Obtain the relation between electric field and electric potential.

As shown in figure consider two closely space equipotential surfaces A and B with potential values $V$ and $V+\delta V$ where $\delta V$ is the change in $V$ in the direction of the electric field $\vec{E}$.

Let $P$ be a point on the surface $B . \delta l$ 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 $\mathrm{A}$ against the electric field. The work done in this process is $|\overrightarrow{\mathrm{E}}| \delta l$.

But work done,

$\mathrm{W}=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$ $\therefore|\overrightarrow{\mathrm{E}}| \delta l=\mathrm{V}-(\mathrm{V}+\delta \mathrm{V})$

$\therefore|\overrightarrow{\mathrm{E}}| \delta l=-\delta \mathrm{V}$

$\therefore|\overrightarrow{\mathrm{E}}|=-\frac{\delta \mathrm{V}}{\delta l}$

$\therefore|\overrightarrow{\mathrm{E}}|=\left|\frac{\delta \mathrm{V}}{\delta l}\right|$

$\therefore \mathrm{E}=\frac{\mathrm{V}}{l}$

Hence negative value of potential gradient is equal to the magnitude of electric field. $\frac{\delta \mathrm{V}}{\delta l}$ is known as potential gradient. Its unit is $\mathrm{Vm}^{-1}$.

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.