A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d₁ and dielectric constant k₁ and the other has thickness d₂ and dielectric constant k₂ as shown in Figure. This arrangement can be thought as a dielectric slab of thickness d =(d₁+d₂) and effective dielectric constant k. The k is :

Capacitance of a parallel plate capacitor filled with dielectric of constant k₁ and thickness d₁ is,
${\mathrm{C}}_1=\frac{{\mathrm{k}}_1{\mathrm{\epsilon }}_0 \mathrm{A}}{{\mathrm{d}}_1}$
Similarly, for other capacitance of a parallel plate capacitor filled with dielectric of constant k₂ and thickness d₂ is,
${\mathrm{C}}_2=\frac{{\mathrm{k}}_2{\mathrm{\epsilon }}_0 \mathrm{A}}{{\mathrm{d}}_2}$
Both capacitors are in series so equivalent capacitance C is related as :
$$\begin{gathered} \frac{1}{\mathrm{C}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}=\frac{{\mathrm{d}}_1}{{\mathrm{k}}_1{\mathrm{\epsilon }}_0 \mathrm{A}}+\frac{{\mathrm{d}}_2}{{\mathrm{k}}_2{\mathrm{\epsilon }}_0 \mathrm{A}} \\ =\frac{1}{{\mathrm{\epsilon }}_0 \mathrm{A}}\left[\frac{{\mathrm{k}}_2{\mathrm{d}}_1+{\mathrm{k}}_1{\mathrm{d}}_2}{{\mathrm{k}}_1{\mathrm{k}}_2}\right] \\ \text{ So, }\mathrm{C}=\frac{{\mathrm{k}}_1{\mathrm{k}}_2{\mathrm{\epsilon }}_0\mathrm{A}}{\left({\mathrm{k}}_1{\mathrm{d}}_2+{\mathrm{k}}_2{\mathrm{d}}_1\right)} .....\left(\mathrm{i}\right) \\ {\mathrm{C}}^{\text{'}}=\frac{{\mathrm{k\epsilon }}_0\mathrm{A}}{\mathrm{d}}=\frac{{\mathrm{k\epsilon }}_0\mathrm{A}}{\left({\mathrm{d}}_1+{\mathrm{d}}_2\right)} .....\left(\mathrm{ii}\right) \end{gathered}$$
where; d = {d₁+d₂)
Comparing eqns. (i) and (ii), the dielectric constant of new capacitor is :
$\mathrm{k}=\frac{{\mathrm{k}}_1{\mathrm{k}}_2\left({\mathrm{d}}_1+{\mathrm{d}}_2\right)}{\left({\mathrm{k}}_1{\mathrm{d}}_2+{\mathrm{k}}_2{\mathrm{d}}_1\right)}$