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(B) The initial capacitance of a parallel plate capacitor is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric con

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$\frac{14}{11}$

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(B) The initial capacitance of a parallel plate capacitor is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced, the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.Given $t = \frac{2d}{3}$ and $C' = \frac{7}{6} C_0$, we substitute these values into the equation:$\frac{7}{6} \left( \frac{\epsilon_0 A}{d} \right) = \frac{\epsilon_0 A}{d - \frac{2d}{3} + \frac{2d}{3K}}$.Canceling $\epsilon_0 A$ from both sides, we get $\frac{7}{6d} = \frac{1}{d/3 + 2d/3K}$.$\frac{7}{6} = \frac{1}{1/3 + 2/3K} = \frac{3}{1 + 2/K}$.$7(1 + 2/K) = 18$.$1 + 2/K = \frac{18}{7}$.$2/K = \frac{18}{7} - 1 = \frac{11}{7}$.$K = \frac{14}{11}$.

The capacitance of a capacitor becomes $\frac{7}{6}$ times the original value if a dielectric slab of thickness $t = \frac{2d}{3}$ is introduced between the plates, where $d$ is the distance of separation between the plates. The dielectric constant of the slab is

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