SPH4U
Notes
Unit 7.2 Electric Forces: Coulomb’s Law
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Physics Tool box

Coulomb’s Law –The magnitude of the electric force between two point
charges is directly proportional to the product of the charges and inversely
proportional to the square of the distance between them.

F k

k is the Coulomb proportionality Constant

k

Coulomb’s Law applies when the charges on the two spheres are very small,
and the two spheres are small compared to the distance between them.
There are similarities between Coulomb’s Law and Newton’s Law of universal
Gravitation:
o Both are inverse squares laws that are proportional to the product of
quantities that characterize the body involved
o The gravitational force only attracts, while the electric force and
attract or repel.
o The universal gravitational constant is very small while the Coulomb
constant is very large.

q1q2
.
r2
1
N  m2
.
 8.988 109
4 0
C2
o
FE 
kq1q2
,
r2
Fg 
Gm1m2
r2
The electric force between two point charges also depends on the quantity of charge on each body,
which is denoted by q. Coulomb found that the forces that two point charges q1 and q2 exert on each
other are proportional to each charge and therefore proportional to the product q1q2 of the two
charges.
Coulomb’s Law –The magnitude of the electric force between two point charges is directly
proportional to the product of the charges and inversely proportional to
the square of the distance between them.
F k
q1q2
r2
Where k is the Coulomb Proportionality Constant whose numerical value depends on the
system of units used. The absolute value bars are used because the charges q1 and q2 can
be either positive or negative, while the force magnitude F is always positive.
k
1
N  m2
 8.988 109
4 0
C2
0 (epsilon-nought) is permittivity of free space = 8.854 1012
C2
N  m2
SPH4U
Unit 7.2 Electric Forces: Coulomb’s Law
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Plastic rods and fur (real or fake) are typically used to demonstrate electrostatics.
We now that there are exactly two types of charges negative and positive
The Laws of Electric Charges – Two positive charges or two negative charges repel each other. A
positive charge and a negative charge attract each other
Example
The magnitude of the electrostatic force between two point like charged objects is 4.0 106 N .
Calculate the force for each of the following situations:
a) The distance between objects is doubled
b) Distance remains same but charge on one object tripled and charge on other is cut by 5.
c) Both a and b occur.
Solution:
kq1q2
2
 r1 
F2
r22

 
a)
F1 kq1q2  r2 
r12
r 
F2  F1  1 
 r2 
2
1
  4.0 10 N   
2
 1.0  106 N
6
2
SPH4U
Unit 7.2 Electric Forces: Coulomb’s Law
Page 3 of 6
The magnitude of the force is now 1.0 106 N
1
k 3q1 q2
5
2
F
3
r2
b) 2 

kq1q2
F1
5
2
r1
3
5
F2  F1
3
  4.0 106 N   
5
7
 2.4 10 N
The magnitude of the force is now 2.4 107 N
1
k 3q1 q2
5
2
F2
3 1 
r22
c)

  
kq1q2
F1
5 2
2
r1
F2  F1
3
20
 3 
  4.0 106 N   
 20 
 6.0 107 N
The magnitude of the force is now 6.0 107 N
Example
Three objects with charges A: 6.0C , B: 5.0C , and C: 4.0C are placed in a line.
Determine the net electric force on charge A.
A
+
Solution:
0.70m
The magnitude of the force exerted on A by C
FAC 
kq AqC
2
rAC

m2 
9
9.0

10
N

6.0 106 C  4.0 106 C 

2 
C


2
  0.70  0.30  m 
 0.216 N
B
C
-
+
0.30m
SPH4U
Unit 7.2 Electric Forces: Coulomb’s Law
Therefore the force on A by C is 0.216 N [left]
The magnitude of the force exerted on A by B
FAB 
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Charges are same sign so
the forces repels.
kq AqB
2
rAB

m2 
9
9.0

10
N

6.0 106 C  5.0 106 C 

2 
C 

2
 0.7m 
 0.5510 N
Charges are different
sign so the forces attract.
Therefore the force on A by B is 0.551N [right]
So the net force on A is :
F  F
AC
 F AB
 0.216 N [left ]  0.551N [right ]
 0.33N [right ]
Example
A neutral metal sphere A (0.10kg), hangs from an 2.0 m insulating wire. An identical
sphere B, with charge –q, is brought into contact with sphere A. The spheres repel each
other so that the angle of the wire is 12 . What was the initial charge on B
12
B
A
Solution:
First we must determine what the forces are that holds a 0.10 kg sphere at an angle of
12 .
So let’s draw a force diagram
FT
12
Now T cos 12   mg
FE
T sin 12   FE
So we can solve for
FE
Fg
SPH4U
Unit 7.2 Electric Forces: Coulomb’s Law
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FE sin 12 

mg cos 12 
FE  mg tan 12 

N
  0.10kg   9.8   tan 12  
kg 

1
 2.08 10 N
Before we can apply
F k
q1q2
, we need the distance (r value)
r2
r
 sin 12 
2.0m
r  2.0m sin 12 
12
 0.416m
2.0m
r
The charge on each sphere after touching are now identical so
original charge q on B
Now,
 1  1 
 q  q 
2  2 
FE  k 
r2
4 FE r 2
2
q 
k
q

4 FE r 2
k
4  2.0 101 N   0.416m 
9.0 109 N 
 3.9 106 C
6
The initial charge on B is 3.9 10 C
m2
C2
2
q1  q2 , therefore half of the
SPH4U
Unit 7.2 Electric Forces: Coulomb’s Law
Extra Notes and Comments
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