Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho =k{r}^4$, 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

Answer: 2

$\rho =k{r}^2$
$E\left(r=\frac{R}{2}\right)=\frac{1}{8}E(r=R)$
$\frac{q_{\text{ enclosed}}}{4\pi s_o(R/2{)}^2}=\frac{1}{8}\frac{Q}{4\pi {\epsilon }_0{R}^2}$
$32q_{\text{enclosed }}=Q$
$q_{\text{enclosed}}=\underset{0}{\overset{n/2}{\int }}k{r}^{2}4\pi r^{2}dr=\frac{4\pi k}{(a+3)}{\left(\frac{R}{2}\right)}^{(a+3)}$
$Q=\frac{4\pi k}{(a+3)}{R}^{(2+3)}$
$\frac{Q}{q_{\text{andeod }}}=2^{2+3}$
$2^{2+3}=32$
$a=2$