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(C) Let the initial distance between the plates be $d$. The capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.When a dielec

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$5$

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(C) Let the initial distance between the plates be $d$. The capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t = 3 \, mm$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.To maintain the same potential difference,the effective capacitance must remain the same. The problem states that the distance between the plates is increased by $\Delta d = 2.4 \, mm$,so the new distance is $d' = d + 2.4$.The new capacitance with the slab is $C' = \frac{\varepsilon_0 A}{d' - t + \frac{t}{K}} = \frac{\varepsilon_0 A}{d + 2.4 - 3 + \frac{3}{K}} = \frac{\varepsilon_0 A}{d - 0.6 + \frac{3}{K}}$.For the capacitance to remain unchanged,we must have $d = d - 0.6 + \frac{3}{K}$.This simplifies to $0.6 = \frac{3}{K}$.Therefore,$K = \frac{3}{0.6} = 5$.

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