**Approach 1**: We can think of this setup as two capacitors ($C_1$ and $C_2$) in parallel where

$C_1 = \cfrac{K_1 \epsilon_0 A_1}{d}$, and

$C_2 = \cfrac{K_2 \epsilon_0 A_2}{d}$

Now for two capacitors $C_1$ and $C_2$ in parallel, equivalent capacitance, $C_{eq}$ is given by $C_{eq} = C_1 + C_2$

Or, $C_{eq} = \cfrac{\epsilon_0}{d} (K_1 A_1 + K_2 A_2)$

**Approach 2**: Now let’s work out the effective capacitance from the basics

$C_{eq} = \cfrac{Q}{V}$ $=\cfrac{Q_1 + Q_2}{V}$, where $Q_1$ is the magnitude of charge on part of the capacitor plates with dielectric of dielectric strength $K_1$ sandwiched between them, $Q_2$ is the magnitude of charge on part of the capacitor plates with dielectric of dielectric strength $K_2$ sandwiched between them, and $V$ is the potential difference between two plates

Now, $|V| = \int \overrightarrow{E}.\overrightarrow{dl}$ $= E_1 d$ $= E_2 d$

where $E_1$ is the electric field in the region with dielectric $K_1$ and $E_2$ is the electric field in the region with dielectric $K_2$

Now, $E_1 = \cfrac{\sigma_1}{K_1 \epsilon_0}$ $=\cfrac{Q_1}{K_1 A_1 \epsilon_0}$, and

$E_2 = \cfrac{\sigma_2}{K_2 \epsilon_0}$ $=\cfrac{Q_2}{K_2 A_2 \epsilon_0}$

So, $V = \cfrac{Q_1 d}{K_1 A_1 \epsilon_0}$ $= \cfrac{Q_2 d}{K_2 A_2 \epsilon_0}$

Or $Q_1 + Q_2 = V \cfrac{\epsilon_0 (A_1 K_1 + A_2 K_2)}{d}$

Or $C_{eq} =$ $\cfrac{Q_1+Q_2}{V}$ $= \cfrac{\epsilon_0 (A_1 K_1 + A_2 K_2)}{d}$