Chapter 21: Electric Charge
Example Questions & Problems
e  1.602  1019 C
1
4
 k  8.99  109 N  m2 / C2
0
F
q1 q2
1
4
0
r2
rˆ
Example 21.1
a. Why does a plastic ruler that has been rubbed with a cloth have the ability to pick up
small pieces of paper?
b. When you pull transparent plastic tape off a roll and try to position it precisely on a
piece of paper, the tape often jumps over and sticks where it is not wanted. Why?
c. A balloon is negatively charged by rubbing and then clings to a wall. Does this mean
that the wall is positively charged? Why does the balloon eventually fall?
d. If you rub a coin briskly between your fingers, it will not become charged. Why not?
Example 21.2
Imagine the charges in 1 gram of hydrogen are separated so that all of the electrons and
all of the protons are separated by a distance equal to the earth’s diameter. What would
be the (i) force of attraction between the electrons and the protons and its (ii)
acceleration (compare this value to the acceleration due to gravity)?
(NA = 6.02 × 1023 atoms/mol ≈ e/gram)
Example 21.3
Earth’s atmosphere is constantly bombarded by cosmic ray protons that originate
somewhere in space. If the protons all passed through the atmosphere, each square
meter of Earth’s surface would intercept protons at the average rate of 1500 protons per
second. What would be the electric current intercepted by the total surface area of the
planet?
21-1
Example 21.4
The charges and coordinates of two charged particles held fixed are q1 = +3 C, (3.5
cm, 0.5 cm), and q2 = 4 C, (2.0 cm, 1.5 cm).
a. (i) Draw a diagram of the situation and (ii) determine the magnitude and direction of
the electrostatic force on particle 2 due to particle 1?
b. At what x and y coordinates should a third charge q3 = 4 C be placed such that
the net electrostatic force on particle 2 due to particles 1 and 3 is zero? Draw a
diagram of the situation.
Example 21.5
Particles 1 and 2 of charge q1 = q2 = 2e = 3.2  1019 C are on the
y axis at distance d = 17.0 cm from the origin. Particle 3 of charge
q3 = 4e = 6.4 × 1019 C is moved gradually along the x axis from x
= 0 to x = 5.0 m. At what values of x will the magnitude of the
electrostatic force on the third particle from the other two particles
be (a) minimum and (b) maximum? What are these (c) minimum
and (d) maximum magnitudes?
21-2
Example A
In the figure, what are the (a) magnitude and (b) direction of the net
electrostatic force on particle 4 due to the other three particles? All four
particles are fixed in the xy plane, and (q1, q2, q3, q4) = (3.20, +3.20,
+6.40, +3.20)  1019 C, 1 = 35.00, d1 = 3.00 cm, and d2 = d3 = 2.00 cm.
Solution
Summing the electric forces acting on charge 4 (according to Newton’s 2nd law), a freebody diagram (FBD) indicates that the net force should be in the third quadrant. Now to
determine the individual magnitudes and directions of the forces, we start with the force
between charges 1 & 4 is given by F14. The magnitude of the force F14 is

| q || q |
F14  k 1 4  8.99  109 N  m2 / C2
d1

3.20  10
19

C 3.20  1019 C
 0.03 m
2
  1.02  10
24
N
and the angle is 14 = 1800 + 350 = 2050. For forces F24 and F34, note that the magnitude
of force F34 is twice as big as F24 since q3 = 2q2. Following the same type of calculation
as for F14, the answers are
q3q4

 4.60  1024 N
F34  k
d3
F34  
  180
 34
q2q4

 2.30  1024 N
F24  k
d2
F24  
  270
 24
Summing the forces on charge 4 gives
Fx  F14x  F24x  F34x  F14 cos(205)  F34  5.4  1024 N
Fnet  (Fx , Fy )  
F  F14y  F24y  F34y  F14 sin(205)  F24  2.9  1024 N

 y
The net force acting on charge-4 is
Fnet 
 Fx 
2

 Fy

2

 5.4  10
24
N
   2.9  10
2
24
N

2
 6.2  1024 N  Fnet
 2.9  1024 N 
  tan1 
 208  
24 
 5.4  10 N 
Example B
In the figure, three identical conducting spheres initially have the
following charges: (QA, QB, QC) = (4Q, -6Q, 0). Spheres A and B are
fixed in place, with a center-to-center separation that is much larger
than the spheres. Two experiments are conducted. In experiment 1,
sphere C is touched to sphere A and then (separately) to sphere B,
and then it is removed. In experiment 2, starting with the same initial
states, the procedure is reversed: Sphere C is touched to sphere B
and then (separately) to sphere A, and then it is removed. What is the ratio of the
electrostatic force between A and B at the end of experiment 2 to that at the end of
experiment 1?
Solution
In experiment 1, sphere C first touches sphere A, and they divided up their total charge
(4Q plus 0) equally between them. Thus, sphere A and sphere C each acquired charge
2Q. Then, sphere C touches B and those spheres split up their total charge (2Q –6Q) so
21-3
that B ends up with charge equal to -2Q. The force of repulsion between A and B is
therefore
F1  k
(2Q)(2Q)
d2
at the end of experiment 1. Now, in experiment 2, sphere C first touches B which leaves
each of them with charge -3Q. When C next touches A, sphere A is left with charge Q/2.
Consequently, the force of repulsion between A and B is
F2  k
(3Q)( 21 Q)
d2
at the end of experiment 2. The ratio is
F2 (3  21 )

 0.875
F1 (2  2)
21-4