A parallel-plate capacitor is half-filled with dielectric constants ε₁ and ε₂ in two different configurations shown in the original figure. In configuration 1 the two dielectric layers divide the plate separation equally; in configuration 2 they divide the plate area equally. If the corresponding capacitances are C₁ and C₂, then C₁/C₂ is:

Question

G1 Physics · Accepted Answer

Configuration 1: the dielectrics are stacked along the electric-field direction, so they behave as two capacitors in series, each of thickness d/2 and area A. The equivalent dielectric factor is 2ε₁ε₂/(ε₁+ε₂). Hence, C₁ = [2ε₁ε₂/(ε₁+ε₂)](ε₀A/d). Configuration 2: the dielectrics occupy equal half-areas, so they behave as two capacitors in parallel. Thus, C₂ = ε₀ε₁(A/2)/d + ε₀ε₂(A/2)/d = [(ε₁+ε₂)/2](ε₀A/d). Therefore, C₁/C₂ = [2ε₁ε₂/(ε₁+ε₂)] / [(ε₁+ε₂)/2] = 4ε₁ε₂/(ε₁+ε₂)². Correct option: A.