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Chapter 2 Electrostatics: Problem 2.11
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Problem 2.11: Use Gauss’s law to find the electric field inside and outside a spherical
shell of radius R that carries a uniform surface charge density σ. Compare your
answer to Prob. 2.7.
Solution:
Consider a gaussian spherical surface of radius $r$ where $r\ge R$.
According to Gauss Law,
$$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = \frac{\sigma\times 4\pi R^2}{\epsilon_0}$$
Since the charge distribution is symmetrical, $\vec{E}$ and $\vec{da}$ point in the same direction and the direction is $\hat{r}$
$$E\times 4\pi r^2 = {\sigma\times 4\pi R^2 \over \epsilon_0}$$
$$\vec{E} = {\sigma\times R^2 \over \epsilon_0\times r^2}\hat{r}$$
Now, For $r<R, Q_{enc} = 0$
$$\therefore \oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = 0$$
Since, $\vec{da}\ne 0$ and both $\vec{E}$ and $\vec{da}$ points in the same direction.
$$\therefore \vec{E} = 0$$
Hence ,
For $r<R, \vec{E}=0$ and
For $r\ge R$, $\vec{E} = {\sigma\times R^2 \over \epsilon_0\times r^2}\hat{r}$
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