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Physics 112
Homework 2 (Ch16)
Due June 18
Questions:
1. 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.
2. Consider a small positive test charge located on an electric field line at some point, such as
point “P” in figure below. 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.
3. 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.
Physics 112
Homework 2 (Ch16)
Due June 18
4. 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:
5. 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
 

 F 8.4 N   ˆj
By definition: E  
 9.5 105 N / C ˆj
q  8.8 10 6 C

The electric field is directed up. ( ĵ - unit vector directed up).
6. 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.
Solution
Method1: 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.
E1  k
E2  k
E3  k
Q
d
 E1x  k
2
Q
2d
Q
2
d2
Q
d
2
 E2 x  k
E3
, E1 y  0
Q
2d
cos45  k
2Q
o
2
 E3 x  0 , E1 y  k
4d
2
, E2 y  k
2Q
4d
Q1
E  Ex2  E y2  k

Q
d
d
Q
d2
Q2
  tan 1
Ey
k
2Q
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  10 N  m C
9
2
2
2

E1
2
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
E2
 2.25 10 C  
6
1.00 m 
2
 45o from the x-direction.
1
4
 2    3.87  10 N C
2

Ex
Method2: Use symmetry. See similar problem in the Lecture 1 (slide 10).
Q3
Physics 112
Homework 2 (Ch16)
Due June 18
7. 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.
8. A point charge Q rests at the center of an uncharged thin spherical conducting shell. (See
figure below) 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.
9. 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 the figure below, with the charge 12 cm above the lowest (vertical)
position. If the field points to the right in the figure below, determine the magnitude and sign
of the point charge.
Physics 112
Homework 2 (Ch16)
Due June 18
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

F

F
cos


mg

0
F

y
T
T


cos 
Q


mg
1.0 10 3 kg 9.8m / s 2
tan  
E
1.2 10 4 N / C

2593
 6.5 10 7 C  0.65C
43