Explain why, for any charge configuration, the equipotential surface through a point is normal to the electric field at that point.

Question

(i) Explain why, for any charge configuration, the equipotential surface through a point is normal to the electric field at that point.

Draw a sketch of equipotential surfaces due to a single charge (– q), depicting the electric field lines due to the charge.

(ii) Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side ‘a’ as shown alongside.

Answer 1

The work done in moving a charge from one point to another on an equipotential surface is zero. If the field is not normal to an equipotential surface, it would have a nonzero component along the surface. This would imply that work would have to be done to move a charge on the surface which is contradictory to the definition of equipotential surface

Mathematically Work done to move a charge dq, on a surface, can be expressed as

dW = dq \((\overrightarrow E.\overrightarrow {dr})\) 

But dW = 0 on an equipotential surface

\(\therefore\overrightarrow E\perp\overrightarrow{dr}\)

Equipotential surfaces for a charge –q is shown alongside.