First let us consider a uniform field, for which the lines of force are parallel and equidistance from each other. Let A and B be two points in a uniform electric field of intensity E.Let W_{AB}, be the work done to carry a test charge q_{0} from A to B. Let the potential at A and B be V_{A} and V_{B}, respectively. We have already seen that W_{AB}/q₀ = V_{B} – V_{A}

The work done to carry the test charge from A to B is W_{AB} = q₀EAB

q₀EAB/q₀ = V_{B} – V_{A}

E = (V_{B} – V_{A})/AB

Let the potential difference V_{B} – V_{A} = V and AB = l

Then E = V/l

The electric field between the two plates separated by a distance d is uniform. If the potential difference between the plates is V. then E = v/ dTo show that E = – dV/dx

Let us consider a non-uniform electric field. The electric intensity varies from one point to another. Let A and B be two points which are very close to each other so that the electric intensity is almost constant.

Let OB = x, AB = dx and OA = x + dx. The potential at B is V and that at A is V – dV. Consider test charge q_{0 }kept at A where electric intensity is E. The force acting on q_{0} charge at A is F = q_{0}E. The work done to carry q_{0} charge from A to B through an infinitesimally small displacement dx is dW = F (-dr)

dW = – q₀ Edx, dW/q₀ = -Edx … (1)But dW/q₀ is the work done in carrying unit test charge from A to B which is the potential difference between the points A and B.

dW/q₀ = V_{B} – V_{A} = V – (V – dV) = dV

From equation (1) dW/q₀ = – Edx

Hence dV = – Edx

E = – dV/dx

dV/ dx is the rate of change of electrostatic potential with distance and is called the potential dx gradient. Thus the electric intensity at a point in an electric field is the negative potential gradient.