In a parallel plate capacitor set up, the plate area of capacitor is 2 m² and the plates are separated by 1 m. If the space between the plates are filled with a dielectric material of thickness 0.5 m and area 2 m² (see fig.) the capacitance of the set-up will be __________εₒ. (Dielectric constant of the material = 3.2) (Round off to the Nearest Integer)

Solution

Physics sir · 4 steps

The slab and the air gap sit one after the other between the plates, so treat them as two series capacitors.

1. Two layers. The slab fills half the gap, t = 0.5 m, across the full plate area. The remaining d - t = 0.5 m is air. A slab of thickness t behaves like an air gap of only t/K.

   C = (ε₀ A)/((d - t) + t/K)

2. Put in numbers. The denominator is 0.5 + 0.5/3.2 = 0.5 + 0.15625 = 0.65625 m. So C = 2ε₀/0.65625 ≈ 3.05 ε₀.
3. Series check. As two capacitors: the slab gives (3.2 × 2ε₀)/0.5 = 12.8 ε₀ and the air gives 2ε₀/0.5 = 4 ε₀. In series, (12.8 × 4)/(12.8 + 4) ε₀ = 51.2/16.8 ε₀ ≈ 3.05 ε₀, the same value.
4. So. The capacitance is about 3.05 ε₀, which rounds to 3.

The common slip: 1. Multiplying the slab thickness by K instead of dividing gives 2/(0.5 + 1.6) ≈ 0.95, which rounds to 1, not 3.

The idea to keep. A slab parallel to the plates adds in series with the air gap. It acts like an air gap of thickness t/K, so the capacitance is ε₀ A / (d - t + t/K).

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