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(B) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\varepsilon_0 A}{d}$.When a metal plate of t

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$2$

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(B) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\varepsilon_0 A}{d}$.When a metal plate of thickness $t = d/2$ is inserted between the plates,the effective distance between the plates becomes $d_{eff} = d - t = d - d/2 = d/2$.Since the inserted plate is a metal,it acts as a short circuit across the region it occupies. The effective capacitance $C^{\prime}$ of the system is equivalent to a parallel plate capacitor with an air gap of $d/2$.Thus,$C^{\prime} = \frac{\varepsilon_0 A}{d - d/2} = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.Comparing this with the initial capacitance $C$,we get $C^{\prime} = 2C$.Therefore,the ratio $\frac{C^{\prime}}{C} = 2$.

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