and $\sigma_{1}=\sigma_{2} \quad$ [Surface charge density are cqual $]$
$\therefore \frac{q_{1}}{r \pi r^{2}}=\frac{q_{2}}{4 \pi R^{2}}$
So, $q_{1}=\frac{Q r^{2}}{R^{2}+r^{2}}$ and $q_{2}=\frac{Q R^{2}}{R^{2}+r^{2}}$
Now, potential, $V=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}}{r}+\frac{q_{2}}{R}\right]$
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{Q r}{R^{2}+r^{2}}+\frac{Q R}{R^{2}+r^{2}}\right]$
$=\frac{Q(R+r)}{R^{2}+r^{2}} \frac{1}{4 \pi \varepsilon_{0}}$



