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) $\dfrac{8}{5}C_0$
Concept. $C = \dfrac{\epsilon_0A}{d - t + t/K}$.
Step 1. Substitute. $d - \dfrac d2 + \dfrac d8 = \dfrac{5d}{8}$: $C = \dfrac85C_0$.
Answer. $\dfrac85C_0$ (Option B).
Solution by Sreeraj P, M.Sc Physics