Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness d/2 and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

Solution

Physics sir · 3 steps

A metal sheet has no field inside, so it just takes away half the gap the field must cross.

1. Field inside metal. Inside a conductor in electrostatics the field is zero, so there is no potential drop across the sheet. The whole voltage now falls across the air that is left.
2. Effective gap. Air left = d - d/2 = d/2. It does not matter where the sheet sits, because the two air gaps on either side always add up to d/2.

   C' = (ε₀ A)/(d - t) = (ε₀ A)/(d/2) = 2 (ε₀ A)/d = 2C

3. So. C' : C = 2 : 1, which is option (a).

The idea to keep. A metal slab of thickness t only shortens the gap: C = (ε₀ A)/(d - t). It is the dielectric-slab formula with K → ∞.

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