A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant K=4. The thickness of the dielectric material is x, where x < d.
Let C₁ and C₂ be the capacitance of the system for χ=1/3 d and X=(2 d)/3, respectively. If C₁=2 μ F the value of C₂ is __________ μ F

Solution

Physics sir · 4 steps

A slab of thickness x acts like air of thickness x/K, so find the effective gap and compare.

1. Effective gap. The slab and the air gap are two capacitors in series. A dielectric of thickness x behaves like air of thickness x/K, so the effective air gap is d - x + x/K.

   C = (ε₀ A)/(d - x + x/K) = (ε₀ A)/(d - 3x/4) (K = 4)

2. First case. For x = d/3, the gap is d - d/4 = 3d/4, so C₁ = 4/3(ε₀ A)/d = 2 μF. This gives (ε₀ A)/d = 1.5 μF.
3. Second case. For x = 2d/3, the gap is d - d/2 = d/2, so C₂ = 2(ε₀ A)/d = 2 × 1.5 μF.
4. So. C₂ = 3 μF, so the answer is 3.

The idea to keep. For a partly filled capacitor, replace each dielectric layer of thickness t by air of thickness t/K. Then C = ε₀ A divided by the effective gap.

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