Capacitance of capacitor becomes $\large\frac{4}{3}$ times its original value if a dielectric slab of thickness $\large\frac{d}{2}$ is inserted b/w the plates (d= seperation between the plates) . The dielectric constant of the slab is:

Question

$(A)\;2 \\ (B)\;4 \\ (C)\;6 \\ (D)\;8$