.
C
3. A particle with charge 5.0-C is placed at the corner of a cube. The total
electric flux in N  m2/C through all sides of the cube is:
A) 0
B)
C)
D)
E)
7.1  104
9.4  104
1.4  105
5.6  105
Solution:
Surface of cube centered at origin = sum of surfaces of 8 “quadrant” cubes
For q at corner of cube, no flux goes through the 3 sides that share the corner.
1
5.0  106 C
 
 7  104 N  m 2 / C
8 9  1012 C 2 / Nm 2
Ans: B
4. A round wastepaper basket with a 0.15-m radius opening is in a uniform electric
field of 300 N/C, perpendicular to the opening. The total flux through the sides and
bottom, in N  m2/C, is:
A) 0
B) 4.2
C) 21
D) 280
E) can't tell without knowing the areas of the sides and bottom
Solution:
No flux through the sides.
   300 N / C    3  0.22 m 2   36 N  m 2 / C
Ans: C
5. A 3.5-cm radius hemisphere contains a total charge of 6.6  10–7 C. The flux
through the rounded portion of the surface is 9.8  104 N  m2/C. The flux through the
flat base is:
A) 0
B) +2.3  104 N  m2/C
C) –2.3  104 N  m2/C
D) –9.8  104 N  m2/C
E) +9.8  104 N  m2/C
Solution:

7  107 C
 10  104 N  m2 / C  3  104 N  m2 / C
12
2
2
9  10 C / Nm
Ans: C
6. A conducting sphere of radius 0.01 m has a charge of 1.0  10–9 C deposited on it.
The magnitude of the electric field in N/C just outside the surface of the sphere is:
A)
B)
C)
D)
E)
zero
450
900
4500
90,000
Solution:

1.0  109 C
4  3   0.01 m
2
2
 106 C / m2
E
Ans:

106 C / m 2

 105 N / C
12 2
2
 0 9  10 C / Nm
E
7. Two large insulating parallel plates carry charge of equal magnitude, one positive
and the other negative, that is distributed uniformly over their inner surfaces. Rank
the points 1 through 5 according to the magnitude of the electric field at the points,
least to greatest.
A) 1, 2, 3, 4, 5
B)
C)
D)
E)
5, 4, 3, 2, 1
1, 4, and 5 tie, then 2 and 3 tie
2 and 3 tie, then 1 and 4 tie, then 5
2 and 3 tie, then 1, 4, and 5 tie
Solution:
Fields from each plate:
 |+| 
Total fields:
0
;
|+|


|-|
|-|

0
Ans: C
8. A solid insulating sphere of radius R contains a positive charge that is distributed
with a volume charge density that does not depend on angle but does increase with
distance from the sphere center. Which of the graphs below correctly gives the
magnitude E of the electric field as a function of the distance r from the center of the
sphere?
A) A
B) B
C) C
D) D
E) E
Solution:
For r  R :
E   4 r 2  
 = const,

  4 3 
r
 0  3 

E r
Er
 increases with r 
E  rn
with n  1
Ans: D
9. Positive charge Q is distributed uniformly throughout an insulating sphere of
radius R, centered at the origin. A particle with a positive charge Q is placed at x = 2R
on the x axis. The magnitude of the electric field at x = R/2 on the x axis is:
A)
B)
C)
D)
E)
Q/40R2
Q/80R2
Q/720/R2
11Q/180R2
none of these
Solution:
Field due to the sphere:
r
R
:
2
E
E   4 r 2  
Q
8 0 R 2
 4 3 
r 
 0 4 R 3  3 
3
1
Q
E
Qr
4 0 R 3
Total field:
Q
E
Field due to the point charge:
R

4 0  2 R  
2

Q
Q
Q
E


2
2
8 0 R
9 0 R
72 0 R 2
2

Q
9 0 R 2
Ans: C
10. Positive charge Q is placed on a conducting spherical shell with inner radius R1
and outer radius R2. A point charge q is placed at the center of the cavity. The
magnitude of the electric field at a point outside the shell, a distance r from the center,
is:
A)
Q / 4 0 R12
B)
Q / 4 0 ( R12  r 2 )
C) q/40r2
D) (q + Q)/40r2
E)
(q  Q) / 4 0 ( R12  r 2 )
Solution:
E
Field due to the sphere:
Ans: D
E
q
4 0 r
2

4 0 r 2
E
Field due to the point charge:
Total field:
Q
Q
4 0 r
2

q
4 0 r 2
qQ
4 0 r 2