The figure shows a parallelplate capacitor of plate area A

The figure shows a parallel-plate capacitor of plate area A = 11.2 cm^2 and plate separation 2d= 7.20 mm. The left half of the gap is filled with material of dielectric constant K_1 = 26.3; the top of the right half is filled with material of dielectric constant k_2 = 43.6; the bottom of the right half is filled with material of dielectric constant K_3 = 61.9. What is the capacitance?

Solution

C₁ = k₁ * epsilon₀ * A / 4 d = (26.3 * 8.85 * 10-12 * 11.2 * 10-4) / (4 * 3.6 * 10-3)

= 18.1 pF

C₂ = k₂ * epsilon₀ * A / 2 d = (43.6 * 8.85 * 10-12 * 11.2 * 10-4) / (7.20 * 10-3)

= 60 pF

C₃ = k₃ * epsilon₀ * A / 2 d = (61.9 * 8.85 * 10-12 * 11.2 * 10-4) / (7.20 * 10-3)

= 85.2 pF

C₂ and C₃ are in series

C₂₃ = 60 * 85.2 / (60 + 85.2) = 35.21 pF

C₁ and C₂₃ are in parallel

C = 18.1 + 35.21

= 53.31 pF = 53.31 * 10^-12 F