If the electric field were not normal to the equipotential surface, it would have non-zero component along the surface.

To move a unit test charge against the direction of the component of the field work would have to be done.

But this is in contradiction to the definition of an equipotential surface because the potential difference between any two points on the surface is $\Delta \mathrm{V}$ is zero so in

$\therefore$ Work $\mathrm{W}=q \Delta \mathrm{V}, \Delta \mathrm{V}=0, \therefore \mathrm{W}=0$

and work done in electric field $\overrightarrow{\mathrm{E}}$ with displacement $\vec{d} l$

$\mathrm{W}=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}=\mathrm{E} d l \cos \theta$ $0=\mathrm{E} d l \cos \theta$ $\therefore 0=\cos \theta \quad[\because \overrightarrow{\mathrm{E}} \neq 0, \overrightarrow{d l} \neq 0]$ $\therefore \quad \theta =\frac{\pi}{2}$

`Hence, the electric field $\overrightarrow{\mathrm{E}}$ is normal to the equipotential surface at that point.