EP025 Note 1

CHAPTER 1 : ELECTROSTATICS

1.1 Coulomb’s Law.
1.2 Electric Field.
1.3 Electric Potential.
1.4 Charge In A Uniform Magnetic Field.

SLT (HOURS)

Lecture :2

Tutorial :7

Practical :0

Ind. Learning : 9

TOTAL : 18

CHAPTER 1 :
ELECTROSTATICS

CONCEPTUAL MAP

1.0 INTRODUCTION

Electrostatics refers to the
study of electric charge at rest.

Generally, a particle (atom)
consists of positive charged
nucleus (which consists of
neutrons and protons) and
surrounded by negative
charged electrons.

Electric charges that occur in
nature are always multiples of an
elementary charge :

where :
e = -1.602x10-19C
p = +1.602x10-19C

In the SI system, the unit of charge
is coulomb (C).

1C = 1A x 1s

A common method to charge a
neutral body is by :

• rubbing with another neutral body.
• touching with a charged body.



1.1 COULOMB’S LAW
LEARNING OUTCOMES (LO) :

1.1 COULOMB’S LAW

“Two point charges Q and q repel OR attract one

another with force F, which is directly proportional to
the product of the charges and inversely proportional to

the square of the distance r between them”.

where k = = 9.0x109Nm2C-2

with o is a universal constant known as ‘permittivity of
free space’ with value of 8.85x10-12C2N-1m-2

Two identical charges repel each other while two
opposite charges attract one another with force F.

The charge which experienced the force is called
the test charge, q.

The unit for force is kg m s-2 or Newton (N).

If a test charge, q is exerted
by  2 forces due to the
present of some point
charges, Q then the
magnitude and the direction
of the resultant force can
be determined as follow :

Furthermore, an appropriate free body diagram (FBD)
must be used.

&

QUESTION 1

In a hydrogen atom, the electron is at a distance of
5x10-11 m from the nucleus which consists of one
proton. Calculate (a) the force between the two
charges (b) the ratio of the electrostatic force to the

gravitational force between the charges.

ANSWERS
(a) F = 9.2x10-8N
(b) FE/Fg = 2.3x1039

&

QUESTION 2

Three charges each of +3 C are fixed along the
x-axis at x = 0, 0.30 m and 0.50 m. Calculate the
resultant force on the charge in the middle due to the

other two charges.

ANSWERS
FT = 1.125N (towards q3)

1.2 ELECTRIC FIELD
LEARNING OUTCOMES (LO) :

1.2 ELECTRIC FIELD

Physically, electrostatic field (or electric field), E refers
to the electric field at any point (in 3-D) around a

stationary charged body (electric charge) where the
electrostatic force can be experienced.

The direction of E is depending on the types of the
charge (Q+ : outwards, Q- : inwards)

Electrostatic field strength at
EA a point around a point charge, Q

refers to the mathematical calculation
• A of the electrostatic force, F per unit
rA test charge, q.

rC rB or

•C •B where r = the distance between the
point charge to the point.
EC EB

The SI unit for electrostatic field is N C-1 and
sometimes written as V m-1.

The electrostatic field strength is
represented graphically by means
of field lines which has the following
properties :

• tangents at each point.
• do not intersect each other.
• the concentration lines indicates

its strength.

Examples of Electric Field Patterns

Isolated point charges Two point charges of the same magnitude

Uniformly charged Two point charges of different
parallel plate magnitude

If a point experiences  2
electric field due to the present
of some point charges, Q then
the magnitude as well as the
direction of the resultant
electric field can be determined
as follow :

Again, free body diagram (FBD) must be used.

&

QUESTION 1

Two parallel metal plates separated 2 cm are
connected to a 120 V battery. Calculate (a) the electric
field intensity between the plates (b) the force exerts

on an electron in between the plates.

ANSWERS
(a) E = 6kVm-1
(b) F = 9.6x10-16N

&

QUESTION 2

Two point charges of +5 C and -2 C respectively are
separated by a distance of 5.0 cm. Determine the
electric field at the midpoint between the charges.

