A capacitor has capacitance 5μF when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness d/2 is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ μF.

Solution

Physics sir · 4 steps

The slab fills only half the gap, so split the capacitor into two in series: an air part and a slab part.

1. Split the gap. The slab covers the full plate area but only half the thickness. So the capacitor behaves like an air capacitor of gap d/2 in series with a dielectric capacitor of gap d/2.
2. Each part. With air across the full gap, C₀ = (ε₀ A)/d = 5 μF. Halving the gap doubles it, so the air part is C₁ = 10 μF. The slab part also has gap d/2, and K = 1.5 multiplies it, so C₂ = 1.5 × 10 = 15 μF.

   1/C = 1/10 + 1/15 = 5/30 = 1/6

3. Check with formula. The slab formula gives the same result: C = (ε₀ A)/(d - t + t/K) = (ε₀ A)/(d/2 + d/3) = 6/5·(ε₀ A)/d = 6/5 × 5 μF.
4. So. The capacitance with the slab is 6 μF.

The common slip: 7.5. Multiplying 5 by K = 1.5 treats the slab as filling the whole gap and gives 7.5. Here the slab fills only half the gap.

The idea to keep. A slab of thickness t acts like an air gap of only t/K. Replace t by t/K in the gap, or treat the air and slab layers as capacitors in series.

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