Physics - Modern Physics Question with Solution | TestHub

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, $\mathrm{dV}=4\pi {\mathrm{r}}^2\mathrm{dr}$ 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, $\begin{array}{l}\begin{array}{rl}{\mathrm{Q}}_{\mathrm{S}}& ={\int }_0^{{\mathrm{R}}_1}\mathrm{d}\mathrm{Q}=4\pi {\rho }_0{\int }_0^{{\mathrm{R}}_1}\mathrm{rdr}\\ & =4\pi {\rho }_0\frac{{\mathrm{R}}_1^2}{2}=2\pi {\rho }_0{\mathrm{R}}_1^2\end{array}\\ \therefore {\mathrm{Q}}_{\mathrm{S}}=2\pi {\rho }_0{\mathrm{R}}_1^2...(1)\end{array}$ Now, the total charge on the hollow sphere, ${\mathrm{Q}}_{\mathrm{h}}=-\left(4\pi {\mathrm{R}}_2^2\right)\sigma$ ...(2) By question, $Q_s+Q_h=0$ $\begin{array}{l}\Rightarrow 2\pi {\rho }_0{\mathrm{R}}_1^2=4\pi {\mathrm{R}}_2^2{}^{\mathrm{h}}\\ \Rightarrow {\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)}^2=\frac{{\rho }_0}{2\sigma }\Rightarrow \frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}=\sqrt{\frac{{\rho }_0}{2\sigma }}\end{array}$