ELECTROMAGNETISM
ELECTROSTATICS
"Electrostatic" pertains to electric charges at rest or
to fields or phenomena produced by stationary
charge(s).
Electric charges.
There are two types of charges called positive
and negative.
Like charges repel each other.
Unlike charges attract each other.
Charges within atoms.
An atom has a heavily charged positive nucleus
at its center surrounded by tiny negatively
charged electrons which cause the atoms to be
electrically neutral and therefore the number of
positive charges is equal to the number of
negative charges. Each electron carries a
charge equal to –e, the neutron has no net
charge and each proton has a charge +e.
Therefore
Number of protons = Number of electrons
NOTE: The numerical value of e = 1.6 x10-19 C
Conductors / Insulators.
It can happen that when certain substances
come in to contact some outer shell electrons
transfer from one substance to the other. In this
case one substance looses electrons and the
other gains electrons.
Negatively charged bodies have an excess of
electrons and
Positively charged bodies have an electron
deficiency.
If an electron becomes free of its atom we call it
a conduction electron and the conductivity of a
material is determined by
1. The number of free electrons in the material
and
2. The ability of these electrons to move
through the material.
Electric conductors are materials with many
free electrons i.e. materials where very little
energy is required to free some of electrons
from the atoms outer shells.
Conductors usually metals will have enough
energy at ordinary temperatures to produce free
electrons.
Conductors also allow the movement of electric
charges through them.
Insulators are materials that contain very few
free electrons and do not easily allow the
electric charges to move through the material.
Insulators are materials in which the atoms keep
a strong hold on their electrons.
Electric insulators are materials which have a
negligible number of charges that are free to
move.
INDUCED CHARGES.
When a negative charge is placed near a
conductor, free electrons in the conductor are
repelled to the far end of the conductor which
then has an excess of electrons and is therefore
negatively charged.
The deficiency of electrons at the near end is
equivalent to a positive charge.
If the conductor is then cut in two these induced
charges are separated.
ELECTRIC FIELDS.
An electric charge sets up an electric field in the
space surrounding it and an electric force is
exerted on any charged body placed in this
field. Electric fields may be represented by
electric field lines.
Electric field lines are mapped out with lines of
force, and the direction of the line of force is
taken as the direction in which a single isolated
positive charge would move. An electric field
exists at a point if a positive test charge placed
there experiences a force. The strength and
direction of the electric field is represented by
the electric field vector.
Single isolated charges
Two unlike charges facing each other.
Two like charges facing each other.
Two parallel charged metal plates
Symbol used for charge is Q measured in
S.I. Units called Coulombs symbol C.
Typically charge is of the order of C i.e.
X 10-6 C
Electric Field Intensity. E
Intensity of an Electric Field at a particular
point in a measure of the magnitude of the field
at that point. The electric field intensity at a
point is defined as the force experienced by a
unit positive test charge placed at that point and
takes the direction of the force.
Symbol E and is defined as the force per unit
charge acting at that point.
F
E
Q
Where F is the force acting on a charge of
magnitude Q Coulombs placed at the point.
S.I. Units
N / C  N C 1
COULOMB’S LAW
Coulombs law states that the force acting
between two charges Q1 and Q2 at a distance
r(m) apart is directly proportional to the product
of the magnitudes of the charges and inversely
proportional to the square of the distance
between them
Q1Q2
F 2
r
The constant of proportion ality k
Q1Q2
F k 2
r
NOTES:
F is a vector quantity therefore we require a
magnitude and a direction so it is actually more
correct to say that
Q1 Q2
F k
r2
and
Direction is decided by the relative signs on the
charges using the fact that Like charges repel and
Unlike charges attract
2.
k is a constant and was found to be
k
1
4

Whereis a constant for the medium in which the
charges are located, and is known as the
permittivity of the medium.
Medium
 C2 / N . m2
Vacuum/air
8.85 x 10-12
Paper
2 x 10-11
Glass
6.195 x 10-11
Water
7.08 x 10-10
Sometimes the value given for a particular medium
is its relative permittivity r which is defined as the
permittivity of the material relative to the
permittivity of a vacuum or air and therefore the
permittivity of the material m is equal to
m = r . 

In most cases the medium is air and therefore
k
1
4o
medium

1
9
2 2

9
x
10
Nm
C
12
4 8.85 x10


relative permittivity
paper
2.25
glass
7
water
80
We can use Coulomb’s law to calculate the
intensity of an electric field due to an isolated
point charge Q1 at a point A a distance r(m)
from the charge
Q1
A
r (m)
Step 1. Place a second charge Q2 at the point
A. Therefore Q2 is in the electric field of Q1
and it is a distance of r(m) from Q1.
Q1
Q2
r(m)
Step 2. Use Coulomb’s Law to calculate the
Force on Q2 due to the charge Q1 a distance
r(m) away
Q1 Q2
F k
r2
Step 3. Calculate the intensity at the point A using
the definition of intensity of force per unit charge at
a point.
F
Q1 Q2
Q1
E
k
k 2
2
Q2
Q2 r
r
This then is the equation for the value of the
intensity of the electric field at a point A a
distance of r metres from a point charge Q1
S.I. Units N / C
E is a vector quantity therefore we need to
specify both a magnitude and a direction.
The above equation is for the magnitude of the
intensity and
Direction is defined in the direction of the
electric field lines i.e. the direction that a +ve
charge would move if placed at the point A.
Therefore if Q1 is a positive charge then the
direction is away from Q1
If Q1 is a –ve charge then the direction is
towards Q1.
Since the quantities are vectors the resultant
electric force or field due to all charges is the
vector sum of the forces/fields due to each
individual charge
PROBLEM SHEET
Question 2. Two equal charge of 10C are
placed 15 cm apart.
Calculate the force acting on the charges if they
are (a) placed in air (b) placed in water
Question 2.(a) Calculate the electric field
intensity associated with a +ve charge of
5C at a point 5 cm from the charge. (b) If the
charge was – 5 C calculate the electric field
intensity.
Question 3.(a) Two charges +6C and +0.66C
Are placed 16 cm apart. If a charge of – 8C is
placed between them and in line with them where
along the line will the net force on the negative
charge equal zero.
(b) Calculate the electric field intensity midway
between the two +ve charges.