Two charged spherical conductors of radii R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres ${\mathrm{\sigma }}_1/{\mathrm{\sigma }}_2$ is:

detailed solution

Correct option is B

Potential at the surface of spherical conductor of radius R carrying charge Q, VS=Q4πε0RLet Q1 and Q2 are the charges on two spherical conductors of radii R1 and R2 respectively. When these two charged spherical conductors connected by a wire, the potential at their surfaces becomes equal.∴ VS1=VS2Q14πε0R1=Q24πε0R2⇒Q1Q2=R1R2           ....(i)Surface charge density, σ= Charge  Area ∴σ1σ2=Q14πR12Q24πR22=Q1Q2×R22R12=R1R2×R22R12=R2R1