Obtain the relation between electric field and electric potential.

(N/A) As shown in the figure,consider two closely spaced 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 $A$ against the electric field. The work done in this process is $|\vec{E}| \delta l$.
But work done,$W = V_{A} - V_{B}$.
Therefore,$|\vec{E}| \delta l = V - (V + \delta V)$.
$|\vec{E}| \delta l = -\delta V$.
$|\vec{E}| = -\frac{\delta V}{\delta l}$.
Hence,the negative value of the potential gradient is equal to the magnitude of the electric field. $\frac{\delta V}{\delta l}$ is known as the potential gradient. Its unit is $V \cdot m^{-1}$.
From this,there are two important conclusions:
$(1)$ The electric field is in the direction in which the potential decreases most steeply.
$(2)$ The magnitude of the electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.