Question

A spherically symmetric charge distribution is characterised by a charge density having the following variation:
$\rho(r) = \rho_0\Bigl(1 - \frac{r}{R}\Bigr)\quad \text{for }r<R,$
$\rho(r) = 0\quad \text{for }r \ge R.$
Where $r$ is the distance from the centre of the charge distribution and $\rho_0$ is a constant. The electric field at an internal point $(r<R)$ is:

• A. $\displaystyle \frac{\rho_0}{3\epsilon_0}\Bigl(\frac{r}{3}-\frac{r^2}{4R}\Bigr)$
• B. $\displaystyle \frac{\rho_0}{4\epsilon_0}\Bigl(\frac{r}{3}-\frac{r^2}{4R}\Bigr)$
• C. $\displaystyle \frac{\rho_0}{\epsilon_0}\Bigl(\frac{r}{3}-\frac{r^2}{4R}\Bigr)$
• D. $\displaystyle \frac{\rho_0}{12\epsilon_0}\Bigl(\frac{r}{3}-\frac{r^2}{4R}\Bigr)$

Answer 1

Answer: (C)

$\displaystyle \frac{\rho_0}{\epsilon_0}\Bigl(\frac{r}{3}-\frac{r^2}{4R}\Bigr)$

Answer 2

Gauss’s law approach

1. Compute the enclosed charge for $r<R$:

$$Q_{\rm enc}(r) = \int_0^r \rho_0\Bigl(1-\frac{r'}{R}\Bigr)\,4\pi r'^2\,dr' =4\pi\rho_0\Bigl[\frac{r^3}{3}-\frac{r^4}{4R}\Bigr].$$

2. Apply Gauss’s law:

$$E\cdot 4\pi r^2 = \frac{Q_{\rm enc}(r)}{\epsilon_0} \;\;\Longrightarrow\;\; E(r) =\frac{1}{4\pi\epsilon_0}\frac{Q_{\rm enc}(r)}{r^2} =\frac{\rho_0}{\epsilon_0}\Bigl(\frac{r}{3}-\frac{r^2}{4R}\Bigr).$$