(N/A) If the electric field $\overrightarrow{E}$ were not normal to the equipotential surface,it would have a non-zero component along the surface.
To move a unit test charge against the direction of this component,work would have to be done.
However,this contradicts the definition of an equipotential surface,which states that the potential difference $\Delta V$ between any two points on the surface is zero.
Since work $W = q \Delta V$ and $\Delta V = 0$,the work done $W$ must be $0$.
Also,the work done in an electric field $\overrightarrow{E}$ for a small displacement $\overrightarrow{dl}$ on the surface is given by $W = \overrightarrow{E} \cdot \overrightarrow{dl} = E dl \cos \theta$.
Since $W = 0$,we have $0 = E dl \cos \theta$.
Given that $\overrightarrow{E} \neq 0$ and $\overrightarrow{dl} \neq 0$,it follows that $\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$ $(90^{\circ})$.
Hence,the electric field $\overrightarrow{E}$ is normal to the equipotential surface at that point.