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 radial 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 (answer upto two decimal places)

Answer 1

Given $\rho ={Kr}^a$

$E(r=\frac{r}{2})=\frac{1}{8}E(r=R)$
$\frac{q_{enclose{d}_{R/2}}}{4\pi {ϵ}_0(R/2{)}^2}=\frac{1}{8}\frac{Q}{4\pi {ϵ}_0(R{)}^2}$
$⟹32q_{enclose{d}_{R/2}}=Q...\left(1\right)$

$q_{enclosed}=\int \rho dv=\int K{r}^{a}4\pi r^{2}dr$
$q_{enclose{d}_{R/2}}={\int }_0^{R/2}K{r}^{a}4\pi r^{2}dr=\frac{4K\pi }{(a+3)}(\frac{R}{2}{)}^{(a+3)}$
$Q=q_{enclose{d}_R}={\int }_0^{R}K{r}^{a}4\pi r^{2}dr=\frac{4K\pi }{(a+3)}(R{)}^{(a+3)}$
$\frac{Q}{q_{enclose{d}_{R/2}}}=2^{a+3}$
putting in equation (1),
$2^{(a+3)}=32$
$\therefore a=2$