# If the magnitude of the Electric Field at any point on a surface is 0, then is the net charge enclosed guaranteed to be zero?

I got this question on a recent Physics exam. The answer was Yes.

I understand the basic logic using Gauss's Law: $$\phi=\frac{Q_\text{enclosed}}{\varepsilon_0}=\iint_S\vec{E}.d\vec{A}$$ So if there is no electric field, there is necessarily no enclosed charge.

However, my question is, if given a surface enclosing a net charge, is it possible to arrange numerous point charges outside of the surface so that the electric field cancels out to zero at every point? Yet there is still a net charge inside? Note that the question just said "surface" and did not mention symmetry or charges outside the surface.

You cannot arrange the charges outside of a surface to cancel the charges inside the surface for the exact same reason you cannot arrange charges outside of a surface to create a net surface flux when there are no enclosed charges.

That may sound silly, but assume you could do what you are asking. Since that net charge makes a $$Q_{enc}/\epsilon_{0}$$ net surface flux, the exterior charge must be making a $$-Q_{enc}/\epsilon_{0}$$ surface flux to get the hypothetical cancellation you are asking about. Now remove the net charge that is enclosed. With the enclosed charge removed, the exterior flux remains and is now uncanceled. Therefore, there is a net surface flux with zero charge enclosed, which violates the law.

In essence, the question you are asking amounts to being equivalent to the normal way we think of the law, and so if it is true one way it is true the other. The law amounts to the fact that the divergence of the E field is the charge itself, and so is as true as that can be experimentally verified.

• makes sense thanks! Commented Mar 2, 2023 at 4:28