A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants ε₁ and ε₂, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are C₁ and C₂ respectively, then C₁/C₂ is:
First Configuration
Second Configuration

Solution

Physics sir · 4 steps

Layers stacked one above the other form a series pair; slabs side by side form a parallel pair.

1. First configuration. The two layers lie one above the other, each of thickness d/2 and full area A. Charge must pass through both in turn, so they act as two capacitors in series, each C = (2ε ε₀ A)/d.
2. Series result. Combining (2ε₁ ε₀ A)/d and (2ε₂ ε₀ A)/d in series gives C₁ = (2ε₀ A)/d · (ε₁ ε₂)/(ε₁ + ε₂).
3. Second configuration. The slabs sit side by side, each of area A/2 and full gap d. Both see the same voltage, so they are in parallel and their capacitances add: C₂ = (ε₀ A)/2d(ε₁ + ε₂).

   C₁/C₂ = ((2ε₀ A)/d · (ε₁ ε₂)/(ε₁ + ε₂))/((ε₀ A)/2d(ε₁ + ε₂)) = (4ε₁ ε₂)/(ε₁ + ε₂)²

4. So. The factor (ε₀ A)/d cancels and the ratio is (4ε₁ ε₂)/(ε₁ + ε₂)², which is option (c).

The idea to keep. Ask how the field crosses each dielectric. Passing through one after the other means series, so halve the thickness. Sitting side by side means parallel, so halve the area. A ratio of capacitances must have no units.

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