Capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness t = 2d/3 is introduced in between the plates. ‘d’ is the separation between the plates. The dielectric constant of the dielectric slab is

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$\frac{11}{14}$

$\frac{14}{11}$

$\frac{7}{11}$

$\frac{11}{7}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{7}{6} \frac{{Aε}_{0}}{d} = \frac{{Aε}_{0}}{d - \frac{2 d}{3} [1 - \frac{1}{K}]} \\ \frac{7}{6} = \frac{1}{1 - \frac{2}{3} + \frac{2}{3 K}} \Rightarrow K = \frac{14}{11} \end{array}$

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