A composite parallel plate capacitor is made up of two different dielectric materials with different thickness (t₁ and t₂) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducting foil is V.

Solution

Physics sir · 4 steps

The foil splits this into two capacitors in series, so the 100 V divides in the inverse ratio of capacitances.

1. Two capacitors. The foil is a conductor, so it acts as a shared plate. The top layer and the bottom layer form two capacitors in series, each with plate area A.
2. Find each capacitance. Use C = (εᵣ ε₀ A)/t. Top: C₁ = (3ε₀ A)/(0.5 mm) = 6k. Bottom: C₂ = (4ε₀ A)/(1 mm) = 4k, where k = (ε₀ A)/(1 mm).
3. Split the voltage. In series both capacitors carry the same charge Q, so V = Q/C is larger across the smaller capacitor. The drops are in the ratio V₁ : V₂ = C₂ : C₁ = 4 : 6 = 2 : 3.

   V₁ = 2/5× 100 = 40 V, V₂ = 3/5× 100 = 60 V

4. So. The longer battery plate is on top, so the top plate is at 100 V and the bottom plate at 0 V. The foil sits V₂ above the bottom plate, so its voltage is 60 V.

The common slip: 40. Taking the drop across the top layer instead of the foil's potential above the grounded bottom plate gives 40, not 60.

The idea to keep. A conducting foil inside a capacitor splits it into capacitors in series. The voltage divides inversely as the capacitances, so the thicker, lower-capacitance layer takes the bigger share.

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