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

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$6: 5$

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(C) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.For the first case,$t = \frac{d}{2}$ and $K = 4$:$C_1 = \frac{\epsilon_0 A}{d - \frac{d}{2} + \frac{d/2}{4}} = \frac{\epsilon_0 A}{\frac{d}{2} + \frac{d}{8}} = \frac{\epsilon_0 A}{\frac{5d}{8}} = \frac{8 \epsilon_0 A}{5d}$.For the second case,$t = \frac{d}{3}$ and $K = 4$:$C_2 = \frac{\epsilon_0 A}{d - \frac{d}{3} + \frac{d/3}{4}} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{12}} = \frac{\epsilon_0 A}{\frac{9d}{12}} = \frac{12 \epsilon_0 A}{9d} = \frac{4 \epsilon_0 A}{3d}$.Taking the ratio $C_1 : C_2$:$\frac{C_1}{C_2} = \frac{8 \epsilon_0 A}{5d} 	imes \frac{3d}{4 \epsilon_0 A} = \frac{24}{20} = \frac{6}{5}$.Thus,$C_1 : C_2 = 6: 5$.

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