```Unit I - Worksheet 3: Coulomb's Law Key
q1q2
2 , where
r
, describe in words the relationship among electric force, charge, and
1. Given the mathematical representation of Coulomb’s Law, F  k
2
9
k  9.0 10 Nm
2
C
distance.
The electric force is proportional to the product of the charges and is inversely proportional to
the square of the distance between the charges.
2. By how much does the electric force between a pair of charged bodies diminish when their
separation is doubled? tripled?
k
k
k
F2 
 2
2
2
r
(2r)
4r
The force is ¼th as much.
F1 
F2  41 F1
k
k
k
F2 
 2 F2  91 F1
2
2
r
(3r)
9r
The force is 1/9th as much.
F1 
3. Two positive charges of 6.0 x 10-6 C are separated by 0.50 m. Draw a force diagram for each of
the charges, considering only electrostatic forces. What is the magnitude of the force between
the charges? Is this force repulsive or attractive?
Fe
+6
+6
Fe
9.0  10
qq
F  k 12 2 F 
r
9 Nm2
C2
(6.0  10 -6 C)(6.0  10 -6 C)
(.50m) 2
 1.3N
Like charge repells, so the force is repulsive.
4. A negative charge of 2.0 x 10-4 C and a positive charge of 8.0 x 10-4 C are separated by 0.30 m.
What is the magnitude of the force between the charges? Is this force repulsive or attractive?
9.0  10
qq
F  k 12 2 F 
r
9 Nm2
C2
(-2.0  10 -4 C)(8.0  10 -4 C)
(.30m) 2
 1600N
Opposite charges attract, so the force is attractive.
5. A young man accumulates a charge q1 of +2.0 x 10-5 C while sliding out of the front seat of a
car. His girlfriend, who had been waiting in the wind, has picked up some extra electrons and
now has a charge q2 of -8.0 x 10-5 C.
Draw a sketch of the situation. Estimate the magnitude of the electrical force that each person
exerts on the other when separated by a distance of 6.0 m. Is the force attractive or repulsive?
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Fe Fe
B
9.0  10
qq
F  k 12 2 F 
r
G
9 Nm2
C2
(2.0  10 -5 C)(-8.0  10 -5 C)
(6.0m) 2
 .40N
Opposite charges attract, so the force is attractive.
6.
Suppose the two people in the previous problem move toward each other. Calculate the
magnitude of the electrical force of one on the other when their separation is reduced by a factor of
10.
9.0  10
qq
F  k 12 2 F 
r
Or F1 
k
r2
F2 
k

( 101 r) 2
k
2
1
100 r
9 Nm2
C2
(2.0  10 -5 C)(-8.0  10 -5 C)
(.6m) 2
 40N
F2  100F1 , so the force is 100 times larger.
7. The figure below shows three point charges that lie along the x axis. Determine the magnitude
and direction of the net electrostatic force on charge q1.
F3on1
9.0  10
qq
 k 12 2 F 
r
F2on1
9.0  10
qq
 k 12 2 F 
r
9 Nm2
C2
9 Nm2
C2
(-7.0  10 -6 C)(6.0  10 -6 C)
(.15m) 2
(-4.0  10 -6 C)(6.0  10 -6 C)
(.2m) 2
 16.8N to the right
 5.4N to the left
Making to the right positive we add -5.4N and 16.8N to get 11.4N to the right.
7. Two pith balls shown in the diagram below have a mass of 1.0 g each and have equal charges.
One pith ball is suspended by an insulating thread. The other charge is brought to within 3.0 cm.
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of the suspended ball (r = 0.03 m). The suspended pith ball is deflected from its rest position
until the thread forms an angle of 30˚ with the vertical. At this angle, the ball is in equilibrium.
a) Draw a force diagram depicting the forces acting on the suspended ball.
Calculate (b) mg
(c) FE
(d) the charge on the pith balls
FT
30o
30
Fe
Fg
0.03 m
b) Fg = mg =.001kg x 9.8N/kg = .0098N or .001kg x 10 N/kg = .010N
c) Since the system is not moving we know the gravitational force is equal to the vertical component
of the tension and the electric force is equal to the horizontal component of the tension.
Therefore the triangle below can be drawn. We know the gravitational force and the angle.
Fg
FT
tan30  
Fe
, Fe  tan30  (.0098N)  .0057N or .0058N if g  10N/kg
Fg
Fe
qq
F r2
(.0058N)(. 030m) 2
 5.8  10 -16 C 2 , q  2.4  10 -8 C
d) F  k 1 2 2 , q2  e 
9 Nm
k
r
9.0

10
9. Three charges are placed as shown below.
Determine the magnitude and direction of the net
C
electrostatic force on charge q1. As part of the solution, include a force diagram.
2
2
9.0  10 9 Nm
(1.8  10 -5 C)(4.5  10 -6 C)
q1q 2
C2
k 2 F
 .405N to right and down
r
( (3m) 2  (3m) 2 ) 2
2
F3on1