Q. A solid sphere of radius $R$ has a charge $\mathcal{Q}$ 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 $a.$
A) 2
B) 3
C) 2.5
D) 0.2
Explanation
(a) From Gauss’ theorem,
$E \propto \frac{q}{r^{2}}\ \ \ \ \ \ \ \ \ \ (q = charge\ enclosed)$
$\frac{E_{2}}{E_{1}} = \frac{q_{2}}{q_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}$
$8 = \frac{\int_{0}^{R}{\left( 4\pi r^{2} \right)kr^{a}dr}}{\int_{0}^{R/2}{\left( 4\pi r^{2} \right)kr^{a}dr}} \times \frac{\left( \frac{R}{2} \right)^{2}}{(R)^{2}}$
Solving this equation we get, $a = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.