Question

Let there be a spherically symmetric charge distribution with charge density varying as $\rho \left(r\right)={\rho }_o(\frac{5}{4}-\frac{r}{R})$ where $r$ is the distance from the origin. The electric field at a distance $r(r<R)$ from the origin is given by

• A. $\frac{4\pi {\rho }_{o}r}{3{ϵ}_o}(\frac{5}{3}-\frac{r}{R})$
• B. $\frac{{\rho }_{o}r}{4{ϵ}_o}(\frac{5}{3}-\frac{r}{R})$
• C. $\frac{4{\rho }_{o}r}{3{ϵ}_o}(\frac{5}{4}-\frac{r}{R})$
• D. $\frac{{\rho }_{o}r}{3{ϵ}_o}(\frac{5}{3}-\frac{r}{R})$

Answer 1

The correct option is B $\frac{{\rho }_{o}r}{4{ϵ}_o}(\frac{5}{3}-\frac{r}{R})$

Apply Shell theorem, the total charge upto distance $r$ can be calculated as follows,
$dq=4\pi r^2.dr.\rho$
$=4\pi r^2{\rho }_o[\frac{5}{4}-\frac{r}{R}]dr=4\pi {\rho }_o[\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr]$
$\Rightarrow \int dq=q=4\pi {\rho }_o{\int }_0^r(\frac{5}{4}{r}^{2}dr-\frac{r^3}{R}dr)$
$\Rightarrow q=4\pi {\rho }_o(\frac{5}{4}\frac{r^3}{3}-\frac{1}{R}\frac{r^4}{4})$
Therefore, electric field
$E=\frac{kq}{r^2}=\frac{1}{4\pi {ϵ}_o}\frac{1}{r^2}.4\pi {\rho }_o[\frac{5}{4}\left(\frac{r^3}{3}\right)-\frac{r^4}{4R}]$
$E=\frac{{\rho }_{o}r}{4{ϵ}_o}[\frac{5}{3}-\frac{r}{R}]$