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(C) The capacitance of a parallel plate capacitor with dielectric constant $K$ is given by $C = \frac{K \epsilon_0 A}{d}$.For the two capacitors in se

Vedclass · Accepted Answer

$\frac{3 d}{4}$

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(C) The capacitance of a parallel plate capacitor with dielectric constant $K$ is given by $C = \frac{K \epsilon_0 A}{d}$.For the two capacitors in series,$C_1 = \frac{2 \epsilon_0 A}{d}$ and $C_2 = \frac{4 \epsilon_0 A}{d}$.The equivalent capacitance $C_{eq}$ for capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the values: $\frac{1}{C_{eq}} = \frac{d}{2 \epsilon_0 A} + \frac{d}{4 \epsilon_0 A} = \frac{2d + d}{4 \epsilon_0 A} = \frac{3d}{4 \epsilon_0 A}$.Thus,$C_{eq} = \frac{4 \epsilon_0 A}{3d}$.For an equivalent air capacitor $(K=1)$ with area $A$ and separation $d'$,the capacitance is $C_{eq} = \frac{\epsilon_0 A}{d'}$.Equating the two expressions: $\frac{\epsilon_0 A}{d'} = \frac{4 \epsilon_0 A}{3d}$.Solving for $d'$,we get $d' = \frac{3d}{4}$.

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