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,
We thus arrive at two important conclusions concerning the
relation between electric field and potential:

(i)
Electric field is in the
direction in which the potential decreases steepest

(ii)
Its magnitude is given by the
change in the magnitude of potential per unit displacement normal to the
equipotential surface at the point.

let a point P at which a test charge is situated with same fig A and B having potential of V nd V+ dv respectively which are at r and r+dr from the test charge then

ReplyDeletev + dv = v + dw/q.

this implie

dv = dw/q. ......... (1)

now word done in movinge test chare with a infinitesimally small distance dr dr

then

dw = fdrcos180 ( moving in opp dir of E)

=-q.Edr

dw/q = -Edr .........(2)

now comparing both we get

dv = -Edr

E = -dv/dr

therefore electric field is eqal to the neg of potential gradient