A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ 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
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$C_1 = \frac{K_1 \varepsilon_0 \frac{A}{2}}{\left(\frac{d}{2}\right)} = \frac{K_1 \varepsilon_0 A}{d}$
$C_2 = \frac{K_2 \varepsilon_0 \frac{A}{2}}{\left(\frac{d}{2}\right)} = \frac{K_2 \varepsilon_0 A}{d}$
and $C_3 = \frac{K_3 \varepsilon_0 A}{\left(\frac{d}{2}\right)} = \frac{2 K_3 \varepsilon_0 A}{d}$
$\frac{1}{C_{eq}} = \frac{1}{C_1 + C_2} + \frac{1}{C_3}$ $= \frac{1}{\frac{\varepsilon_0 A}{d} (K_1 + K_2)} + \frac{1}{\frac{\varepsilon_0}{d} \times 2K_3}$
$\frac{1}{c_{eq}} = \frac{d}{\varepsilon_0 A} \left[ \frac{1}{K_1 + K_2} + \frac{1}{2K_3} \right]$
$c_{eq} = \left[\frac{1}{k_1 + k_2} + \frac{1}{2k_3}\right]^{-1} \cdot \frac{\varepsilon_0 A}{d}$
So $K_{eq} = \left[\frac{1}{K_1 + K_2} + \frac{1}{2 K_3}\right]^{-1}$
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