A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. 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:

Here the capacitance, C = Aκε₀/d
The above three capacitors C1, C2, C3 are in parallel and then it is in series with C4
Here, C1 = (A/3)k1ε₀d/2 = 2k1/3C, C2 = (A/3)k2ε₀d/2 = 2k2/3C, C3 = (A/3)k3ε₀d/2 = 2k1/3C, C4 = (A)k4ε₀d/2 = 2k4/kC
Now the equivalent capacitance for the combination of four capacitors is
1/Ceq = 1/(C1+C2+C3) + 1/C4
or 1/C = 3k/2C[1/k1+1/k2+1/k3] + k/2k4C (as Ceq = C)
or 2k = 3k1+k2+k3+1k4