Question

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

Hint:

When two conductors are connected by a wire, they share charges until their potentials become equal. This principle is fundamental in electrostatics and is used to determine the final charges and potentials in connected systems.

Answer 1

Final Potential of Shell A
Initially, the charge on shell A is $Q_A = \sigma (4\pi R^2)$.
Initially, the charge on shell B is $Q_B = \sigma (4\pi (2R)^2) = \sigma (16\pi R^2)$. When the two shells are connected by a wire, they will have the same potential, say $V$.
Let the final charges on A and B be $Q'_A$ and $Q'_B$ respectively. The potential of a spherical shell with charge Q and radius R is $V = \frac{kQ}{R}$ where $k = \frac{1}{4\pi \epsilon_0}$. Since they are connected by a wire, their potentials become equal: $$V_A = V_B$$ $$\frac{kQ'_A}{R} = \frac{kQ'_B}{2R}$$ $$Q'_A = \frac{Q'_B}{2}$$ $$Q'_B = 2Q'_A$$ Also, the total charge is conserved: $$Q'_A + Q'_B = Q_A + Q_B$$ $$Q'_A + 2Q'_A = \sigma (4\pi R^2) + \sigma (16\pi R^2)$$ $$3Q'_A = \sigma (20\pi R^2)$$ $$Q'_A = \frac{20\pi R^2 \sigma}{3}$$ The final potential of shell A is: $$V_A = \frac{kQ'_A}{R} = \frac{1}{4\pi \epsilon_0} \frac{20\pi R^2 \sigma}{3R} = \frac{20 \pi R^2 \sigma}{12 \pi \epsilon_0 R} = \frac{5R\sigma}{3\epsilon_0}$$ Therefore, the final potential of shell A is $\frac{5R\sigma}{3\epsilon_0}$. Final Answer: The final answer is $\boxed{\frac{5R\sigma}{3\epsilon_0}}$