Question

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

$\rho= kr ^{2}$
$E\left(r=\frac{R}{2}\right)=\frac{1}{8} E(r=R)$
$\frac{ q _{\text{ enclosed}}}{4 \pi s _{ o }( R / 2)^{2}}=\frac{1}{8} \frac{ Q }{4 \pi \varepsilon_{0} R ^{2}}$
$32 q_{\text {enclosed }}=Q$
$q_{\text{enclosed}} =\int\limits_{0}^{n / 2} kr ^{2} 4 \pi r ^{2} dr =\frac{4 \pi k }{( a +3)}\left(\frac{ R }{2}\right)^{( a +3)}$
$Q =\frac{4 \pi k }{( a +3)} R ^{(2+3)}$
$\frac{ Q }{ q _{\text {andeod }}}=2^{2+3}$
$2^{2+3}=32$
$a =2$