I've read through the standard explanation of the electric field due to a spherical shell with uniform charge density. This explanation argues that because a Gaussian surface inside the shell encloses zero charge, by gauss' law, the electric field must be zero here. However, shouldn't the statement actually be that the flux is zero here? For example, a Gaussian sphere enclosing no charge above an infinite plane has zero net flux through it, and so it satisfies Gausss' theorem, but inside the sphere, the electric field is still the standard field due to an infinite sheet, no? So doesn't the argument re the spherical shell only rule out that the flux thru a gaussian surface inside the shell will be zero, and not that the field there is zero?
1 Answers
You are right, Gauss's law in its integral form generally just relates the charge enclosed by the Gaussian surface to the flux of the electric field through our Gaussian surface
$$\oint\mathbf E\cdot\text d\mathbf a=\frac{Q_\text{encl}}{\epsilon_0}$$
i.e., if there is no enclosed charge, then we know there is no flux through the surface.
However, as pointed out in the comments, if we have chosen the Gaussian surface carefully such that symmetry allows us to argue that if there is a field, if must point perpendicularly to the surface and must have a constant magnitude on this surface, then it must be the case that $$\oint\mathbf E\cdot\text d\mathbf a=\oint E\ \text d a=E\oint\ \text d a=\frac{Q_\text{encl}}{\epsilon_0}$$
And so now we can make the argument that since $\oint\ \text d a\neq0$, if $Q_\text{encl}=0$ it must be that $E=0$.
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