ANSWERS
ET = 100.8x106 NC-1

1.4 CHARGE IN A UNIFORM MAGNETIC FIELD
LEARNING OUTCOMES (LO) :

1.4 CHARGE IN A UNIFORM MAGNETIC FIELD

A uniform electrostatic field is one
which lines are parallel and which
the magnitude of the force exerted
on a test charge is constant at
every point of the field.

If a test charge, q is placed or
moving in the field, electrostatic
force will be produced on the
charge and change its state of
motion.

CASE 1 : STATIONARY CHARGE

+ u=0 Based on equation :

u=0 - F = qE

When a test charge, q is
placed in the uniform
electric field, E then
electrostatics force, F will
exerted on the charge.

+F Therefore, the charge is
then moving in a straight
F- line towards the plate.

The direction of motion
depends on the type of the
test charge :

q+ : towards –ve plate.
q- : towards +ve plate.

CASE 2 : MOVING CHARGE PERPENDICULARLY TO FIELD

+ Based on equation :
v
F = qE
v
When a test charge, q is
- moving perpendicularly in a
uniform electric field, E then
electrostatics force, F will
exerted on the charge.

+F Therefore, the charge is
v then deflecting in a
parabolic trajectory
v towards the plate.

F- The direction of motion
depends on the type of the
test charge :

q+ : deflects to –ve plate.
q- : deflects to +ve plate.

CASE 3 : MOVING CHARGE PARALLEL TO FIELD

v Based on equation :
+
v F = qE

+ When a test charge, q is
moving parallel in a uniform
electric field, E then
electrostatics force, F will
exerted on the charge.

v Therefore, if q initially move
+F towards opposite sign plate,
v then it will accelerate
(speeding up).
+F
if q initially move towards
same sign plate, then it will
decelerate (slowing down)
before turning back to
accelerate towards
opposite sign plate.

1.3 ELECTRIC POTENTIAL
LEARNING OUTCOMES (LO) :

1.3 ELECTRIC POTENTIAL

Electric potential, V refers to the
•  work performed (or required) to bring

a unit electric charge from infinity to a
point (for example : point A) in an
electric field.

•A

where r = distance between the point and Q.

• Unit electric charge refers a particle
•A of charged 1C.

Electric potential is a scalar
quantity. The SI unit for electric
potential is J C-1 or normally known
as volt, V.

Work (or energy) is required for two
identical charged bodies since
they’re repel each other.

The total electric potential, VT at a
certain point (eg. point O) is given
by the following equation :

where r1, r2 and r3 respectively are the
distance from the each point charge to
point O.

Note :
The magnitude also depends on the type of charges.

The potential difference, V
between two points (for example :
• A A and B) in an electric field is the
difference in the values of the
electric potentials at the two
points.

•B In other word, the potential difference
refers to the work done in moving unit

charge from one point the other.

•A
•B

RELATIONSHIP POTENTIAL DIFFERENCE-ELECTRIC FIELD

•  If the charge is moved from
infinity to a fixed point, the work
done can be written as :

Since F = qE and
• A Then :

or

• Potential energy, U refers to the
•A work performed (or energy required)
to bring an electric charge, q from
infinity to a point (for example : point
A) in an electric field.

W = qV

or

• Based on the Work-Energy Theorem :
•A W = Uf – Ui
= UA – U ; U = 0

Then :

where r = distance between
the point and Q.

Therefore, the potential energy, U
•  can be related with potential, V as :

Potential energy is a scalar quantity
with unit of joule, J or kg m2 s-2.

•A

The total electric potential energy,
UT at a certain point (eg. point O)
is given by the following equation :

qo
+

where r1, r2 and r3 respectively are the
distance from the each point charge to
point O.

Note :
The magnitude also depends on the type of charges.

EQUIPOTENTIAL SURFACE

An equipotential surface is a
set of points which have the
same magnitude of electric
potential, V.

The work performed when a
charge is moved within the
equipotential surface :

WAB = UB – UA
= q(VB – VA)
=0

&

QUESTION 1

Calculate the electric potential at a point (a) 15 cm and
(b) 50 cm from a point charge of 3C. (c) Calculate the

work done to move a charge of 0.5C from a point
50cm from the point charge 3C to a point 15cm from

the point charge.

ANSWERS
(a) V = 180kV
(b) V = 54kV
(c) W = 63mJ