Physics II (SHE1225)
Chapter 4.
Electric Charge and Electric Field
Example of electric charge in application:
a. Lightning –which is the flow of electric charge in the atmosphere, is
Allah most spectacular display of electricity in nature.
b. Rubbing a plastic sheet- electrostatic charge (charging by friction)
c. Walking on a carpet on a dry day – charging by friction
d. Photostat machine and laser printing machine
e. Inkjet Printer
f. Lighting rod (Invented by Benjamin Franklin to avoid damage to
building due to lightning strike)
g. Electrostatic dust collector (use electrostatic effect to collect dust)
h. Chimney purifier and many more (use electrostatic effect to collect dust
particle
i.
Atomic Force Microscope
j.
Van de Graff Generator
Electric Charge
Like mass, electric charge is a fundamental property of matter. The electron is
considered as orbiting a nucleus, a core containing most of the atom’s mass
in the form of proton and electrically neutral particle called neutrons.
Particle
Electric Charge
Mass
Electron
-1.602 x 10-19
9.109 x 10-31 kg
Proton
+1.602 x 10-19
1.672 x 10-27 kg
0
1.674 x 10-27 kg
Neutron
The electric charge of an object is an integral multiple of fundamental
charges:
q=ne
n is an integer (This arise from quantum mechanic)
SI unit for charge is Coulomb (Charles Coulomb)
Electron comes from a Greek word of Amber (600BC). Electrostatic charge
effect was observed when amber is rubbed using cloth or paper.
The earliest investigator was an Englishmen William Gilbert (1544-1603)
From the Law of conservation of charge – Charge can neither be created nor
destroyed.
Method of charging.
a. Charging by friction – The transfer of charge due to contact. (Glass
(+ve) Amber (-ve)
b. Charge by contact or conduction – excessive charge will flow to the
material
c. Charge by induction – the charge is induced (no contact)
d. Charge by polarization
Coulomb Law:
The law was discovered by Charles Coulomb (1736-1806) A French engineer
and physicist. He used a torsion balanced to prove the theory.
+q
+q
He found out that the electrostatic force depends on the product of charges
and inverse to square distance
Fq
F  1/r2
and
so
F k
q1 q 2
r2
where k is a constant (9.09 x 109 Nm2/C2)
The force is directed along the line joining the two forces
Notation: (Please remember)
F1,2 = The force exerted on charge q1 to q2
F2,1 = The force exerted on charge q2 by q1
After extended experiment
k=
1
4 o
where o is permitivity of free space.
The value of
o =
1
C2
 8.854 x10 12
4k
Nm 2
So Coulomb Law can be written
F=
1
q 1 q2
4 o r 2
1 e = 1.602 x 10 –19 C
1C = 6.24 x 10 18 e-
SUPERPOSITION OF ELECTRIC CHARGES
The principle of superposition of electrical forces states that the combined
effect of many forces acting simultaneously is the same as the vector some of
the individual forces:
Resultant force  F =  F1 + F2 + F3 …….FN
However, remember that Force is a vector. The vector property of the force
has to be taken into consideration.
Examples:
1. Two point charges of magnitude 3.0 x 10–9 C and 6.0 x 10–9 C are
separated by a distance of 0.30 m. Find the electric force of repulsion
between them.
Ans:
F=
kq1q2
(3.0  10–9 C)(6.0  10–9 C)
= (8.99  109 Nm2/C2)
= 1.8  10–6 N
2
r
(3.0  10–1 m)2
2. The total charge on two spheres is 600 µC. The two spheres are
placed 0.900 m apart and the force of repulsion between the two is
30.0 N. What is the charge on each sphere?
Ans:
We are given that Q1 + Q2 = 6.00  10–4 C
Also, we know F = 30.0 N =
(1)
kQ1Q2
r2
From which, we find
Q1Q2 =
(30.0 N)(0.900 m)2
= 2.70  10–9 C2
(8.99  109 Nm2/C2)
(2)
We solve (1) and (2) simultaneously to find
Q1 = 4.53  10–6 C
and
Q2 = 5.95  10–4 C
3. Two point charges of equal magnitude repel each other with a force of
2.00 N when separated by 5.00 cm. Find the magnitude of the charge
on each
Ans:
Let us call Q the charge on each. The force of repulsion between them is F =
Q=
Fr2
=
k
kQ2
. From which,
r2
(2.00 N)(5.00  10–2 m)2
= 7.45  10–7 C
(8.99  109 Nm2/C2)
The Electric Field
The Electric Field E at any point in space is define as the electric force F per
unit charge exerted on a small positive test charge qo place at that point.
E=
F kQ

qo r 2
Where E= Electric Field F= Electric Force
+
Superposition of Electric Field
The electric field resulting from several point charges is just the superposition
of their individual fields:
n
F=F1 +F2 +F3….
F =
F
i 1
i
So E = E1 +E2 +E3 +E4……EN
In other words, the electric field resulting from several point charges is just the
superposition of their individual fields.
Examples:
1. The electric force on a point charge of 5.0 x 10–9 C at some point is
3.8 x 10–3 N in the positive x direction. What is the magnitude of the
electric field at this location?
Ans:
E=
F 3.8  10–3 N
=
= 7.6  105 N/C (in +x direction)
q 5.0  10–9 C
2. The magnitude of the electric field at a certain location is 500 N/C and
the field is directed east to west. (a) Find the magnitude and direction
of the force acting on a proton placed at this point. (b) Repeat for an
electron located at the point.
Ans;
(a) F = qE = (1.60  10–19 C)(500 N/C) = 8.00  10–17 N (force is in westward direction)
(b) The magnitude of the force is the same, but the direction is eastward for the electron.
3. A proton is shot vertically upward with a speed of 2.00 x 10 5 m/s in a
downward-directed electric field of 500 N/C. How high will it rise?
Ans:
The force on the proton is downward, and its magnitude is
F = qE = (1.60  10–19 C)(500 N/C) = 8.00  10–17 N
and the acceleration is also downward of magnitude
a=
F
8.00  10–17 N
=
= 4.79  1010 m/s2
m 1.67  10–27 kg
To find the maximum height, H, reached, we use
vy2 = voy2 + 2ay
0 = (2.00  105 m/s)2 – 2(4.79  1010 m/s2)H
H = 4.18  10–1 m = 41.8 cm
Electric Flux and Gauss Law
The Gauss law
The net amount of electric flux that passes through a closed surface is directly
related to the amount of electric charge inside the surface.
Flux – is the rate at which something passes through a surface
Exp.: the rate at which water moves through a cross section of a pipe may be
described as flux.
Electric Flux:
Number of Electric Field lines that passes through a given surface.
Gauss law for electrostatic.
The net amount of electric flux through any (real or imaginary) closed surface
is directly proportional to the net electric charge enclosed within that surface.
At equilibrium, any access charge placed on a conductor resides entirely on
its surface.
E=0
A quantitative approached to Gauss law
The electric flux E through a small portion of surface area A is the product
of the magnitude of the Electric field |E|, the magnitude of the surface area
A, and the cosine of the angle  between the direction of the field and the
direction of the normal
E =EA cos 
The flux through a surface depends on the amount of charge Q within the
surface. The amount of charge outside the surface has no effect on the flux.
E =
Q
