I am reading Berkeley Physics Course, Volume 2 (Electricity and Magnetism
by Edward M. Purcell). I am in chapter $3$, page $92$, and the book discusses conductors.
The following is from the book:
Because the surface of a conductor [in Fig $3.2$] is necessarily
a surface of constant potential, the electric field, which is $-\nabla \varphi$, must be perpendicular to the surface at every point on the surface
I have omitted the picture because it is not relevant.
What is the reasoning?
I understand that the potential, $\varphi$, is a continuous function,
and since $E=0$ inside the conductor and since $E=-\nabla\varphi$
I get that $\varphi=0$ inside and on the surface (from continuity)
of the conductor.
However, I don't understand the reason the book gives for explaining
why the field is perpendicular to every point on the surface.
Best Answer
Loosely speaking, the gradient of a scalar field (such as the electrostatic potential) points in the direction of that field's greatest change. Since no change occurs in the field when you go along the surface, the gradient shouldn't have a component in that direction.
Here is another intuitive explanation: Imagine for a moment that the electric field was not perpendicular to the surface. That means it has a component along the surface. Now, electric fields exert a force on charges, so now we have a force on the charges in the conductor along the surface of the conductor. This force isn't balanced by anything else, so it will then move the charge around. But that means that our system wasn't yet in equilibrium, since charge was moving around. In equilibrium, the charges must be at rest, and that can only be the case when there is no electric force along the surface, i.e., when it's perpendicular to it.
Note: You say that $\varphi = 0$ inside and on the surface of the conductor. That is not true. $\varphi$ is constant inside and on the surface of the conductor.