Question
Using Gauss’s theorem, show mathematically that for any point outside the shell, the field due to a uniformly charged thin spherical shell is the same as if the entire charge of the shell is concentrated at the centre. Why do you expect the electric field inside the shell to be zero according to this theorem?

Answer

$\phi =\oint\limits_{s}\overrightarrow{\text{E}}.\overrightarrow{\text{d}}\text{s} = \frac{\text{q}}{\varepsilon_{\circ}}$
Derivation: $\text{E}\times4\pi\text{r}^{2} =\frac{\sigma}{\varepsilon_{\circ}}4\pi\text{R}^{2}$
$\therefore \text{E} = \frac{\sigma\text{R}^{2}}{\varepsilon_{\circ}\text{r}^{2}}$
where $\text{q} = 4\pi\text{R}^{2}\sigma$ is the total charge on the spherical shell. Electrostatic field is zero, since total charge inside the shell is zero or charge reside on the surface of the shell.

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