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(B) Let $d$ be the initial distance between the plates. The initial capacitance is $C = \frac{A \epsilon_0}{d}$.When a dielectric plate of thickness $

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$5$

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(B) Let $d$ be the initial distance between the plates. The initial capacitance is $C = \frac{A \epsilon_0}{d}$.When a dielectric plate of thickness $t = 2 \,mm$ is inserted, the new capacitance $C_1$ is given by $C_1 = \frac{A \epsilon_0}{d - t + \frac{t}{k}}$.When the distance between the plates is increased by $x = 1.6 \,mm$ to maintain the same potential difference (which implies the same capacitance $C_2 = C$), the new distance becomes $d' = d + x$.The new capacitance is $C_2 = \frac{A \epsilon_0}{d + x - t + \frac{t}{k}}$.Since $C_2 = C$, we have $d + x - t + \frac{t}{k} = d$.Simplifying this, we get $x - t + \frac{t}{k} = 0$, or $t - x = \frac{t}{k}$.Substituting the values $t = 2 \,mm$ and $x = 1.6 \,mm$:$2 - 1.6 = \frac{2}{k}$$0.4 = \frac{2}{k}$$k = \frac{2}{0.4} = 5$.

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

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