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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
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5
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 $\mathrm{C}_2=\mathrm{C}$
$\begin{aligned} & \therefore \mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}=\mathrm{d} \\ & \therefore \mathrm{t}-\mathrm{x}=\frac{\mathrm{t}}{\mathrm{k}} \\ & \therefore 2-1.6=\frac{2}{\mathrm{k}} \\ & \therefore 0.4=\frac{2}{\mathrm{k}} \quad \therefore \mathrm{k}=5\end{aligned}$
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 $\mathrm{C}_2=\mathrm{C}$
$\begin{aligned} & \therefore \mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}=\mathrm{d} \\ & \therefore \mathrm{t}-\mathrm{x}=\frac{\mathrm{t}}{\mathrm{k}} \\ & \therefore 2-1.6=\frac{2}{\mathrm{k}} \\ & \therefore 0.4=\frac{2}{\mathrm{k}} \quad \therefore \mathrm{k}=5\end{aligned}$
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