Two isolated metallic solid spheres of radii R and 2 R are charged such that both have same charge density σ. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ′. The ratio σ′/σ is :

Solution

Physics sir · 4 steps

Connected spheres reach the same potential, so they share the total charge in proportion to their radii.

1. Charge before. Charge is σ × area. The small sphere has q₁ = σ · 4π R² and the big one has q₂ = σ · 4π (2R)² = 16π R² σ. The total is 20π R² σ, and the wire conserves it.
2. Equal potential. After connecting, charge flows until both spheres are at the same potential. Since V = kQ/r, equal V means Q ∝ r, so Q₁ : Q₂ = R : 2R = 1 : 2.

   Q₂ = 2/3 × 20π R² σ = 40/3π R² σ

3. New density. Divide by the big sphere's area: σ' = (40π R² σ / 3)/(16π R²) = 5/6σ.
4. So. σ'/σ = 5/6, which is option (b).

The idea to keep. When two conductors are joined by a wire, set their potentials equal and keep the total charge fixed. For spheres, charge goes as r, so surface density goes as 1/r.

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