A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constants K₁ and K₂, each of area A/2 and thickness d/2, are inserted in the space between the plates [the two slabs are stacked one above the other over one half of the plate area,…

Question

A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constants K₁ and K₂, each of area A/2 and thickness d/2, are inserted in the space between the plates [the two slabs are stacked one above the other over one half of the plate area, so that half the plate area faces a plain air gap of width d while the other half faces K₁ and K₂ in series]. The capacitance of the capacitor will be given by

Accepted Answer

(1) (ε₀A)/d(1/2+(K₁K₂)/(K₁+K₂)). Split the capacitor down the middle into two side-by-side halves, each of plate area A/2; being on the same plates, they are in parallel. Air half: Cₐ=(ε₀(A/2))/d=(ε₀A)/(2d). Filled half: the two slabs are stacked across the gap, so they are in series. C₁=(K₁ε₀(A/2))/(d/2)=(K₁ε₀A)/d and likewise C₂=(K₂ε₀A)/d. C_b=(C₁C₂)/(C₁+C₂)=(ε₀A)/d·(K₁K₂)/(K₁+K₂). C=Cₐ+C_b=(ε₀A)/d(1/2+(K₁K₂)/(K₁+K₂)).

Step-by-step solution

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