Chapter 21: Electric Charge
Example Questions & Problems
e  1.602  1019 C
1
4ò0
 k  8.99  109 N  m2 / C2
F
1
q1 q2
4ò0
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. Find 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 particle of charge q3 = 4 C be placed
such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?
Solution
In your thinking, try to picture and be as specific as possible in looking at the
force between these two charges. Now you will need to translate this into a
vector picture onto paper. What force does charge q2 feel from q1? That is, tell
me something about the vector force; i.e. what is its magnitude and the angle of
force F12. From Coulomb’s law, we know that the force acts along straight lines and
q2 should feel a force pulling it towards the center of q1. In this case, the force is to the
right and downward below the horizontal axis.
2
The electric force between the two charges is F12  (kq1q2 / r12 ) rˆ12 where r12 is the
distance between the two charges and a unit vector in the direction of r12. Since this
force acts along a straight line (from one CM to the other), the direction of F12 is also the
same direction as r12. So the important thing we need to determine r12  r1  r2 , which then
allows us to get the magnitude and direction of the electric force.
a. The distance between these two points can either be thought of as a
displacement vector or simply vector subtraction. The coordinates of the two
charges are r1 = (0.035, 0.005) m and r2 = (0.02, 0.015) m. Using vector
subtraction, the magnitude of r12 is r12
r12  r1  r2 
x
1
 x 2    y1  y 2  
2
2
 0.020  0.035    0.015  0.005 
2
2
 0.056 m
The magnitude of the force exerted by q1 on q2 is
F12  k
| q1q2 |
 8.99 10  3.0 10  4.0 10   35 N  F

9
6
6
12
r122
(0.056)2
The angle of r12 is directed towards q1 and makes an angle 12 with the +x axis, where
 y1  y 2 
1  1.5  0.5 
  tan 
  10.3  10  12
 2.0  3.5 
 x1  x 2 
12  tan1 
b. Question: Where should the third charge be placed so that the net force
on charge-3 is zero? Since there is already one force acting on q2, the
third charge should clearly be equal and opposite to F12, which I will call
F32. That is, means, F12 = F32 and the direction differs by 180o:
F32  F12  35N and 32  180  10  170
What conditions does this equilibrium imply?
qq
qq
q
4C
F32  F12 
 k 22 3  k 12 2 
 r32  r12 3  0.056
 0.0645m
q1
3C
r32
r12
I now know both the magnitude and direction of the vector r32, which means that I am
able to determine the components of r3. From vector addition (see the
diagram on the right),
r3x  r2x  r32 cos170
r3  r2  r32  
r3y  r2y  r32 sin170
Consequently,
21-2
 x3  x 2  r32cos32  –2.0 cm  6.45 cm  cos 170   –8.4 cm  x3



 y3  y 2  r32 sin32  1.5 cm  6.45 cm  sin 170   2.7 cm  y 3
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 = 6e = 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?
Solution
Are there an symmetric charges? Yes – the two charges are symmetric about
the x-axis, so all of the y-components will sum to zero, regardless of charge3’s location. Since there is no symmetry about the y-axis, the x-components
of these two forces will not cancel out. Applying Newton’s 2nd law, we get
 Fx  F1x  F2x  2F1 cos 
F
2
 F1y  F2y  0
The distance from the charges (1 & 2) to charge 3 is x. Using right triangle
ideas to find the angle , we get
x
cos  
2
x  d2
Putting this all together, we have
Fnet  2Fcos  
2  q1  q3
40  (x 2  d2 )
x
2
x d
2

2
q1  q2  2e
q3  4e
2e  4e  x
0  (x 2  d2 )3/ 2
(a, b & c)
We are interested in determining minimum and maximum force locations. Since you will
find that this course is fairly mathematical, it is extremely important to connect physical
intuition with mathematical intuition. So let’s interpret the mathematics.
 Physically, where will a minimum occur when charge-3 is moved anywhere along the
x-axis? Since the y-components always cancel out, we only need to focus on finding
out where the x-components cancel out. There are two locations, at the origin and at
x = ∞. Why are these zero? Does your mathematical derivation of the force agree
with this? Yes.
Fnet 

2e  4e  x
0  (x 2  d2 )3/2
x 0

0
x 
Physically, where will the maximum occur? This is much harder to see this because
there has to be a sweet spot between the inverse square law behavior and the sum
of the x-components of the two forces. However, mathematically, this is well defined
– we take the derivative wrt x and set it to zero, then we find that critical point.
To find the maximum, we set the derivative of the force to zero and finding the critical
point:
21-3

dF d 
8e 2 x

0

2
2 3/2 
dx dx  40  (x  d ) 

d 
x
  1  (x 2  d2 )3/2  x  32 (x 2  d2 )1/2  2x 

0

dx  (x 2  d2 )3/2  
(x 2  d2 )3

 (x 2  d2 )3/2  3x 2  (x 2  d2 )1/2
 (x 2  d2 )2  3x 2

x
1
2
d
1
2
 17cm  12 cm
Focusing on the numerator to find the critical point, in order for this to be zero the
numerator must be zero:
 (x 2  d2 )3/2  3x 2  (x 2  d2 )1/2 
(x 2  d2 )3/2  3x 2  (x 2  d2 )1/2

  0 
2
2 3
(x

d
)


(x 2  d2 )2  3x 2 or
x
1
2
d
1
2
 17cm  12 cm
d. The value of the net force at x = d/ 2 is Fnet = 4.9  1026 N.
Fmax 
4
4e 2 x
  8.99  10 N  m / C
2
9
4 0  (x 2  d2 )3/ 2
2


16  1.6  10 19 C
   0.12 m 
2
(0.122  0.17 2 )3/ 2  m3/2
Fmax  4.9  10 26 N
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
q2q4

 2.30  10 24 N
F24  k
d2
F24  
  270
 24
q3 q4

 4.60  10 24 N
F34  k
d3
F34  
  180
 34
Summing the forces on charge 4 gives
21-4
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
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)
21-5