Obtain the relation between electric field and electric potential .

Explanation

As shown in figure consider wo closely space equipotential surfaces A and B with potential values V and V + `deltaV` where `deltaV` is the change in V in the direction of the electric field `vecE` .

Let P be a point on the surface B.`deltal` is the perpendicular distance of the surface A from P.
Suppose that a unit positive charge is moved along the perpendicular from the surface B to the surface A against the electric field. The work done in this process is `|vecE|deltal`.
But work done ,
`W=V_(A)-V_(B)`
`:.|vecE|deltal=V-(V+deltaV)`
`:.|vecE|deltal=-deltaV`
`:. |vecE|=-(deltaV)/(deltal)`
`:. |vecE|=|(deltaV)/(deltal)|`
`:. E=(V)/(l)`
Hence negative value of potential gradient is equal to the magnitude of electric field. `(deltaV)/(deltal)` is known as potential gradient . Its unit is `Vm^(-1)` .
From this there are two important conclusions are as below.
(1) Electric field is the direction in which the potential decreases steepest.
(2) The magnitude of electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.