Electric Charge
and
Coulomb’s Law
Fundamental Charge: The charge
on one electron.
e = 1.6 x 10
-19
C
Unit of charge is a Coulomb (C)
Two types of charge:
Positive Charge: A shortage of electrons.
Negative Charge: An excess of electrons.
Conservation of charge – The net charge of a
closed system remains constant.
Nucleus
-
-
n + n
+ +
n
+
n
n
+
+ n
-
-
-
-
Negative
Neutral Atom
Atom
Positive
Atom
Number
Numberof
ofelectrons
electrons><=Number
Numberof
ofprotons
protons
Number
of
electrons
Number
of
protons
-19
-2e
=
-3.2
x
10
CC
+2e = +3.2 x 10-19
Electric Forces
Like Charges - Repel
F
+
+
Unlike Charges - Attract
-
F
F
+
F
Coulomb’s Law – Gives the electric force
between two point charges.
q1q2
F k 2
r
Inverse Square
Law
k = Coulomb’s Constant = 9.0x109 Nm2/C2
q1 = charge on mass 1
q2 = charge on mass 2
r = the distance between the two charges
The electric force is much stronger than the
gravitational force.
Example 1
Two charges are separated by a distance r and have a force
F on each other.
qq
F k
F
1 2
2
r
q2
q1
F
r
If r is doubled then F is :
¼ of F
If q1 is doubled then F is :
2F
If q1 and q2 are doubled and r is halved then F is : 16F
Example 2
Two 40 gram masses each with a charge of 3μC are
placed 50cm apart. Compare the gravitational force
between the two masses to the electric force between the
two masses. (Ignore the force of the earth on the two
masses)
3μC
40g
3μC
40g
50cm
m1m2
Fg  G 2
r
 6.67 10
11
(.04)(.04)
2
(0.5)
 4.27 10
q1q2
FE  k 2
r
6
6
(
3

10
)(
3

10
)
9
 9.0 10
(0.5) 2
13
N
 0.324 N
The electric force is much greater than the
gravitational force
Example 3
Three charged objects are placed as shown. Find the net
force on the object with the charge of -4μC.
F k
- 5μC
45º
202  202  28cm
20cm
q1q2
r2
(5 106 )(4 106 )
F1  9 10
 4.5N
2
(0.20)
9
(5 106 )(4 106 )
F2  9 10
 2.30 N
2
(0.28)
9
F1 45º
- 4μC
5μC
20cm
F2
F1 and F2 must be added together as vectors.
F1
2.3cos45≈1.6
45º
F2
2.3sin45≈1.6
F1 = < - 4.5 , 0.0 >
+ F2 = < 1.6 , - 1.6 >
Fnet = < - 2.9 , - 1.6 >
- 1.6
- 2.9
29º
θ
3.31
Fnet  2.9 2  1.6 2  3.31N
  1.6 

  tan 

29

  2.9 
1
3.31N at 209º
Example 4
Two 8 gram, equally charged balls are suspended on earth
as shown in the diagram below. Find the charge on each
ball.
20º
L = 30cm
FE
q
10º 10º
L = 30cm
30sin10º
q
r
r =2(30sin10º)=10.4cm
2
q1q2
q
FE  k 2  k 2
r
r
FE
Draw a force diagram for one charge and treat as an
equilibrium problem.

T sin 80  .08
T
FE
q
.08
T
 .081N

sin 80
Tsin80º
80º
Tcos80º
Fg = .08N
FE  T cos 80
q2

k

(.
081
)
cos
80
.104 2
.014
2
2
q 
(.104)
k
q  1.3 10 7 C