If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will
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(b) $C = \frac{{K{\varepsilon _0}A}}{d} \propto \frac{K}{d}$
Hence, $\frac{{{C_1}}}{{{C_2}}} = \frac{{{K_1}}}{{{K_2}}} \times \frac{{{d_2}}}{{{d_1}}} = \frac{K}{{2K}} \times \frac{{d/2}}{d} = \frac{1}{4}$
Therefore, $ C_2 = 4C_1$
Hence, $\frac{{{C_1}}}{{{C_2}}} = \frac{{{K_1}}}{{{K_2}}} \times \frac{{{d_2}}}{{{d_1}}} = \frac{K}{{2K}} \times \frac{{d/2}}{d} = \frac{1}{4}$
Therefore, $ C_2 = 4C_1$
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