Chap 1: Positive charge

ELECTRIC force

and

electric field

JQB 22903

(ENGINEERING SCIENCE)

Bell Ringer Activity

• Using only the PVC, move

the soda can, but you
cannot touch the can with
the PVC or blow on the
can.

• Explain how you did it.

1.1 INTRODUCTION

What is electrostatics?

•

Electrostatics is the study of electrical charges
in static condition.

•

It involves;

i.

The influence of a charge to other
charges,

ii.

The forces between the charges,

iii. The area of influence (electric field)

and energy produced and loss in
certain operations.

Simple model of the atom.

• In the middle of each atom is a

"nucleus."

• The nucleus contains two kinds of tiny

particles,

called

"protons"

and

"neutrons".

• Orbiting around the nucleus are even

smaller particles called "electrons".

Protons  positive (+) charge.

Electrons  negative (-) charge.

Neutrons no charge, (neutral).

Draw

Draw a positive charge.

1.2 CURRENT AND ELECTRON

MOVEMENT

The important characteristic of charge is that electric

charge is always conserved.

LAW OF CHARGES : Like charges repel each other

Unlike charges attract each others

+

-

Repulsive force

Attractive force

+

-

-

+

Draw

Draw a positive charge.

Conservation of Charge

•No case of the creation or destruction of net
electric charge has ever been found

•Electrons are always transferred in whole –
they cannot be divided into fractions of
electrons

Electrons are never created nor

destroyed, but are simply transferred

from one material to another

Example

• Initially, when two objects (rod and silk) are charged by

being rubbed together, charge is not created in the process.

• The objects become charged because negative charge is

transferred from one object to another.

• When a glass rod is rubbed with silk, some electrons from rod

are transferred to the silk; leaving behind the same amount
of positive charges.

INSULATORS AND CONDUCTORS

• Substance can be classified according to their ability to

conduct electric charges

• Conductors : Electron in conductors can move freely between

atoms (metal, copper)

• Insulator : Electron in insulators are bound to the atoms and

have very low chance to move (eg: rubber, glass, plastic)

• Semiconductors : materials between conductors and insulator

in terms of their ability to conduct electric charges (eg:
Silicon, GaAs)

Figure shows the resistivity and the conductivity of conductor, semiconductor and

insulator.

1.3 Coulomb’s law

(Pronounced KOO-lum)

• There must be forces between the two charges. (Repulsive or

attractive force).

• This law can be said as:
‘If there are two point charges or spherically distributed charges

q1 and q2 at a distance r apart there will be an attractive
force (unlike charges) or a repulsive force (like charges) on
each charge or on the sphere of charge’

2

2

1
r

qqk
F 
Where F= force between two charges

r= separation between two charges
k= coulomb’s constant,

22

9109

C

Nm

x

 o

k
4

1


21

2121085.8







mNC

x
o


= permeability of free space
q1
q2

r

1.3 Coulomb’s law – continued

• Simplified:

Coulomb's law states that the magnitude of the
electrostatic force between two point charges is directly
proportional to the product of the magnitudes of the
charges and inversely proportional to the square of the
distance between them.

• 𝐹 =𝑘𝑄1𝑄2

𝑟2

• The proportionality constant k is called the electrostatic

constant and has the value:

9.0 X 109N⋅𝑚2⋅𝐶−2in free space

r

q1
q2

F21
F12

r

q1
q2

F21
F12

The electric force is directed between the centers of the two point
charges.

Attractive force
between q1 and q2.

Repulsive force
between q1 and q2.

Example

1. Calculate the magnitude of force on charges 5µ C and 2µ C. Both

are 4 cm apart from each other.

[56.25 N]

2.

The magnitude of force on each charge of two equal charges
separated 1 m apart is 1.8 x 1010N. Compute the magnitude of
each charge.

[1.41 C]

Example:
What is the net force on the charge q1 due to the other two
charges? q1 = +1.2 C, q2 = 0.60 C, and q3 = +0.20 C.

The net force on q1 is Fnet = F21 + F31

F31

F21



The components of the net force are:

N

10

0.2

cos

3

21

31

,21

,31

,net

















F

F

F

F

F
x

x

x

N

10

4.1

sin

0

3

21

,21

,31

,net

















F

F

F

F
y

y

y

Where from the figure

38.0
m1.3

m

5.0
sin

92.0
m1.3

m

2.1
cos













The magnitude of the net force is:

N

10

4.2

3

2

,net

2

,net

net










y

x
F

F

F









35

70.0

tan

,net

,net





x

y

F

F

The direction of the net force is:

1.4 ELECTRIC FIELD

There is a force on each charge between two charges placed at

a distance apart.

This phenomenon could explain that;

 there is an influence of one charge on another and
There is an area of influence around the charge

The influence is greater as the distance apart is smaller.
The area of influence of a charge is called electric field.
An electric field is said to exist in the region of space around a

charge.

An electric field is a vector quantity that has magnitude and

direction.

What is the difference between the electric field lines due
to positive and negative charge?

Electric field is radially

outward

Electric field is radially

inward

Direction of electric field around the charge Q can be
visualized by placing small positive charge into the field
and displaying the force vector acting on each.

+

+

+

+

-

+

+

+

Like charges (+ +) Opposite charges (+ -)

Electric field strength

Electric field strength, E at a
point is defined as the force
acting

on

a

unit

positive

charge at that point.

r

Q1
Q2

+

+

Derivation:

Force on test charge Q2 at a
distance r from a point charge
Q2 is

Electric field strength

Hence,

2

2

1
r

Q

Qk
F 

2

2

2

21

2

1

r
kQ

Q

r

QkQ

Q
F
E









2

1

r

Qk
E 

Unit: N/C

Example:
Calculate the magnitude and direction of electric field at a
point P in Figure below which is 30 cm to the right of a point
charge Q = -3.0 x 10-6C.

P

-

Q = -3.0 x 10-6C

P

-

Solution

Step 1:
Draw the direction of the electric field on P due to Q.

Ep

Step 2:
Calculate the magnitude of the electric field on P.

2

1

r

Qk
E 




2

6

9

3.0

10

0.3

10

9











C

N

EP
/

10

0.3

5




Therefore,

Step 3:
Write the vector component.

P
Ep

Direction of E is to
the –x direction

Final answer:

Ep = 3.0 x 105N/C, to the –x axis (to the left)

or

Ep = -3.0 x 105i N/C

REFERENCES

1. Ts. Azzafeerah Mahyuddin

2. Open Textbook. Physical Sciences Grade 11. Electrostatics
Accessed on October 2022.https://www.siyavula.com/read

3. Basic Electronics 3rdEdition. Sean Westcott and Jean Riescher
Westcott, 2020. Mercury Learning and Information.