The distance between two plates of a capacitor is d and its capacitance is C₁, when air is the medium between the plates. If a metal sheet of thickness (2 d)/3 and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes C₂. The ratio C₂/C₁ is

Solution

Physics sir · 3 steps

A metal sheet carries no field inside it, so it just eats up gap: only d/3 of air is left.

1. Metal inside. Inside a conductor the electric field is zero, so there is no potential drop across the sheet. The sheet behaves like a stretch of the gap that is no longer there.
2. Effective gap. The air left between the plates has thickness d - 2d/3 = d/3. This is true wherever the sheet sits, as long as it is parallel to the plates.

   C₂ = (ε₀ A)/(d - t) = (ε₀ A)/(d/3) = 3 (ε₀ A)/d = 3C₁

3. So. C₂/C₁ = 3 : 1, which is option (b).

The idea to keep. A metal slab of thickness t acts like a dielectric with K = ∞, so C = (ε₀ A)/(d - t). Just subtract its thickness from the gap.

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