Space between the plates of a parallel plate capacitor of plate area 4 cm² and separation of 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is _ pF.
(Given ε₀ = 8.85 × 10⁻¹² F/m)

Solution

Physics sir · 4 steps

The two slabs are stacked between the plates, so they act as two capacitors in series.

1. Air capacitance first. Work out C₀ = (ε₀ A)/d = (8.85 × 10⁻¹² × 4 × 10⁻⁴)/(1.77 × 10⁻³) = 2 × 10⁻¹² F = 2 pF. Every other value can be built from this.
2. Each layer. Each slab covers the full plate area but has thickness d/2, so it is a capacitor of value (kε₀ A)/(d/2) = 2kC₀. That gives C₁ = 2 × 5 × 2 = 20 pF and C₂ = 2 × 3 × 2 = 12 pF.

   Cₛₑᵣᵢₑₛ = (C₁ C₂)/(C₁ + C₂) = (20 × 12)/(20 + 12) = 7.5 pF

3. Add the parallel one. Capacitors in parallel add directly: 7.5 + 7.5 = 15 pF.
4. So. The effective capacitance of the combination is 15 pF.

The common slip: 15.5. Treating the two slabs as side by side (parallel) gives 8 pF for the dielectric capacitor and a total of 15.5, not 15.

The idea to keep. Slabs stacked one above the other along the gap are in series. Slabs placed side by side, each filling the full gap, are in parallel. Find ε₀ A/d once and scale it for each piece.

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