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 $r=\frac{R}{2}is\frac{1}{8}$ times that at r = R, find the value of a.

Answer 1

Let $E$ be the electric field due to solid sphere at any point $r$.
Let $q_{in}$ be the charge enclosed in the solid sphere of radius $r=\frac{R}{2}$
From the given condition
$E(r=\frac{R}{2})=\frac{1}{8}E(r=R)$
$⟹\frac{k{q}_{in}}{(\frac{R}{2}{)}^2}=\frac{1}{8}\frac{kQ}{(R{)}^2}$
$⟹\frac{Q}{q_{in}}=32$-----(i)
Let us consider a elementary solid sphere of radius $r$ inside the solid sphere of radius $R$
Consider the thickness of this small sphere be $dr$.Let $dV$ be the volume of this small sphere.
Volume $dV$ can be given by
$dV=4\pi r^{2}dr$----(ii)
The charge $dq$ on this volume can be given as
$dq=\rho 4\pi r^{2}dr$
$⟹dq=K{r}^{a}4\pi r^{2}dr$
$⟹dq=4\pi K{r}^{(a+2)}dr$-----(iii)
To get $q_{in}$ we need to integrate $dr$ from 0 to $\frac{R}{2}$
$⟹{\int }_0^{q_{in}}=4\pi K{\int }_0^{\frac{R}{2}}{r}^{a+2}dr$
$q_{in}=\frac{4\pi K{R}^{a+3}}{(a+3)2^{a+3}}$
IIy on integrating equ-(iii) from 0 to $R$ we get
$Q=\frac{4\pi K{R}^{a+3}}{a+3}$
On putting values of $Q$ and $q_{in}$ in (i) we get
$2^{a+3}=32$
$\therefore a=2$