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 the sphere. The ratio $R_{2}/R_{1}$ is
A) $\frac{\sigma}{\rho_{0}}$
B) $\sqrt{\frac{2\sigma}{\rho_{0}}}$
C) $\sqrt{\frac{\rho_{0}}{2\sigma}}$
D) $\frac{\rho_{0}}{\sigma}$
Explanation
(c) The total positive charge on the inner solid sphere
$\int_{0}^{R_{1}}{4\pi\ r^{2}\ dr\ .}\frac{\rho_{0}}{r} = \ + 4\pi\frac{R_{1}^{2}}{2}\rho_{0}$
The total negative charge on the outer shell $= - 4\pi R_{2}^{2}\sigma$. But both are equal in magnitude as the sum is zero.
$\frac{R_{2}^{2}}{R_{1}^{2}} = \frac{\rho_{0}}{2\sigma}\ \ \ \therefore\frac{R_{2}}{R_{1}} = \ \sqrt{\frac{\rho_{0}}{2\sigma}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.