A solid sphere of radius $R_{1}$ and volume charge density $\rho=\frac{\rho_{0}}{\mathrm{r}}$ 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 $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ is

• A

  $\frac{\sigma}{\rho_{0}}$
• B

  $\sqrt{2 \sigma / \rho_{0}}$
• Correct answer
  C

  $\sqrt{\rho_{0} /(2 \sigma)}$
• D

  $\frac{\rho_{0}}{\sigma}$

Solution

Let us first calculate the total charge on the soild sphere. Let us consider a concentric sphere of radius $r$ and thickness dr. Then volume of thesphere, $\mathrm{d} \mathrm{V}=4 \pi \mathrm{r}^{2} \mathrm{dr}$ Given, the volume charge density of the sphere $=\rho=\frac{\rho_{0}}{\mathrm{r}}$
$\therefore$ Charge on this sphere, $\mathrm{dQ}=\rho \cdot \mathrm{dV}=\frac{\rho_{0}}{\mathrm{r}} \cdot 4 \pi \mathrm{r}^{2} \mathrm{dr}=4 \pi \rho \cdot \mathrm{rdr}$ $\therefore$ Total charge on the whole solid sphere, $\begin{array}{l} \begin{aligned} \mathrm{Q}_{\mathrm{S}} &=\int_{0}^{\mathrm{R}_{1}} \mathrm{~d} \mathrm{Q}=4 \pi \rho_{0} \int_{0}^{\mathrm{R}_{1}} \mathrm{rdr} \\ &=4 \pi \rho_{0} \frac{\mathrm{R}_{1}^{2}}{2}=2 \pi \rho_{0} \mathrm{R}_{1}^{2} \end{aligned} \\ \therefore \mathrm{Q}_{\mathrm{S}}=2 \pi \rho_{0} \mathrm{R}_{1}^{2} ...(1) \end{array}$ Now, the total charge on the hollow sphere, $\mathrm{Q}_{\mathrm{h}}=-\left(4 \pi \mathrm{R}_{2}^{2}\right) \sigma$ ...(2) By question, $Q_{s}+Q_{h}=0$ $\begin{array}{l} \Rightarrow 2 \pi \rho_{0} \mathrm{R}_{1}^{2}=4 \pi \mathrm{R}_{2}^{2}{ }^{\mathrm{h}} \\ \Rightarrow\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}=\frac{\rho_{0}}{2 \sigma} \Rightarrow \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\sqrt{\frac{\rho_{0}}{2 \sigma}} \end{array}$

Hence, the correct answer is option: 'C'.