Relation between Electric Intensity and Potential

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 q₀ 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/ d

To 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 q₀ kept at A where electric intensity is E. The force acting on q₀ charge at A is F = q₀E. The work done to carry q₀ charge from A to B through an infinitesimally small displacement dx is dW = F (-dr)

dW = - q₀ Edx, dW/q₀ = -Edx … (1)

But dW/q₀is 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.