Q. Let $P(r) = \frac{Q}{\pi R^{4}}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q.$ For a point $'p'$ inside the sphere at distance $r_{1}$ from the centre of the sphere, the magnitude of electric field is
A) 0
B) $\frac{Q}{4\pi \in_{0}r_{1}^{2}}$
C) $\frac{Qr_{1}^{2}}{4\pi \in_{0}R^{4}}$
D) $\frac{Qr_{1}^{2}}{3\pi \in_{0}R^{4}}$
Explanation
(c) Charge enclosed by the sphere of radius $r_{1}$ is
$Q^{'} = \int_{0}^{r_{1}}{\delta\left( 4\pi r^{2} \right)dr =}\int_{0}^{r_{1}}{\frac{Q}{\pi R^{4}}r(4\pi r^{2})dr}$
$Q^{'} = \frac{Q}{\pi R^{4}}(4\pi)\left\lbrack \frac{r^{4}}{4} \right\rbrack_{0}^{r_{1}} = \frac{Qr_{1}^{4}}{R^{4}}$
$\therefore E.F.$ at $p$ is $E = \frac{1}{4\pi\varepsilon_{o}}.\frac{Q'}{r_{1}^{2}} = \frac{1}{4\pi\varepsilon_{o}}.\frac{Qr_{1}^{4}/R^{4}}{r_{1}^{2}}$
$E = \frac{Qr_{1}^{2}}{4\pi\varepsilon_{o}R^{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.