3.1.2 In Class or Homework Exercise
1. How many excess electrons are on a ball with a charge of 4.00 1017 C?
q  4.00 1017 C
e  1.60 1019 C / electron
4.00 1017
 250 electrons
1.60 1019 C / electron
2. Two electrons in an atom are separated by 1.5 1010 m, the typical size of an
atom. What is the force between them?
r  1.5 1010 m
q1  1.60 1019 C
q2  1.60 1019 C
k  8.99 109 Nm 2 / C 2
Fe  ?
kq1q2
r2
(8.99 109 )(1.60 1019 ) 2

(1.5 1010 ) 2
Fe 
 1.0 108 N
Since the electrons are both negatively charged, the force is repulsive.
3. Two charged bodies exert a force of 0.145 N on each other. If they are moved
so that they are one fourth as far apart, what force is exerted?
F1  0.145N
F1 
kq1q2
r12
F2 
kq1q2
r2 2

kq1q2
( 14 r1 ) 2
 16
kq1q2
r12
 16 F1
 16(0.145)
 2.32 N
4. Two charges, q1 and q2 , are separated by a distance d and exert a force F .
What new force will exist if
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a.
q1 is doubled?
kq q
F  12 2
d
k 2q1q2
d2
kq q
 2 12 2
d
 2F
F2 
b.
q1 and q2 are cut in half?
kq q
F  12 2
d
k 12 q1 12 q2
d2
1 kq1q2

4 d2
F2 

c.
1
4
F
d is tripled?
kq q
F  12 2
d
kq1q2
(3d ) 2
1 kq1q2

9 d2
F2 

d.
1
9
F
d is cut in half?
kq q
F  12 2
d
F2 
kq1q2
( 12 d ) 2
kq1q2
d2
 4F
4
e.
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q1 is tripled and d is doubled?
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F
kq1q2
d2
k 3q1q2
(2d ) 2
3 kq1q2

4 d2
F2 

3
4
F
5. The hydrogen atom contains a proton, mass 1.67 1027 kg, and an electron,
mass 9.111031 kg. What is the ratio of the magnitude of the average
electrostatic force of attraction between them to the gravitational force of
attraction between them?
q p  1.60 1019 C
qe  1.60 1019 C
m p  1.67 1027 kg
me  9.111031 kg
k  8.99 109 Nm 2 / C 2
G  6.67 1011 Nm 2 / kg 2
kqe q p
2
Fe
 r
Fg Gme m p
r2
kqe q p

Gme m p

(8.99 109 )(1.60 1019 ) 2
(6.67 1011 )(9.1110 31 )(1.67 10 27 )
 2.27 1039
So the electric force between an electron and a proton is 2.27 1039 times
larger than the gravitational force between them. In general, electric forces are
much larger than gravitational forces. In fact, gravitational forces are usually
only significant when at least one of the objects is extremely large, such as a
planet.
6. Two electrons are “arranged” so that one is above the other. The bottom
electron is resting on a table. How high will the second electron ``float'' above
this bottom electron? In other words, at what height will the electrical force of
repulsion be equal and opposite to the gravitational force of attraction of the
earth?
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me  9.111031 kg
qe  1.60 1019 C
g  9.80m / s 2
k  8.99 109 Nm 2 / C 2
r ?
Drawing a free body diagram for the top electron we get
As can be seen in the free body diagram, the two forces must have equal
magnitudes. Notice that we can ignore the force of gravitational attraction
between the two electrons since it is so small (as shown in the previous
problem for an electron and a proton)
Fe  Fg
kqe qe
 mg
r2
(8.99 109 )(1.601019 ) 2
 (9.111031 )(9.80)
2
r
r  5.08m
7. Three particles are placed in a line. The left particle has a charge of -67 C ,
the middle +45 C , and the right -83 C . The middle particle is 72 cm from
each of the others.
a. Find the net force on the middle particle.
q A  67  C  6.7  105 C
qB  45 C  4.5 105 C
qC  83 C  8.3 105 C
rAB  72cm  0.72m
rBC  72cm  0.72m
FnetB  ?
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FAB 
kq A qB
rAB 2
(8.99 109 )(6.7 105 )(4.5 105 )
(0.72) 2
 52 N

FBC 
kqB qC
rBC 2
(8.99 109 )(4.5 105 )(8.3 105 )
(0.72) 2
 65 N

Since A is oppositely charged to B, it will be attracting B to the left.
Since C is oppositely charged to B, it will be attracting it to the right.
Using the right as positive,
Fnet  FAB  FBC
 52  65
 13 N
Since the net force is positive, it is to the right.
b. Find the net force on the right particle.
q A  67  C  6.7  105 C
qB  45 C  4.5 105 C
qC  83 C  8.3 105 C
rAC  144cm  1.44m
rBC  72cm  0.72m
FnetC  ?
FAC 
kq A qC
rAC 2
(8.99 109 )(6.7 105 )(8.3 105 )
(1.44) 2
 24 N

