A parallel plate air capacitor is charged up to \(100 \mathrm{~V}\). A plate \(2 \mathrm{~mm}\) thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by \(1.6 \mathrm{~mm}\). The dielectric constant of the thick plate is

If \(d\) is the initial distance between the plates then capacitance is given by \(\mathrm{C}=\frac{\mathrm{kA} \in_0}{\mathrm{~d}}\)
When a plate of thickness \(\mathrm{t}=2 \mathrm{~mm}\) is inserted between the plates the capacitance,
\(
C_1=\frac{k A \epsilon_0}{\left(d-t+\frac{t}{k}\right)}
\)
If distance between the plates is increased by \(\mathrm{x}=1.6 \mathrm{~mm}\) Capacitance become, \(\mathrm{C}_2=\frac{\mathrm{kA} \in_0}{\left(\mathrm{~d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right)}\) But \(\mathrm{C}_2=\mathrm{C}\)
\(\begin{aligned} & \therefore \mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}=\mathrm{d} \\ & \therefore \mathrm{t}-\mathrm{x}=\frac{\mathrm{t}}{\mathrm{k}} \\ & \therefore 2-1.6=\frac{2}{\mathrm{k}} \\ & \therefore 0.4=\frac{2}{\mathrm{k}} \quad \therefore \mathrm{k}=5\end{aligned}\)