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

For the small elemental shell considered of width $dr$ and radius r inside the sphere as shown in figure below, charge on the element is given as
 $dq=\rho dV\Rightarrow dq=k{r}^{\alpha }4\pi r^{2}dr$
 

 $\Rightarrow q=\int dq=\int k{r}^{\alpha }4\pi r^{2}dr\Rightarrow q=\frac{k4\pi r^{\alpha +3}}{\alpha +3}$
Electric field due to this charge at a distance r distance r inside the sphere is given as

$$\begin{gathered} E_r=\left[\frac{K4\pi r^{\alpha +3}}{\alpha +3}\right]\frac{1}{r^2}\Rightarrow E_R=\left[\frac{K4\pi r^{\alpha +3}}{\alpha +3}\right]\frac{1}{R^2} \\ \Rightarrow \frac{E_r}{E_R}=\frac{1}{8}=\frac{{(R/2)}^{\alpha +1}}{R^{\alpha +1}}\Rightarrow \alpha =2 \end{gathered}$$