When
we looked at gravity fields, we saw that gravitational potential energy is the
energy needed to bring an object from infinity to a certain point.
We also saw that gravitational potential was the energy per unit mass.
There are two similar quantities in electric fields, potential
energy and electric potential.
Suppose we moved a charge form one point to another in an electric field. We would have to do a job of work against the field (or get work out if it were with the field), so there is an energy change. We can show this as a graph:
The
area under the graph represents the energy
transformed. Using the
mathematical trick of calculus, we can show that the energy is given by the
equation:
We will look at the calculus treatment at the end of tutorial.
A positive charge of 6 nC attracts a negative charge of 5 nC from a distance of 4 mm. What is the work done? What does the sign mean? 
Electric Potential
We
can define the electrical potential as the energy
per unit charge. It has the code
V_{e}
and is given by the equation:
The units are Joules per Coulomb (JC^{1}) or Volts (V). Indeed the volt is defined as the energy transformed when unit charge is moved between two points. We can show the relationship between the electric field and potential in this graph. Potential is the area under the graph:
Worked Example If a hydrogen atom consists of a proton and an electron at an average separation of 5.0 × 10^{11 }m, what is the electric potential at this distance and the potential energy of the electron? What would the field strength be? Give your answer to an appropriate number of significant figures. (Electronic charge = 1.60 × 10^{19} C) 
Answer Use V_{e} = __1__ Q = 9 × 10^{9} m F^{1} × 1.6 × 10^{19} C^{ }= 28.8 V = 29 V (2 s.f) 4pe_{0} r 5.0 × 10^{11} m

Potential energy = charge × potential = 28.8 V × 1.60 ×10^{19} C = 4.61 × 10^{18} J = 4.6 × 10^{18} J (2 s.f) 
Field
Strength = potential ÷ separation =
28.8 V ÷ 5.0 × 10^{11} m = 5.76 × 10^{11}
N C^{1} = 
Equipotentials
All points within an electric field that have the same potential are called equipotentials, rather like contours on a map. In a radial field, the equipotentials are concentric circles. In a uniform field, they are parallel lines equally spaced.
Suppose we have the positive plate in a uniform field at +300 V, and the negative plate is at 0 V. Suppose we have 10 equipotentials equally spaced in the uniform field. Each equipotential has a potential difference of + 30 V. So there are equipotentials at 0 V, 30 V, 60 V, 90 V, etc.
When you travel along a line of equipotential, the work done is zero.
The pattern for a radial field around a positive charge is like this:
Notice how the equipotential lines are at 90 degrees to the field lines. Notice also that the closer you get towards the charge, the higher the potential (the work done per coulomb) becomes. Viewed from the side, our potential hill would be something like this:
Why are equipotentials not equally spaced in a radial field? 
The pattern for the equipotentials for the field between two unlike charges is like this:
Image from TAP (IoP)
Note how the equipotential line half way between the two charges is straight.
For two like charges, the pattern is like this:
Image from TAP (IoP)
Note the pattern of equipotentials around the neutral point.
A tiny negatively charged oil drop is held stationary in an electric field between two horizontal parallel plated as shown. The mass is 2.5 ´ 10^{10} kg.
(a)
What are the two forces acting on the drop and in which direction do they
act?
(b)
The drop is stationary. What
can be said about the two forces? (c) What is the charge on the oil drop? 
The pattern for a uniform field is like this:
When we move along a line of equipotential, we do no work at all. Contour lines on a map are lines of equipotential.
The equation for the electric field is:
When we move between any two equipotentials, we move through a potential difference, or a difference in energy per unit charge. We have met this before in current electricity.
Comparing
Electric and Gravitational Fields
There
are many analogies that can be drawn between electric fields and gravity fields.
Theoretical physicists would go as far as saying that the two are
possibly different manifestations of the same thing.
Let us compare the two:
Feature 
Electric
Field 
Gravity
Field 
Quantity
susceptible to the force 
Charge 
Mass 
Constant
of Proportionality 
__1__
4pe_{0}
where
e_{0}
is the permittivity of free space. The
value of
e
can
be changed by adding a material. 
G
The
value of
G, the universal gravity
constant has the same value for all media, including a vacuum.

Relationship
with distance 
Inversely
proportional to
r^{2}. 
Inversely
proportional to
r^{2}. 
Force
Equation 
F = __1__ Q_{1}Q_{2 } 4pe_{0}_{ } r^{2} 
F = G m_{1}m_{2} r^{2 } 
Direction
of force 
Can
be attractive or repulsive 
Always
attractive 
Relative
strength 
Strong
at close range 
Weak.
Can only be felt with massive objects 
Range 
Infinite 
Infinite 
The
gravitational attraction between particles in an atom is so small as to be
negligible. The nucleus and its
electrons are held together entirely by electrostatic forces, and these are
involved in chemical reactions.
Gravity
forces hold planets together and hold them in their orbits. Electrostatic forces over the interplanetary distances can be
ignored.
Using Calculus to link Electrostatic Force and Electrostatic Energy (Extension)
The graph below was produced to model the force acting on a positive test charge of 5.0 × 10^{9} C at certain distances from a second charge of 5.0 × 10^{9} C. The distance varies between 1.0 mm and 1.5 mm. The graph shows the force against the distance:
We know that if we were to move the test charge from 4.0 × 10^{3} m to 1.0 × 10^{3} m, we would have to do a job of work, as we are applying a force against the repulsive electric field. On the graph below, the red lines show this and the work done is the area under the graph between the two red lines:
The counting of squares would only give an approximation. Calculus integration would give us a more precise result. So we can write:
We know that:
Where:
So we can write this as:
In this case k, Q_{1}, and Q_{2} are constants. Using the power rule for integration, we can write:
Maths Note Power rule for integration
For differentiation the converse is true:
Yes, I have missed out the constant, C. In this case, C = 0. 
We'll do a worked example.
Worked example A test charge of 5 nC is moved towards a second charge of 5 nC from a position of 4.0 mm to 1.0 mm Calculate the work that needs to be done. 
Answer Use:
W = [(8.99 × 10^{9} m F^{1} × 5.0 × 10^{9} C × 5.0 × 10^{9} C) ÷ 1.0 × 10^{3} m]  [(8.99 × 10^{9} m F^{1} × 5.0 × 10^{9} C × 5.0 × 10^{9} C) ÷ 4.0 × 10^{3} m]
W = 2.25 × 104 J  5.62 × 105 J = 1.69 × 10^{4} J = 1.7 × 10^{4} J (2 s.f.)
The positive sign means that work has to be put in. The work done results in an increase in potential energy of +1.7 × 10^{4} J.

We can apply the same rule to the relationship between electrical field strength and electrical potential. Electrical field strength is force per unit charge and electrical potential is energy per unit charge. This is shown on the graph:
We can therefore write:
The converse is also true. The electrical field strength is the gradient of the graph of electrical potential against distance. So we can write:
Similarly the force is product of the electrical field strength and the charge Q_{2}, so we can write: