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

Answer: 2

From Gauss theorem,
$E\propto \frac{q}{r^2}(q=$ charge enclosed)
Therefore,
$\frac{E_2}{E_1}=\frac{q_2}{q_1}\times \frac{r_1^2}{{\text{r}}_2^2}$
$8=\frac{\underset{0}{\overset{R}{\int }}\left(4\pi r^2\right)k{r}^{a}dr}{\underset{0}{\overset{R/2}{\int }}\left(4\pi r^2\right)k{r}^{a}dr}\times {\left(\frac{\frac{R}{2}}{R}\right)}^2$
$8\times 4=2^{\left(a+3\right)}$
Solving this equation we get, $a=2$