A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K₁ and K₂ of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

Solution

Physics sir · 4 steps

Split the capacitor down the middle: air on the left, two slabs stacked in series on the right.

1. Read the circuit. The left half (area A/2) has only air across the full gap d. The right half has K₁ on top of K₂, so a charge crossing that half passes through both slabs. Those two slabs are in series, and that stack is in parallel with the air half.
2. Air half. C₀ = (ε₀ (A/2))/d = (ε₀ A)/2d.
3. Each slab. Each slab has area A/2 and thickness d/2, so the two halvings cancel: C₁ = (K₁ ε₀ (A/2))/(d/2) = (K₁ ε₀ A)/d, and in the same way C₂ = (K₂ ε₀ A)/d. In series they give (C₁ C₂)/(C₁ + C₂) = (ε₀ A)/d · (K₁ K₂)/(K₁ + K₂).

   C = (ε₀ A)/2d + (ε₀ A)/d·(K₁ K₂)/(K₁ + K₂) = (ε₀ A)/d(1/2 + (K₁ K₂)/(K₁ + K₂))

4. So. Adding the parallel parts gives option (a).

The idea to keep. Dielectrics side by side across the gap are in parallel. Dielectrics stacked one above the other between the plates are in series. Use each slab's own area and thickness in C = Kε₀ A/d.

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