JEE Main 2022 · 25 July · evening · Single correct · +4 / −1
Capacitance of an isolated conducting sphere of radius R₁ becomes n times when it is enclosed by a concentric conducting sphere of radius R₂ connected to earth. The ratio of their radii ( R₂/R₁ ) is :
- (a) n/(n - 1)
- (b) 2n/(2n + 1)
- (c) (n + 1)/n
- (d) (2n + 1)/n
Answer and solution
Answer (a) n/(n - 1)
Solution
Physics sir · 4 steps
Write both capacitances, divide them, and solve that one ratio for R₂/R₁.
- Isolated sphere. Alone, the sphere of radius R₁ has C = 4πε₀ R₁.
- Earthed outer shell. With the earthed concentric shell of radius R₂ around it, the pair is a spherical capacitor: C' = 4πε₀ (R₁ R₂)/(R₂ - R₁).
n = C'/C = R₂/(R₂ - R₁)
- Solve for the ratio. Put x = R₂/R₁, so n = x/(x - 1). Then nx - n = x, which gives x(n - 1) = n.
- So. R₂/R₁ = n/(n - 1), which is option (a).
The idea to keep. An earthed shell close around a conductor raises its capacitance by the factor R₂/(R₂ - R₁). The closer the shell, the bigger the boost.
Answer from the NTA final answer key. Solution by PrepWiser.