Question

A spherically symmetric charge distribution is characterised by a charge density having the following variation : $\rho\left(r\right)=\rho_{0}\left(1-\frac{r}{R}\right)$ for $r < R$ $\rho \left(r\right)=0 \,$ 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. $\frac{\rho_{0}}{4\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)$
• B. $\frac{\rho_{0}}{\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)$
• C. $\frac{\rho_{0}}{3\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)$
• D. $\frac{\rho_{0}}{12\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)$

Answer 1

The Correct Option is B

$\oint \varepsilon . d s=\frac{\Sigma a}{\varepsilon_{0}}$
$\varepsilon\left(4 \pi r^{2}\right)=\rho \cdot \int_{0}^{r}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r$
$4 \pi \rho_{0}\left[\frac{r^{3}}{3}-\frac{r^{2}}{4 R}\right]_{0}^{q}=\varepsilon\left(4 \pi r^{2}\right)$
$\varepsilon=\frac{\rho_{0}}{\varepsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4 R}\right)$