Question

Show that the line integral of the electric field E over a closed path is equal to zero.

Answer 1

The electric field is a negative gradient of potential:

$$\vec{E}=-\nabla\phi.$$

The line integral is

$$\int\vec{E}\text{d}\vec{l}=-\int\nabla\phi\text{d}\vec{l}.$$

A closed path means that the limits of integration are the same, therefore:

$$-\int^a_a\nabla\phi\text{d}\vec{l}=-(F(a)-F(a))=0.$$