Potential Difference in a Uniform Electric Field
Starting with the definition of work, prove that at every point on an equipotential surface, the surface must be perpendicular to the electric field there.
We are asked to prove that at every point on an equipotential surface, the surface must be perpendicular to the electric field there starting from the definition of work.
The charge in potential between two points (A) and (B) on an equipotential surface is given by
$$W_{\mathrm{AB}}=q \Delta V=-q \int_{\mathrm{A}}^{\mathrm{B}} \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}$$
By definition, the change in electric potential $$\Delta V$$ at an equipotential surface is zero:
$$-q \int_{\mathrm{A}}^{\mathrm{B}} \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathrm{S}}=0 $$
$$\int_{\mathrm{A}}^{\mathrm{B}} \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathrm{S}}=0 $$
$$\int_{\mathrm{A}}^{\mathrm{B}} E S \cos \theta=0$$
Therefore, for a non-zero electric field, $$\cos \theta$$ must be zero or $$\theta=90^{\circ} .$$ Hence, on an equipotential surface, the electric field is perpendicular to the surface.