To obtain relation between electric intensity E and electric potential V, LET US consider two equipotential surfaces A and B spaced closely as shown in fig. 1 (c) let the potential of A be VA = V and potential of B be VB = (V + dV) where dV is increase in potential in the DIRECTION of electric intensity E normal to A and B.

Suppose dr is perpendicular distance between the two equipotential surfaces. When a unit POSITIVE charge is taken along this perpendicular distance from the surface B to the surface A against the electric field.

Work done WBA = | E | dr

By definition

WBA = VA – VB = V – (V + dV) = - dV
 |E| dr = – dV

To obtain relation between electric intensity E and electric potential V, let us consider two equipotential surfaces A and B spaced closely as shown in fig. 1 (c) let the potential of A be VA = V and potential of B be VB = (V + dV) where dV is increase in potential in the direction of electric intensity E normal to A and B.

Suppose dr is perpendicular distance between the two equipotential surfaces. When a unit positive charge is taken along this perpendicular distance from the surface B to the surface A against the electric field.

Work done WBA = | E | dr

By definition

WBA = VA – VB = V – (V + dV) = - dV
 |E| dr = – dV