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 $\dfrac{{C}_{2}}{{C}_{1}}$ is

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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 $\dfrac{{C}_{2}}{{C}_{1}}$ is

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

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Correct answer: D. (7)/(3)

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...

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