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(B) The initial capacitance of the parallel plate capacitor with air is given by $C = \frac{\epsilon_0 A}{d} = 5 \mu F$.When a dielectric slab of thic

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$6$

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(B) The initial capacitance of the parallel plate capacitor with air is given by $C = \frac{\epsilon_0 A}{d} = 5 \mu F$.When a dielectric slab of thickness $t = \frac{d}{2}$ and dielectric constant $K = 1.5$ is inserted,the new capacitance $C_{	ext{new}}$ is given by the formula:$C_{	ext{new}} = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$Substituting the given values:$C_{	ext{new}} = \frac{\epsilon_0 A}{d - \frac{d}{2} + \frac{d/2}{1.5}}$$C_{	ext{new}} = \frac{\epsilon_0 A}{\frac{d}{2} + \frac{d}{3}}$$C_{	ext{new}} = \frac{\epsilon_0 A}{\frac{3d + 2d}{6}} = \frac{6 \epsilon_0 A}{5 d}$Since $\frac{\epsilon_0 A}{d} = 5 \mu F$,we have:$C_{	ext{new}} = \frac{6}{5} 	imes 5 \mu F = 6 \mu F$.

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