A parallel plate condenser with plate area A and separation d is filled with dielectrics as shown in fig. The dielectric constants are K₁ and K₂ respectively. The capacitance will be

The combination is equivalent to two capacitors in series, each with plate area A and separation d / 2.
$\begin{array}{r}{\mathrm{C}}_1=\frac{{\mathrm{K}}_1{\mathrm{\epsilon }}_0\mathrm{A}}{(\mathrm{d}/2)}=\frac{2{\mathrm{K}}_1{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\\ {\mathrm{C}}_2=\frac{{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{A}}{(\mathrm{d}/2)}=\frac{2{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\end{array}$

The equivalent capacitance C is given by
$\begin{array}{l}\frac{1}{\mathrm{C}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}\\ \therefore \frac{1}{\mathrm{C}}=\frac{\mathrm{d}}{2{\mathrm{K}}_1{\mathrm{\epsilon }}_0\mathrm{A}}+\frac{\mathrm{d}}{2{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{A}}\\ =\frac{\mathrm{d}}{2{\mathrm{\epsilon }}_0\mathrm{A}}\left[\frac{1}{{\mathrm{K}}_1}+\frac{1}{{\mathrm{K}}_2}\right]=\frac{\mathrm{d}}{2{\mathrm{\epsilon }}_0\mathrm{A}}\left(\frac{{\mathrm{K}}_1+{\mathrm{K}}_2}{{\mathrm{K}}_1{\mathrm{K}}_2}\right)\end{array}$
or     $\mathrm{C}=\frac{2{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\left[\frac{{\mathrm{K}}_1{\mathrm{K}}_2}{{\mathrm{K}}_1+{\mathrm{K}}_2}\right]$