Q 12-02-029NEETJEE MainMedium
A slab of dielectric constant $4$ and thickness $\dfrac{d}{2}$ is placed between the plates of a parallel plate capacitor of separation $d$ and capacitance $C_0$ (air). The new capacitance is
Answer: (B) $\dfrac{8}{5}C_0$
$$C = \frac{\epsilon_0A}{d - t + t/K} = \frac{\epsilon_0A}{d - \frac{d}{2} + \frac{d}{8}} = \frac{\epsilon_0A}{5d/8} = \frac{8}{5}C_0$$
Solution by Sreeraj P, M.Sc Physics