The concentric spheres of radii \(R\) and \(r\) have positive charges \(q_{1}\) and \(q_{2}\) with equal surface charge densities. What is the electric potential at their common centre?

Potential at the centre due to sphere with radius \(R\),
\(V_{1}=\frac{q_{1}}{4 \pi \varepsilon_{0} R}\)
Potential at the centre due to sphere with radius \(r\),
\(V_{2}=\frac{q_{2}}{4 \pi \varepsilon_{0} r}\)
Given, surface charge densities for both spheres
\(\sigma=\sigma_{R}=\sigma_{r}\)
i.e.,
\(\text {i.e., } \frac{q_{1}}{4 \pi \varepsilon_{0} R^{2}} =\frac{q_{2}}{4 \pi \varepsilon_{0} r^{2}} \)
\( \Rightarrow q_{1} =\frac{R^{2}}{r^{2}} q_{2}\)
Total potential at the centre due to both spheres
\(\begin{aligned}
V &=V_{1}+V_{2} \\
&=\frac{q_{1}}{4 \pi \varepsilon_{0} R}+\frac{q_{2}}{4 \pi \varepsilon_{0} r}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{R}+\frac{q_{2}}{r}\right) \\
&=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{R^{2}}{r^{2}} \frac{q_{2}}{R}\right)+\frac{q_{2}}{r}=\frac{q_{2}}{4 \pi \varepsilon_{0}}\left(\frac{R+r}{r^{2}}\right) \\
&=\frac{q_{2}}{4 \pi r^{2} \varepsilon_{0}}(R+r)=\frac{\sigma}{\varepsilon_{0}}(R+r)
\end{aligned}\)