Q 12-02-130JEE MainJEE Main 2021 (27 Jul, Shift 1)Medium
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given the area of the plate $= A$)
Answer: (A) $\dfrac{15}{34}\dfrac{K\varepsilon_0 A}{d}$
The three slabs are in series along the field:
$$C = \frac{\varepsilon_0 A}{\dfrac{d}{K} + \dfrac{2d}{3K} + \dfrac{3d}{5K}} = \frac{K\varepsilon_0 A}{d}\cdot\frac{1}{1 + \frac23 + \frac35}$$
$1 + \dfrac23 + \dfrac35 = \dfrac{15 + 10 + 9}{15} = \dfrac{34}{15}$, so $C = \dfrac{15}{34}\dfrac{K\varepsilon_0 A}{d}$.
Solution by Sreeraj P, M.Sc Physics