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 $r=R/2$ is 1/8 times that at $r=R$, find the value of $a$.

Answer 1

Charge enclosed within the sphere upto radius $r$ is given by:$$q={\int }_0^{r}k{r}^{a}4\pi r^{2}dr$$
$\Rightarrow q=4\pi k\frac{r^{3+a}}{3+a}{|}_0^r=\frac{4\pi k}{(3+a)}{r}^{3+a}........\left(1\right)$

At $r=R/2$ electric field is given as , $E_1=\frac{q}{4\pi {\epsilon }_0(R/2{)}^2}$

Substituting $\left(1\right)$ at $(r=\frac{R}{2})$ in the above equation,

$E_1=\frac{1}{4\pi {ϵ}_0}\times \frac{4\pi k}{(3+a)}\frac{(R/2{)}^{3+a}}{(R/2{)}^2}$

$\Rightarrow E_1=\frac{k}{{ϵ}_o(3+a)}{\left(\frac{R}{2}\right)}^{1+a}$

At $r=R$, electric field is given as ,$E_2=\frac{q}{4\pi {\epsilon }_0{R}^2}$

Substituting $\left(1\right)$ at $(r=R)$ in the above equation,

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{4\pi k}{(3+a)}\frac{(R{)}^{3+a}}{R^2}$

$\Rightarrow E_2=\frac{k}{{ϵ}_0(3+a)}{R}^{1+a}$

Given that , $E_2=\frac{E_1}{8}$

$\Rightarrow {\left(\frac{R}{2}\right)}^{1+a}=\frac{(R{)}^{1+a}}{8}$

$\Rightarrow 1+a=3$

$\Rightarrow a=2$

Accepted answer : $2$