Electrostatic Potential and Capacitance question for JEE Main (JEE Main 2019 (9 Jan, Shift 2)), with solution

Answer: (A) $K = \dfrac{K_1K_2}{K_1+K_2} + \dfrac{K_3K_4}{K_3+K_4}$

Let the plate area be $A$ and separation $d$. Each block has area $A/2$ and thickness $d/2$, so a block of constant $K_i$ has capacitance
$$C_i = \frac{K_i\epsilon_0 (A/2)}{d/2} = K_iC_0,\qquad C_0 = \frac{\epsilon_0A}{d}$$
In each half the two blocks are in series, and the two halves are in parallel:
$$C = C_0\left(\frac{K_1K_2}{K_1+K_2} + \frac{K_3K_4}{K_3+K_4}\right)$$
Since $C = KC_0$, $K = \dfrac{K_1K_2}{K_1+K_2} + \dfrac{K_3K_4}{K_3+K_4}$.

Solution by Sreeraj P, M.Sc Physics