Scientist Michael Faraday brought the concept of electric field in the 19th
century. He used the patterns of electric field lines that shows the existence of the electric field
but these lines are invisible in reality. He explained
that around a charge particle there a space in which another charge
particle experience a force of attraction or repulsion, this region is
called the electric field.

When electric charges acts in certain region of space. They produce
force around it. This is what electric field is! It decides whats the
strength the field carries due to these charge interaction. Hence, the electric field is defined as"Electrical force acting per unit charge exerted on the charged body".
The electric field by a point charge q is given by,Electric field strength = $\frac{Force}{Charge}$
or
E = $\frac{F}{q}$
The electric field strength is expressed in Newton per coulomb (NC^{1}) or volt per meter (Vm^{1}).
Electric field lines (electric flux lines) are the curved or straight paths along which a unit positive charge tends to move in the electric field if it is free to do so. These lines are imaginary but they give us the pictorial visualization of electric field. They show patterns of several lines, which extend between infinity and the source charges.
Pictorial view of electric field lines
The electric field is a region where force acts due to the interaction between the charges. The strength of the field is measured asPictorial view of electric field lines
Electric field strength = $\frac{Force}{Charge}$
or
E = $\frac{F}{q}$
The unit of electric field strength is in newton per coulomb (NC^{1}). It is a vector quantity. It obeys inverse square law. Hence, if a charge q is at a distance r from another charge Q, the force between them is given as
F = $\frac{kqQ}{r^2}$
The charge Q has the electric field strength is given as
E = $\frac{F}{q}$ = $\frac{\frac{kqQ}{r^2}}{q}$ = $\frac{kQ}{r^2}$
This is the formula to calculate the electric field strength.
The Magnitude of the electric field is always positive whereas the direction will be in the way the electrical force acts given as
$\vec{E}$ = $\frac{\vec{F}}{q}$The electric field is given as $\vec{E}$ tells that the electric field has always the vector quantity that rely on charge q (positive or negative).
Let us consider two charges +q and q separated by a small distance. They form a dipole.
E = qd
where,
q is the charge and
d is the distance of separation between them.
A parallel plate capacitor consists of two conducting metal plates X and Y each of area A separated by distance d. The plates can be of any shape. When the plates are connected to a battery, a charge +q appears on plate X and – q appears on plate Y. On each plate the charge is distributed uniformly. As the distance between the plates is small compared to the area of the plate, the electric field is uniform between the plates. The electric field is non uniform at the outer edge of the plates.
The capacitance is
The electric field E acting in a parallel plate capacitors is given by
E = $\frac{\sigma}{\epsilon_o}$ = $\frac{V}{d}$
where $\epsilon_o$ is the permittivity of free space and d is the distance between the platesC = $\frac{\epsilon_o A}{d}$Here A is the area of the plate.
The net electric field is the sum total of the charges acting in the given electric field. If q_{1}, q_{2}, q_{3},...... q_{n} are the charges acting in the given field.
q = q_{1} + q_{2} + q_{3} + q_{4 }+ ...... +q_{n}.
Solved Examples
Question 1: Calculate the magnitude of electric charge $\vec{E}$ at a field point 3 m from the point charge 6 nc?
Solution:
Given:
Distance r = 3 m, Charge q = 6 nc
The magnitude of electric field is
E = $\frac{1}{4 \pi \epsilon_0}$ $\frac{q}{r^2}$
= (9 $\times$ 10^{9} Nm^{2}/C^{2}) $\frac{6 \times 10^{9}}{(3 m)^2}$
= 6 N/C
Solution:
Given:
Distance r = 3 m, Charge q = 6 nc
The magnitude of electric field is
E = $\frac{1}{4 \pi \epsilon_0}$ $\frac{q}{r^2}$
= (9 $\times$ 10^{9} Nm^{2}/C^{2}) $\frac{6 \times 10^{9}}{(3 m)^2}$
= 6 N/C
Question 2: If the charge + 3e and a charge 3 e acts separated by a distance of 4 nm. What is the potential energy of the system?
Solution:
Given:
Charge q = + 3e, Q =  3e, distance r = 4 $\times$ 109 m
The electric potential is given by
U = k $\frac{qQ}{r}$
= (9 $\times$ 10^{9}) $\times$ $\frac{3 \times 1.6 \times 10^{19} \times (3) \times 1.6 \times 10^{19}}{4 \times 10^{9}}$
=  5.065 $\times$ 10^{9} N.
Solution:
Given:
Charge q = + 3e, Q =  3e, distance r = 4 $\times$ 109 m
The electric potential is given by
U = k $\frac{qQ}{r}$
= (9 $\times$ 10^{9}) $\times$ $\frac{3 \times 1.6 \times 10^{19} \times (3) \times 1.6 \times 10^{19}}{4 \times 10^{9}}$
=  5.065 $\times$ 10^{9} N.