Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho =k{r}^a$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, the value of $a$ is

• A. $3$
• B. $5$
• C. $2$
• D. $7$

Answer 1

The correct option is C $2$

Using Gauss's law, we have
$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{1}{{ϵ}_0}\int \left(\rho dv\right)=\frac{1}{{ϵ}_0}{\int }_0^{R}k{r}^a\times 4\pi r^{2}dr$
or $E\times 4\pi R^2=\left(\frac{4\pi k}{{ϵ}_0}\right)\frac{R^{(a+3)}}{(a+3)}$
For $r=\frac{R}{2}.E_2=\frac{k{\left(\frac{R}{2}\right)}^{a+1}}{{ϵ}_0(a+3)}$
Given, $E_2=\frac{E_1}{8}$
or $\frac{k{\left(\frac{R}{2}\right)}^{a+1}}{{ϵ}_0(a+3)}=\frac{1}{8}\frac{k{R}^{(a+1)}}{{ϵ}_0(a+3)}$
$\therefore 2^{\frac{1}{a+1}}=\frac{1}{8}$
or $a=2$.