# Equipotential surface paradox

Consider a point charge +q as shown below. Now consider a equipotential surface and two points in it A and B. now the potential at A is equal to potential at B. If we now integrate (E.ds) from A to B it is equal to zero as A and B are at the same potential. This is true for any path between A and B. Therefore electric field at any point inside the sphere should be zero, but this is not true? Why is that?

• What you say after the "therefore" does not follow from what you say before. Mar 3, 2016 at 18:11

You shouldn't integrate $E.ds$, you integrate $\vec{E}.\vec{ds}$. With your positive point charge, the contribution will be negative while approaching and positive while moving away from the point charge. The complete integral (the sum of the positive and negative contributions) will be zero, without $\vec{E}$ being zero.

Work done is force $\times$ displacement along line of action of the force and the dot product is a way of evaluating the work done $\vec F \cdot d\vec s$

Consider walking along the path $A$ to $H$ in the diagram below.

The walk has been chosen so that it is

• only along an equipotential line when no work is done because the force (field) is at right angles to the equipotential

• at right angles to the equipotential lines when work is done as the line of action of the force (field) is along the direction of the path taken.

So the work done in going from $B$ to $C$, $D$ to $E$ and $F$ to $G$ is zero as those parts of the walk are along equipotential lines.

The work done from $A$ to $B$ plus the work done from $G$ to $H$ is zero as is the work done from $C$ to $D$ plus the work done from $E$ to $F$.
So the net work in going from $A$ to $H$ is zero.

The path, $A$ to $B$, that you have drawn in your diagram can be made up of smaller and smaller steps of walking along an equipotential line and walking at right angles to the equipotential lines to give a net work done of zero.