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(D) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $x$ and dielectric constant $K$ is given by:$C = \frac{\epsilon_

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$3$

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(D) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $x$ and dielectric constant $K$ is given by:$C = \frac{\epsilon_0 A}{d - x + \frac{x}{K}}$Given $K = 4$,the formula becomes:$C = \frac{\epsilon_0 A}{d - x + \frac{x}{4}} = \frac{\epsilon_0 A}{d - \frac{3x}{4}}$For $x = \frac{d}{3}$:$C_1 = \frac{\epsilon_0 A}{d - \frac{3}{4}(\frac{d}{3})} = \frac{\epsilon_0 A}{d - \frac{d}{4}} = \frac{\epsilon_0 A}{\frac{3d}{4}} = \frac{4}{3} \frac{\epsilon_0 A}{d}$Given $C_1 = 2 \mu F$,we have $\frac{4}{3} \frac{\epsilon_0 A}{d} = 2 \mu F \implies \frac{\epsilon_0 A}{d} = \frac{6}{4} = 1.5 \mu F$.For $x = \frac{2d}{3}$:$C_2 = \frac{\epsilon_0 A}{d - \frac{3}{4}(\frac{2d}{3})} = \frac{\epsilon_0 A}{d - \frac{d}{2}} = \frac{\epsilon_0 A}{\frac{d}{2}} = 2 \frac{\epsilon_0 A}{d}$Substituting $\frac{\epsilon_0 A}{d} = 1.5 \mu F$:$C_2 = 2 	imes 1.5 \mu F = 3 \mu F$.

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