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The capacitance of a capacitor becomes $\frac{7}{6}$ times its original value if a dielectric slab of thickness $t=\frac{2}{3}d$ is introduced in between the plates, where d is the separation between the plates. The dielectric constant of the slab is: |
$\frac{11}{7}$ $\frac{7}{11}$ $\frac{11}{14}$ $\frac{14}{11}$ |
$\frac{14}{11}$ |
The correct answer is Option (4) → $\frac{14}{11}$ |
