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For charging the capacitance of given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as plates of capacitor. The thickness of dielectric slab is $\frac{3}{4} d$ , where d is the separation between the plates of parallel plate capacitor. The new capacitance $(C^{'})$ in terms of original capacitance $(C_{0})$ is given by

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

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

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

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

answer is A.

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Detailed Solution

$C_{0} = \frac{A \in_{0}}{d} C^{1} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} o r \frac{1}{C^{1}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} C_{1} = \frac{K A ε_{0}}{(3 d / 4)}$

$C_{2} = \frac{A ε_{0}}{(d / 4)} \frac{1}{C^{1}} = \frac{3 d / 4}{K A ε_{0}} + \frac{d / 4}{A ε_{0}} \Rightarrow C^{1} = \frac{4 K C_{0}}{3 + K}$

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