Question

A spherical charge distribution with volume charge density varying as $\rho \left(r\right)={\rho }_0[\frac{5}{4}-\frac{r}{R}]$, up to $r=R$ and $\rho \left(r\right)=0$ for $r>R$. Here, $r$ is the distance from the centre of sphere. The electric field at a distance $r_o(r_o<R)$ from the centre of sphere will be

• A. $\frac{{\rho }_0{r}_o}{3{ϵ}_0}[\frac{5}{4}-\frac{r_o}{R}]$
• B. $\frac{4\pi {\rho }_0{r}_o}{3{ϵ}_0}[\frac{5}{3}-\frac{r_o}{R}]$
• C. $\frac{{\rho }_0{r}_o}{4{ϵ}_0}[\frac{5}{3}-\frac{r_o}{R}]$
• D. $\frac{4{\rho }_0{r}_o}{3{ϵ}_0}[\frac{5}{4}-\frac{r_o}{R}]$

Answer 1

The correct option is C $\frac{{\rho }_0{r}_o}{4{ϵ}_0}[\frac{5}{3}-\frac{r_o}{R}]$

Let us take a thin shell of thickness $dr$ at distance $r$ from centre of sphere.

Also,
$dq=\rho dv$
$\Rightarrow dq={\rho }_0[\frac{5}{4}-\frac{r}{R}]4\pi r^{2}dr$
Further, net charge enclosed within the sphere of radius $r$.
$q=\int dq=4\pi {\rho }_0{\int }_0^{r_o}[\frac{5r^2}{4}dr-\frac{r^3}{R}dr]$
$\Rightarrow q=4\pi {\rho }_0[\frac{5}{4}\left(\frac{r_o^3}{3}\right)-\frac{1}{R}\left(\frac{r_o^4}{4}\right)]$

Applying Gauss's law at spherical surface,
$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{q_{in}}{{ϵ}_0}$
$EA=\frac{q}{{ϵ}_0}$
$\left(4\pi r_o^2\right)E=\frac{1}{{ϵ}_0}4\pi {\rho }_0[\frac{5}{4}\left(\frac{r_o^3}{3}\right)-\frac{1}{R}\left(\frac{r_o^4}{4}\right)]$
$\left(4\pi r_o^2\right)E=\frac{4\pi {\rho }_0{r}_o^3}{4{ϵ}_0}[\frac{5}{3}-\frac{r_o}{R}]$
$\therefore E=\frac{{\rho }_o{r}_o}{4{ϵ}_0}[\frac{5}{3}-\frac{r_o}{R}]$