Q 12-02-168JEE MainJEE Main 2019 (10 Jan, Shift 1)Easy
A parallel plate capacitor is of area $6\ \text{cm}^2$ and a separation $3\ \text{mm}$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant $K_1 = 10$, $K_2 = 12$ and $K_3 = 14$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:
Answer: (D) $12$
Each slab fills the full gap over one third of the area, so the three act as capacitors in parallel:
$$C = \frac{\epsilon_0(A/3)}{d}(K_1+K_2+K_3) = \frac{\epsilon_0A}{d}\cdot\frac{10+12+14}{3}$$
The equivalent dielectric constant is $K = \dfrac{36}{3} = 12$.
Solution by Sreeraj P, M.Sc Physics