A parallel plate capacitor having plates of area $S$ and plate separation $d$ , has capacitance $C_{1}$ in air. When two dielectrics of different relative permittivities ( $ε_{1} = 2$ and $ε_{2} = 4$ ) are introduced between the two plates as shown in the figure, the capacitance becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is

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5 ϵ 0 d = C 1 In new setup we have 3 different setups i. 5 2 , d 2 , ϵ 1 C ′ = 5 2 ϵ 1 ϵ 0 d 2 = 2 C 1 ii. 5 2 , d 2 , ϵ 2 C ′ ′ = 5 2 ϵ 2 ϵ 0 d 2 = 4 C 1 iii. 5 2 , d , ϵ 1 C ′ ′ ′ = 5 2 ϵ 1 ϵ 0 d = C 1 C 2 = 1 2 C 1 + 1 4 C 1 - 1 + C 1 = 14 6 C 1 = 7 3 C 1 ∴ C 2 C 1 = 7 3

A parallel plate capacitor having plates of area S and plate separation d , has capacitance C 1 in air. When…

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