A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + αx) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :

Solution

Physics sir · 4 steps

k changes across the gap, so treat thin layers as capacitors in series and add up dx/k.

1. Slice the gap. Cut the dielectric into thin layers of thickness dx, parallel to the plates. Each layer is a tiny capacitor dC = (ε₀ k(x) A)/dx. The same field line passes through every layer, so the layers are in series.
2. Add in series. For series capacitors the reciprocals add, so 1/C = ∫₀ᵈ dx/(ε₀ A K(1 + α x)) = ln(1 + α d)/(ε₀ A K α).

   C = (ε₀ A K α)/ln(1 + α d)

3. Use small αd. Since α d ≪ 1, ln(1 + α d) ≈ α d - (α d)²/2 = α d(1 - (α d)/2). Then C ≈ (ε₀ A K)/d(1 - (α d)/2)⁻¹ ≈ (ε₀ A K)/d(1 + (α d)/2).
4. So. The capacitance is (Aε₀ K)/d(1 + (α d)/2), which is option (b). This is the same as using the average value of k across the gap, as expected to first order.

The idea to keep. If k varies along the field (from plate to plate), layers are in series: integrate dx/k. If k varies across the plate area, strips are in parallel: integrate k dA.

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