Question

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

From Gauss theorem,
$E \propto \frac{q}{r^{2}} (q =$ charge enclosed)
Therefore,
$\frac{E_{2}}{E_{1}}=\frac{q_{2}}{q_{1}}\times \frac{r_{1}^{2}}{\text{r}_{2}^{2}}$
$8=\frac{\int\limits _{0}^{R} \left(4 \pi r^{2}\right) k r^{a} d r}{\int\limits _{0}^{R / 2} \left(4 \pi r^{2}\right) k r^{a} d r}\times \left(\frac{\frac{R}{2}}{R}\right)^{2}$
$8\times 4=2^{\left(a + 3\right)}$
Solving this equation we get, $a=2$