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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{K \varepsilon_0 A}{d}$,where $K$ is the dielectric constant,$\varepsilon_0$ i

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Increase by $4$ times

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{K \varepsilon_0 A}{d}$,where $K$ is the dielectric constant,$\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.From the formula,we see that $C \propto \frac{K}{d}$.Let the initial capacitance be $C_1 = \frac{K_1 \varepsilon_0 A}{d_1}$.Given: $d_2 = \frac{d_1}{2}$ and $K_2 = 2K_1$.The new capacitance $C_2$ is given by $C_2 = \frac{K_2 \varepsilon_0 A}{d_2} = \frac{(2K_1) \varepsilon_0 A}{d_1/2} = 4 	imes \frac{K_1 \varepsilon_0 A}{d_1} = 4C_1$.Therefore,the capacity increases by $4$ times.

If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will

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