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Chapter 2 Electrostatics: Problem 2.12
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Problem 2.12 Use Gauss’s law to find the electric field inside a uniformly charged
solid sphere (charge density $\rho$). Compare your answer to Prob. 2.8.
Solution:
Consider a gaussian spherical surface of radius $r$ where $r\ge R$
According to Gauss Law,
$$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = \frac{\rho\times \frac{4}{3}\pi R^3}{\epsilon_0}$$
Since the charge distribution is symmetrical, $\vec{E}$ and $\vec{da}$ point in the same direction $\hat{r}$
$$E\times 4\pi r^2 = \frac{\rho\times \frac{4}{3}\pi R^3}{\epsilon_0}$$
$$\vec{E} = \frac{\rho\times R^3}{ 3 \epsilon_0 r^2}\hat{r}$$
Now, For $r< R, Q_{enc} = \rho \times \frac{4}{3}\pi r^3$
$$E\times 4\pi r^2 = \frac{\rho \times \frac{4}{3}\pi r^3}{\epsilon_0}$$
$$E = \frac{\rho \times r}{3\epsilon_0}\hat{r}$$
Hence,
For $r< R, \vec{E}=\frac{\rho \times r}{3\epsilon_0}\hat{r}$
For $r\ge R, \vec{E}=\frac{\rho\times R^3}{ 3 \epsilon_0 r^2}\hat{r}$
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