Consider two charged metallic spheres S₁ and S₂ of radii R₁ and R₂, respectively. The electric fields E₁ (on S₁) and E₂ (on S₂) on their surfaces are such that E₁/E₂ = R₁/R₂. Then the ratio V₁ (on S₁) / V₂ (on S₂) of the electrostatic potentials on each sphere is :

Solution

Physics sir · 4 steps

On a charged sphere's surface, V is just E times R, so multiply the two ratios together.

1. Field and potential. Outside a charged metallic sphere, the charge Q acts as if it sits at the centre. On the surface, E = kQ/R² and V = kQ/R.
2. Link V to E. Divide the two expressions. V/E = R, so the surface potential is V = ER for any sphere, whatever its charge.

   V₁/V₂ = (E₁ R₁)/(E₂ R₂) = E₁/E₂ × R₁/R₂

3. Put in the ratio. We are given E₁/E₂ = R₁/R₂. So V₁/V₂ = R₁/R₂ × R₁/R₂.
4. So. V₁/V₂ = (R₁/R₂)², which is option (a).

The idea to keep. For a charged sphere, the surface values are tied by V = ER. Whenever a question gives you a ratio of surface fields and radii, multiply them to get the potential ratio.

Answer from the NTA final answer key. Solution by PrepWiser.