Chapter 2 Electrostatics: Problem 2.12

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,

$$\oint \overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{enc}}{{ϵ}_0}=\frac{\rho \times \frac{4}{3}\pi R^3}{{ϵ}_0}$$ 

Since the charge distribution is symmetrical, $\overrightarrow{E}$ and $\overrightarrow{da}$ point in the same direction $\hat{r}$ 

$$E\times 4\pi r^2=\frac{\rho \times \frac{4}{3}\pi R^3}{{ϵ}_0}$$ 

$$\overrightarrow{E}=\frac{\rho \times R^3}{3{ϵ}_0{r}^2}\hat{r}$$ 

Now, For $r<R,Q_{enc}=\rho \times \frac{4}{3}\pi r^3$ 

$$E\times 4\pi r^2=\frac{\rho \times \frac{4}{3}\pi r^3}{{ϵ}_0}$$ 

$$E=\frac{\rho \times r}{3{ϵ}_0}\hat{r}$$

Hence,

For $r<R,\overrightarrow{E}=\frac{\rho \times r}{3{ϵ}_0}\hat{r}$ 

For $r\ge R,\overrightarrow{E}=\frac{\rho \times R^3}{3{ϵ}_0{r}^2}\hat{r}$ 

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