# Gauss’s Law inside the hollow of charged spherical shell

Use Gauss’s Law to prove that the electric field anywhere inside the hollow of a charged spherical shell must be zero.

My attempt:

$$\int \mathbf{E}\cdot \mathbf{dA} = \frac{q_{net}}{e}$$

$$\int E \ dAcos\theta = \frac{q_{net}}{e}$$

$$E \int dA = \frac{q_{net}}{e}$$

$E\ 4\pi r^2 = \frac{q_{net}}{e}$ and since it is a hollow of a charged spherical shell the $q_{net}$ or $q_{in}$ is $0$ so: $E = 0$.

Is my reasoning on this problem correct? Essentially $E$ is $0$ because there is no charge enclosed.

• Spherical symmetry implies that $\mathbf{E}=E(r)\ \hat{e}_r$, where $\hat{e}_r$ is a unit vector in the radial direction. If your charged shell is conducting, then the assumption of spherical symmetry holds. May 12, 2014 at 8:52

You may have forgotten to consider the case where $\vec E \perp\vec A$. Then, also flux is zero. But, it is easy to tell using symmetry that then $\vec E$ would form closed loops which is not permissible. Hence, $E$ has to be zero units.