Consider two closely spaced equipotential surfaces A and B (Fig.) with potential values V and V + 𝛿V, where 𝛿V is the change in V in the direction of the electric field E.
Let P be a point on the surface B.𝛿l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E| 𝛿l.
This work equals the potential difference V(A)–V(B).
Thus, |E|𝛿l = V−(V +𝛿V)= –𝛿V
i.e., |E|= –𝛿V/𝛿l .....(1)
Since 𝛿V is negative, 𝛿V = – |𝛿V|, we can rewrite Eq (1) as
|E|= –𝛿V/𝛿l = +|𝛿V|/𝛿l
Important Points :
1. Electric field is in the direction in which the potential decreases steepest.
2. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.
