A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K₁, K₂ and K₃ as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor then its dielectric constant K is given by :

$K=\frac{K_1 K_3}{K_1+K_2}+\frac{K_2 K_3}{K_2+K_3}$

Applying $C=\frac{\varepsilon_0 A}{d-t_1-t_2+\frac{t_1}{K_1}+\frac{t_2}{K_2}}$, we have

$\frac{\varepsilon_0(A / 2)}{d-d / 2-d / 2+\frac{d / 2}{K_1}+\frac{d / 2}{K_3}}+\frac{\varepsilon_0(A / 2)}{d-d / 2-d / 2+\frac{d / 2}{K_2}+\frac{d / 2}{K_1}} = \frac{K \varepsilon_0 A}{d}$

Solving this equation, we get

$K=\frac{K_1 K_3}{K_1+K_3}+\frac{K_2 K_3}{K_2+K_3}$