Two conducting spherical shells A and B of radii R and $\displaystyle 2$ R are kept far apart and charged to the same charge density $\displaystyle \sigma$. They are connected by a wire. Obtain an expression for final potential of shell A.

Charge on spherical shell A: $\displaystyle q_{A}=4 \pi R^{2} \sigma$ Charge on spherical shell B: $\displaystyle q_{B}=4 \pi(2 R)^{2} \sigma=16 \pi R^{2} \sigma$ After connecting by a wire, their potentials will become equal after sharing of charge: $\displaystyle \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{B}^{\prime}}{2 R}$ $\displaystyle q_{B}^{\prime}=2 q_{A}^{\prime}$ From conservation of charge: $\displaystyle q_{A}+q_{B}=q_{A}^{\prime}+q_{B}^{\prime}$ $\displaystyle 4 \pi R^{2} \sigma+16 \pi R^{2} \sigma=3 q_{A}^{\prime}$ $\displaystyle q_{A}^{\prime}=\frac{20 \pi R^{2} \sigma}{3}$ Final potential of shell A: $\displaystyle V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}^{\prime}}{R}$ $\displaystyle V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{20 \pi R^{2} \sigma}{3 R}$ $\displaystyle V_{A}=\frac{5 \sigma R}{3 \varepsilon_{0}}$Therefore the potential of shell A: $\displaystyle V_{A}=V_{\text {common }}=\frac{q_{A}+q_{B}}{C_{A}+C_{B}}$ $\displaystyle =\frac{4 \pi R^{2} \sigma+16 \pi R^{2} \sigma}{4 \pi \varepsilon_{0} R+4 \pi \varepsilon_{0}(2 R)}$ $\displaystyle =\frac{5 \sigma R}{3 \varepsilon_{0}}$