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(C) The capacitance of a parallel plate capacitor without a dielectric is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and

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$5$

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(C) The capacitance of a parallel plate capacitor without a dielectric is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted,the new capacitance becomes $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.Since the potential difference $V$ remains the same and the charge $Q$ is constant (as the capacitor is disconnected),the capacitance must remain the same. Thus,$C = C'$.$\frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{d' - t + \frac{t}{K}}$$\Rightarrow d = d' - t + \frac{t}{K}$$\Rightarrow d' - d = t(1 - \frac{1}{K})$Given $d' - d = 3.2 \ mm$ and $t = 4 \ mm$:$3.2 = 4(1 - \frac{1}{K})$$0.8 = 1 - \frac{1}{K}$$\frac{1}{K} = 1 - 0.8 = 0.2$$K = \frac{1}{0.2} = 5$.

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