When a unit positive charge is taken along perpendicular distance from the surface B to the surface A against the electric field, the work done is
\(dW=\vec E.\vec{dr}=E\,dr\cos180^0=-Edr\) ........ (1)
This work done equals the potential difference between the surfaces A and B.
\(\therefore dW=V_A-V_B=V-(V-dV)=dV\) ......(2)
From equations (1) and (2),
\(-E\,dr=dV\)
\(E=-\frac{dV}{dr}\)
The negative sign shows that the direction of the electric field \(\vec{E}\) is in the direction of decreasing potnetial.
