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Question

(C) The original capacitance of the parallel plate capacitor is $C_{0} = \frac{\epsilon_{0} A}{d}$.When a dielectric slab of thickness $t = \frac{3}{4

Vedclass · Accepted Answer

$C' = \frac{4K}{K+3} C_{0}$

Vedclass · Answer

(C) The original capacitance of the parallel plate capacitor is $C_{0} = \frac{\epsilon_{0} A}{d}$.When a dielectric slab of thickness $t = \frac{3}{4}d$ is inserted,the system can be modeled as two capacitors $C_{1}$ and $C_{2}$ in series.$C_{1}$ is the capacitance of the part filled with dielectric: $C_{1} = \frac{K \epsilon_{0} A}{3d/4} = \frac{4 K \epsilon_{0} A}{3d}$.$C_{2}$ is the capacitance of the air gap: $C_{2} = \frac{\epsilon_{0} A}{d - 3d/4} = \frac{\epsilon_{0} A}{d/4} = \frac{4 \epsilon_{0} A}{d}$.Since they are in series,the equivalent capacitance $C'$ is given by $\frac{1}{C'} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$.Substituting the values:$\frac{1}{C'} = \frac{3d}{4 K \epsilon_{0} A} + \frac{d}{4 \epsilon_{0} A}$.$\frac{1}{C'} = \frac{d}{4 \epsilon_{0} A} \left( \frac{3}{K} + 1 \right) = \frac{d}{4 \epsilon_{0} A} \left( \frac{3+K}{K} \right)$.Therefore,$C' = \frac{4 K \epsilon_{0} A}{d(3+K)} = \frac{4K}{3+K} C_{0}$.

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