Question

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 $\frac{R_{2}}{R_{1}}$ is

• A. $\frac{\sigma}{\rho_{0}}$
• B. $\sqrt{2\sigma/ \rho_{0}} $
• C. $\sqrt{\rho_{0} /\left(2\sigma\right)} $
• D. $\frac{\sigma }{\rho _{0}} $

Answer 1

Answer: (C)

For solid sphere of radius $R_{1}$
$q_{1}=\int\limits_{0}^{R_{1}} 4 \pi r^{2} d r \rho $
$=\int\limits_{0}^{R_{1}} 4 \pi r^{2} d r \frac{\rho_{0}}{r}$
$q_{1}=4 \pi \frac{R_{1}^{2}}{2} \rho_{0}$
$q_{2}=-4 \pi R_{2}^{2} \sigma$
$q_{1}+q_{2}=0 $
$4 \pi \frac{R_{1}^{2} \,\rho_{0}}{2}-4 \pi R_{2}^{2} \sigma=0$
$\left(\frac{R_{1}}{R_{2}}\right)^{2}=\frac{2 \sigma}{\rho_{0}}$
$\frac{R_{2}}{R_{1}}=\sqrt{\frac{\rho_{0}}{2 \sigma}}$