A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $$K_1$$, $$K_2$$, $$K_3$$, $$K_4$$ arranged as shown in the figure. The effective dielectric constant $$K$$ will be:

Let the total plate area be $$A$$ and the total separation between plates be $$d$$.

From the figure: For each dielectric block, area $$A' = \frac{A}{2}$$ and separation distance $$d' = \frac{d}{2}$$

Capacitance formula for each individual section:

$$C_1 = \frac{K_1 \varepsilon_0 (A/2)}{d/2} = \frac{K_1 \varepsilon_0 A}{d}$$

$$C_2 = \frac{K_2 \varepsilon_0 (A/2)}{d/2} = \frac{K_2 \varepsilon_0 A}{d}$$

$$C_3 = \frac{K_3 \varepsilon_0 (A/2)}{d/2} = \frac{K_3 \varepsilon_0 A}{d}$$

$$C_4 = \frac{K_4 \varepsilon_0 (A/2)}{d/2} = \frac{K_4 \varepsilon_0 A}{d}$$

The upper branch consists of $$C_1$$ and $$C_2$$ in series:

$$C_{\text{upper}} = \frac{C_1 C_2}{C_1 + C_2} = \left(\frac{K_1 K_2}{K_1 + K_2}\right) \frac{\varepsilon_0 A}{d}$$

The lower branch consists of $$C_3$$ and $$C_4$$ in series:

$$C_{\text{lower}} = \frac{C_3 C_4}{C_3 + C_4} = \left(\frac{K_3 K_4}{K_3 + K_4}\right) \frac{\varepsilon_0 A}{d}$$

The upper and lower branches are connected in parallel:

$$C_{\text{total}} = C_{\text{upper}} + C_{\text{lower}} = \left( \frac{K_1 K_2}{K_1 + K_2} + \frac{K_3 K_4}{K_3 + K_4} \right) \frac{\varepsilon_0 A}{d}$$

$$K = \left( \frac{K_1 K_2}{K_1 + K_2} + \frac{K_3 K_4}{K_3 + K_4} \right)$$