Two metal spheres of radius R and 3R have same surface charge density σ . If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ₁ and σ₂ , respectively. The ratio σ₁/σ₂ is

Solution

Physics sir · 4 steps

Touching makes the potentials equal, and on a sphere that forces σ to go as 1/r.

1. Equal potential. Metal spheres in contact form one conductor, so charge flows until both are at the same potential. For a sphere, V = kQ/r.
2. Write V with sigma. Put Q = σ × 4π r² into V = kQ/r. That gives V = 4π k σ r, so equal V means σ r is the same for both spheres.

   σ₁ R = σ₂ (3R) ⇒ σ₁/σ₂ = 3R/R = 3

3. Check with charges. Total charge is σ × 4π(R² + 9R²) = 40πσ R², shared 1:3, so Q₁ = 10πσ R² and Q₂ = 30πσ R². Then σ₁ = 2.5σ and σ₂ = (30πσ R²)/(36π R²) = 5/6σ. Their ratio is 3.
4. So. The smaller sphere ends up with the larger surface charge density: σ₁/σ₂ = 3, option (d). The starting σ does not matter.

The idea to keep. Connected spheres share potential, so charge goes as r and surface charge density goes as 1/r. The smaller sphere always ends up with the higher σ.

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