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FBC 
kqB qC
rBC 2
(8.99 109 )(4.5 105 )(8.3 105 )
(0.72) 2
 65 N

Since A is similarly charged to C, it will be repelling C to the right. Since
B is oppositely charged to C, it will be attracting it to the left.
Using the right as positive,
Fnet  FBC  FAC
 65  24
 41N
Since the net force is negative, it is to the left.
8. A positive charge of 3.0 C is pulled on by two negative charges. One, -2.0
C is 0.050 m to the north and the other, -4.0 C , is 0.030 m to the east.
What total force is exerted on the positive charge?
q A  2.0  C  2.0 106 C
qB  3.0  C  3.0 106 C
qC  4.0  C  4.0 106 C
rAB  0.050m
rBC  0.030m
FnetB  ?
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FAB 
kq A qB
rAB 2
(8.99 109 )(2.0 106 )(3.0 106 )
(0.050)2
 22 N

FBC 
kqB qC
rBC 2
(8.99 109 )(3.0 106 )(4.0 106 )
(0.030)2
 120 N

Since A and B are oppositely charged, A will pull B to the north; since B and C
are oppositely charged, C will pull B to the east
Fnet  FAB  FBC
tan  
Fnet  F  F
2
AB
2
BC
FAB
FBC
22
120
  10.
 (22) 2  (120)2

 120 N
Fnet  120 N ,10. N of E
9. A charged ball has a charge of +16 C . A second ball, located 16 cm to the
east, has a charge of -20. C . A third ball, located 25 cm north of the second
ball, has a charge of +25 C . What is the total force which acts on the first
ball?
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q A  16  C  1.6 105 C
qB  20 C  2.0 105 C
qC  25 C  2.5 105 C
rAB  16cm  0.16m
FnetA  ?
FAB 
kq A qB
rAB 2
(8.99 109 )(1.6 105 )(2.0 105 )
(0.16) 2
 112 N

2
2
rAC  rAB
 rBC
 (0.16)  (0.25)
2
25
16
  57
tan  
2
 0.30m
FAC 
kq A qC
rAC 2
(8.99 109 )(1.6 105 )(2.5 105 )

(0.30) 2
 40.N
Since A and B are oppositely charged, B will pull A to the east; since A and C
are similarly charged, C will push A to the south and west
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Fnet  FAB  FBC
FACx  FAC cos 
 (40.) cos 57
 22 N
Fnetx  FAB  FACx
 112  22
 90.N
Fnet  F
2
netx
FACy  FAC sin 
 (40.)sin 57
 34 N
Fnety  FACy
 34 N
F
2
nety
 (90.)2  (34) 2
 96 N
tan  
Fnetx
Fnety
90.
34
  69

Fnet  96 N , 69 E of S
10. In one model of the hydrogen atom, the electron revolves in a circular orbit
around the proton with a speed of 1.1106 m/s. What is the radius of the
electron's orbit?
qe  1.60 1019 C
me  9.111031 kg
v  1.1106 m / s
r ?
Since the electron is revolving in a circle, there must be a centripetal force. In
this case, the centripetal force will be supplied by the attractive electrical force
between the electron and the proton
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Fc  mac
Fe  me ac
kqe q p
r2
v2
 me
r
(8.99 109 )(1.60 1019 ) 2
 (9.111031 )(1.1106 ) 2
r
r  2.11010 m
11. The two pith balls in the diagram below each have a mass of 1.0 g and equal
charges. One pith ball is suspended by an insulating thread. The other is
brought to 3.0 cm from the suspended ball. The suspended ball is now
hanging (in equilibrium) with the thread forming an angle of 30.0° with the
vertical.
30.0o
3.0 cm
a. Draw a free body diagram for the pith ball on the thread.
b. Calculate the charge on each ball.
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m  1.0 g  0.0010kg
Fg  mg
g  9.80m / s 2
 (0.0010)(9.80)
 0.0098 N
k  8.99 109 Nm 2 / C 2
r  3.0cm  0.030m
q?
Since the pith ball is stationary,
FT  Fg  Fe  0
tan  
Fe
Fg
Fe
0.0098
Fe  5.66 103 N
tan 30.0 
kq1q2
r2
(8.99 109 )q 2
5.66 103 
(0.030) 2
Fe 
q  2.4 108 C
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