Equipotential surfaces and perpendicular field lines

Conceptual: equipotentials and field lines. Explain why electric field lines are always perpendicular to equipotential surfaces, and what would happen if they were not.

Solution

The reasoning. On an equipotential surface, by definition, the potential does not vary: $\Delta V = 0$ for any displacement $\dd{\vv{\ell}}$ that stays on the surface. The link between field and potential gives, for an infinitesimal displacement, $\dd V = -\,\vv{E}\cdot\dd{\vv{\ell}}$ For $\dd V = 0$ along any direction tangent to the surface, it is required that $\vv{E}\cdot\dd{\vv{\ell}} = 0 \quad\Longrightarrow\quad \ev{\vv{E}\perp\dd{\vv{\ell}}}$ i.e. the field is perpendicular to the equipotential at every point.

What would happen otherwise. If the field had a component tangent to the surface, that component would do work moving a charge along the surface, and so the potential would change: but then the surface would no longer be equipotential, contradicting the hypothesis. In other words, the tangential component would spontaneously set the charges in motion along the equipotential until it vanishes. This is why, at equilibrium, field and equipotential surfaces are always orthogonal (in the figure: radial lines of $\vv{E}$ and circles $V = \text{cost}$).

Connections

Topics: Electric field and potential Concepts: Electric field · Electric potential Skills: Conservation of energy Methods: Gradient method Objects: Charged sphere