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:

$\frac{14}{11}$

The correct answer is Option (4) → $\frac{14}{11}$

The effective capacitance of a capacitor is,

$C=\frac{ε_0A}{d-t+\frac{t}{K}}$ [t = Thickness of slab]

When the dielectric slab completely fills the space,

$C'=K.C_0=\frac{Kε_0A}{d}$ [$C_0$ = Original capacitance]

$∴76C_0=\frac{ε_0A}{d-23d+\frac{23d}{K}}$

$⇒76\frac{ε_0A}{d}=\frac{ε_0A}{d-23d+\frac{23d}{K}}$

$⇒K≃1.04$