1. R₁/R₂ 2. R₂/R₁ 3. √(R₁/R₂) 4. R₁²/R₂²

Answer: 2. R₂/R₁ Key idea: connected conductors are at the same potential. V=kQ/R=kσ4π R²/R=4π kσ R . Equal V gives σ₁R₁=σ₂R₂ , so σ₁/σ₂=R₂/R₁ . Option 2.

Topic: Electric potential, potential difference, equipotential surfaces· Electrostatics

Two charged spherical conductors of radius R_1 and R_2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ_1/σ_2) is :

1. \(\frac{R_1}{R_2}\)

2. \(\frac{R_2}{R_1}\)

3. \(\sqrt{\frac{R_1}{R_2}}\)

4. \(\frac{R_1^2}{R_2^2}\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 2. \(\frac{R_2}{R_1}\)

Key idea: connected conductors are at the same potential.

\(V=\dfrac{kQ}{R}=\dfrac{k\sigma4\pi R^2}{R}=4\pi k\sigma R\). Equal V gives \(\sigma_1R_1=\sigma_2R_2\), so \(\dfrac{\sigma_1}{\sigma_2}=\dfrac{R_2}{R_1}\). Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

Two charged spherical conductors of radius R_1 and R_2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ_1/σ_2) is :

1 answer

Your answer