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Homework 2
Ch16: Q 13, 21, 23, 24; P. 25, 34, 41, 43, 50, 67
Questions:
13. Explain why the test charges we use when measuring electric fields must be small.
Solution
If the test charge is too big, it can create strong forces that will relocate charges
responsible for the measured field. As the result of this relocation the electric field will
be different from the original. Because of that, the test charge should be small.
21. Consider a small positive test charge located on an electric field line at some point,
such as point P in Fig. 16–31a. Is the direction of the velocity and/or acceleration of
the test charge along this line? Discuss.
Solution
We assume that there are no other forces (like gravity) acting on the test charge. The
direction of the electric field line gives the direction of the force on the test charge. The
acceleration is always parallel to the force by Newton’s 2nd law, and so the acceleration
lies along the field line. If the particle is at rest initially and then released, the initial
velocity will also point along the field line, and the particle will start to move along the
field line. However, once the particle has a velocity, it will not follow the field line
unless the line is straight. The field line gives the direction of the acceleration or the
direction of the change in velocity.
*23. If the electric flux through a closed surface is zero, is the electric field necessarily

zero at all points on the surface? Explain. What about the converse: If E  0 at all
points on the surface is the flux through the surface zero?
Solution
By definition:  E   EA cos .
Just because the electric flux through a closed surface is zero, the field need not be zero

on the surface. (The sum could be equal to zero even if E  0 .) For instance, if the charge
inside a closed surface is zero ( Q  0) , then according to the Gauss’s law:  E  Q /  0 ,
the total flux is zero. However, charges outside the surface could create a non zero
electric field on the surface. For example, consider a closed surface near an isolated point
charge, and the surface does not enclose the charge. There will be electric field lines
passing through the surface, but the total electric flux through the surface will be zero
since the surface does not enclose any charge. The same number of field lines will enter
the volume enclosed by the surface as leave the volume enclosed by the surface.
On the contrary, if E  0 at all points on the surface, then there are no electric field
lines passing through the surface, and so according to the definition the flux through the
surface is zero.
*24.A point charge is surrounded by a spherical gaussian surface of radius r. If the
sphere is replaced by a cube of side r, will  E be larger, smaller, or the same?
Explain.
Solution
The electric flux depends only on the charge enclosed by the gaussian surface, not on the
shape of the surface.  E will be the same for the cube as for the sphere.
Problems:
25. (I) A downward force of 8.4 N is exerted on a 8.8 C charge. What are the
magnitude and direction of the electric field at this point?
Solution
Use Eq. 16–3 to calculate the electric field.

 F 8.4 N   ˆj
E 
 9.5 105 N / C ˆj - The electric field is directed up.
q  8.8 10 6 C
 

( ĵ - unit vector directed up).
34. (II) Calculate the electric field at one corner of a square 1.00 m on a side if the other
three corners are occupied by 2.25 10 6 C charges.
E3
Solution
The field at the upper right corner of the square is the
vector sum of the fields due to the other three charges.
Let the variable d represent the 1.0 m length of a side of
the square, and let the variable Q represent the charge at
each of the three occupied corners.
Q1
E2
E1
d
Q2
Q3
E1  k
E2  k
E3  k
Q
d
 E1x  k
2
Q
2d
Q
d
2
2
Q
, E1 y  0
d2
 E2 x  k
Q
2d
2
cos45o  k
2Q
4d
2
, E2 y  k
2Q
4d 2
Q
 E3 x  0 , E1 y  k
d2
Add the x and y components together to find the total electric field, noting that Ex  E y .
Ex  E1x  E2 x  E3 x  k
E  Ex2  E y2  k

Q
d
2
k
Ey
Ex
4d
2
0 k
Q
2
1 
  Ey
d 
4 
2
Q
2
Q
1
1 
 2k 2 2 
d 
4 
d 
2
2
 8.988  109 N  m 2 C 2
  tan 1
2Q

 2.25 10 C  
6
1.00 m 
2
1
4
 2    3.87  10 N C
2

 45o from the x-direction
41. (III) An electron (mass m  9.111031 kg ) is accelerated in the uniform field

E ( E  1.45104 N C) between two parallel charged plates. The separation of the
plates is 1.10 cm. The electron is accelerated from rest near the negative plate and
passes through a tiny hole in the positive plate, Fig. 16–60. (a) With what speed
does it leave the hole? (b) Show that the gravitational force can be ignored.
Solution We assume that gravity can be ignored, which is proven in part (b).
(a)
F  qE
W  Fd  qEd 
mv 2

2





v  2qEd / m  2 1.60 10 19 C 1.45 10 4 N / V 1.10 10 2 m / 9.11 10 31 kg  7.49 10 6 m / s
(b) The value of the acceleration caused by the electric field is compared to g.
1.602 10 C 1.45 10

 9.1110 kg 
19
a
qE
a
2.55  1015 m s 2
g

m
4
N C
31
9.80 m s
2
  2.55 10
15
m s2
 2.60  1014
The acceleration due to gravity can be ignored compared to the acceleration caused by
the electric field.
*43. (I) The total electric flux from a cubical box 28.0cm on a side is 1.45103 N  m 2 C .
What charge is enclosed by the box?
Solution
Use Gauss’s law to determine the enclosed charge.
E 
Qencl
o



 Qencl   E o  1.45  103 N  m 2 C 8.85  10 12 C 2 N  m 2  1.28  10 8 C
We assume that 1.45103 N  m 2 C . is the total flux from all six faces of the cub.
*50. (III) A point charge Q rests at the center of an uncharged thin spherical conducting
shell. (See Fig. 16–33.) What is the electric field E as a function of r (a) for r less
than the inner radius of the shell, (b) inside the shell, and (c) beyond the shell? (d)
Does the shell affect the field due to Q alone? Does the charge Q affect the shell?
Solution
(a) Inside the shell, the field is that of the point charge, E  k
Q
r2
.
(b) There is no field inside the conducting material: E  0 .
(c) Outside the shell, the field is that of the point charge, E  k
Q
.
r2
(d) The shell does not affect the field due to Q alone, except in the shell material, where the field
is 0. The charge Q does affect the shell – it polarizes it. There will be an induced charge of – Q
uniformly distributed over the inside surface of the shell, and an induced charge of +Q uniformly
distributed over the outside surface of the shell.
67. A point charge (m  1.0 g) at the end of an insulating string of length 55 cm is
observed to be in equilibrium in a uniform horizontal electric field of 12,000 N C ,
when the pendulum’s position is as shown in Fig. 16–66, with the charge 12 cm
above the lowest (vertical) position. If the field points to the right in Fig. 16–66,
determine the magnitude and sign of the point charge.
FT

43cm
mg

L  55cm
FE
Solution


FE  QE
tan  
552  432
2593

43
43
Since the electric field exerts a force on the charge in the same direction as the electric
field, the charge is positive. Use the free-body diagram to write the equilibrium
equations for both the horizontal and vertical directions, and use those equations to find
the magnitude of the charge.
FE  FT sin 
Fx  FE  FT sin   0

 FE  mg tan  
mg
Fy  FT cos   mg  0
FT 
cos


mg
1.0 10 3 kg 9.8m / s 2
Q
tan  
E
1.2 10 4 N / C

2593
 6.5 10 7 C  0.65C
43