Q. A solid sphere of radius $R_{1}$and volume charge density $\rho = \frac{\rho_{0}}{r}$ is enclosed by a hollow sphere of radius $R_{2}$ with negative surface charge density$\sigma$, such that the total charge in the system is zero,$\ \rho_{0}$ is a positive constant and $r$ is the distance from the centre of sphere. The ratio $\frac{R_{2}}{R_{2}}$ is
A) $\frac{\sigma}{\rho_{0}}$
B) $\sqrt{2\sigma}/\rho_{0}$
C) $\sqrt{\rho_{0}}/2\sigma$
D) $\frac{\rho_{0}}{\sigma}$
Explanation
(c) For solid sphere of radius $R_{1}$
$q_{1} = \int_{0}^{R_{1}}{4\pi r^{2}\ dr\rho}$
$\ \ \ \ \ = \int_{0}^{R_{1}}{4\pi r^{2}d\ r\frac{\rho_{0}}{r}\ }$
$q_{1} = 4\pi\frac{R_{1}^{2}}{2}\rho_{0}$
$q_{2} = - 4\pi R_{2}^{2}\sigma$
$q_{1} + q_{2} = 0$
$4\pi\frac{R_{1}^{2}\rho_{0}}{2} - 4\pi R_{2}^{2}\sigma = 0$
$\left( \frac{R_{1}}{R_{2}} \right)^{2} = \frac{2\sigma}{\rho_{0}}$
$\frac{R_{2}}{R_{1}} = \sqrt{\frac{\rho_{0}}{2\sigma}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.