Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho \left(r\right)=k{r}^a$, 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 $r=R$, find the value of $a$.

Answer 1

Given that,

$\rho =k{r}^a$

$\Rightarrow E(r=\frac{R}{2})=\frac{1}{8}E(r=R)$

$\Rightarrow \frac{q_{enclosed}}{4\pi {\epsilon }_0(\frac{R}{2}{)}^2}=\frac{1}{8}\frac{Q}{4\pi {\epsilon }_0{R}^2}$

On simplifying we get,

$\Rightarrow 32q_{enclosed}=Q$

$\Rightarrow q_{enclosed}=\underset{0}{\overset{\frac{R}{2}}{\int }}k{r}^{a}4\pi r^{2}dr$

$\Rightarrow q_{enclosed}=\frac{4\pi k}{(a+3)}(\frac{R}{2}{)}^{(a+3)}$

$\Rightarrow Q=\frac{4\pi k}{(a+3)}{R}^{(a+3)}$

$\Rightarrow \frac{Q}{q_{enclosed}}=2^{a+3}$

$\Rightarrow 2^{a+3}=32$

$\Rightarrow a=2$