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. $\frac{{\rho }_0}{4{ϵ}_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$
• B. $\frac{{\rho }_0}{{ϵ}_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$
• C. $\frac{{\rho }_0}{3{ϵ}_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$
• D. $\frac{{\rho }_0}{12{ϵ}_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$

Answer 1

Answer: (B)

$\oint \epsilon .ds=\frac{\Sigma a}{{\epsilon }_0}$
$\epsilon \left(4\pi r^2\right)=\rho \cdot {\int }_0^r\left(1-\frac{r}{R}\right)4\pi r^{2}dr$
$4\pi {\rho }_0{\left[\frac{r^3}{3}-\frac{r^2}{4R}\right]}_0^q=\epsilon \left(4\pi r^2\right)$
$\epsilon =\frac{{\rho }_0}{{\epsilon }_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$