Derive a relation between electric field and potential

Explanation

Consider two equipotential surfaces A and B with the potential difference dV between them as shown in figure. Let dl be the perpendicular distance between them and `vec(E )` be the electric field normal to these surfaces

The work done to move a unit positive charge from B to A against the field `vec(E )` through a displacement `vec(dl)` is `dW= vec(E ).vec(dl)= E dl cos pi= - E dl`
This is equal to the potential difference, therefore
`dV= dW rArr dV= - E dl`
`E= - (dV)/(dl)`