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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t$ is inserted,the

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$5$

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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of thickness $t$ is inserted,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d' - t(1 - 1/K)}$,where $d'$ is the new plate separation.Given that the potential difference $V$ remains the same and the charge $Q$ is constant (since the battery is disconnected),the capacitance must remain constant,so $C = C'$.Equating the two expressions: $\frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{(d + 1.6 \ mm) - 2 \ mm(1 - 1/K)}$.This simplifies to: $d = d + 1.6 - 2(1 - 1/K)$.$0 = 1.6 - 2 + 2/K$.$0 = -0.4 + 2/K$.$0.4 = 2/K$.$K = 2 / 0.4 = 5$.

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