Physics - Electrostatics Question with Solution | TestHub

Due to charge $-q$ at the center of the shell, a charge $q$ will be induced on the inner surface and $-q$ on the outer surface. The charge on outer surface will become $Q-q$. Hence surface charge densities will be $\frac{\mathrm{q}}{4\pi {\mathrm{r}}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4\pi {\mathrm{r}}_2^2}$