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(C) The capacitance of a parallel plate capacitor with air is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric co

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$3$

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(C) The capacitance of a parallel plate capacitor with air is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced,the new capacitance $C$ is given by $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.The ratio of the new capacitance to the original capacitance is $\frac{C}{C_0} = \frac{d}{d - t + \frac{t}{K}}$.Given $d = 6\,mm$,$t = 4.5\,mm$,and $K = 9$.Substituting the values: $\frac{C}{C_0} = \frac{6}{6 - 4.5 + \frac{4.5}{9}} = \frac{6}{1.5 + 0.5} = \frac{6}{2} = 3$.Therefore,the capacity becomes $3$ times the original capacity.

$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

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