CBSE 2025, Set 1

Question

Two small solid metal balls A and B of radii $R$ and $2R$ having charge densities $2\sigma$ and $3\sigma$ respectively are kept far apart. Find the charge densities on A and B after they are connected by a conducting wire.

Edvaya · Accepted Answer

The correct answer is Subjective. **Initial charges:** $$Q_A = 4\pi R^2 	imes 2\sigma = 8\pi R^2\sigma$$ $$Q_B = 4\pi (2R)^2 	imes 3\sigma = 48\pi R^2\sigma$$ **Total charge (conserved):** $$Q = Q_A + Q_B = 56\pi R^2\sigma$$ **After connection:** When connected by a conducting wire, charge redistributes until both balls are at the **same potential**: $$V_A = V_B \implies \frac{Q_A'}{4\pi\varepsilon_0 R} = \frac{Q_B'}{4\pi\varepsilon_0 (2R)}$$ $$\frac{Q_A'}{R} = \frac{Q_B'}{2R} \implies Q_A' = \frac{Q_B'}{2}$$ **Using charge conservation:** $$Q_A' + Q_B' = 56\pi R^2\sigma$$ $$\frac{Q_B'}{2} + Q_B' = 56\pi R^2\sigma \implies \frac{3Q_B'}{2} = 56\pi R^2\sigma$$ $$Q_B' = \frac{112\pi R^2\sigma}{3}$$ $$Q_A' = \frac{56\pi R^2\sigma}{3}$$ **Final charge densities:** $$\sigma_A' = \frac{Q_A'}{4\pi R^2} = \frac{56\pi R^2\sigma/3}{4\pi R^2} = \boxed{\frac{14\sigma}{3}}$$ $$\sigma_B' = \frac{Q_B'}{4\pi (2R)^2} = \frac{112\pi R^2\sigma/3}{16\pi R^2} = \boxed{\frac{7\sigma}{3}}$$ > **Key principle:** When two conductors are connected, charge redistributes until they reach the same electric potential, not the same charge density. The surface charge density is inversely proportional to radius for a sphere: $\sigma \propto 1/R$.

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Physics > Electrostatics And Current Electricity > Electrostatic Potential And Capacitance

Keep Practicing

Edvaya Target

Current Question

Community Stats

Track Your Progress

Quick Actions

We Value Your Privacy