A parallel plate capacitor of area A, plate separation d and capacitance C . . .

Question : 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 in figure. If a single dielectric material is to be used to have the same effective capacitance as the above combination then its dielectric constant K is given by :
Doubt by Riddhi 

Solution : 

Let K be the required value of single dielectric constant. 

Here the capacitor C1 and C2 are connected in parallel white the capacitor C3 is connected in series with C1 and C2.

C1=K1ε₀(A/2)/(d/2)
C1=K1ε₀A/d
C1=K1C [∵C=ε₀A/d]

C2=K2ε₀(A/2)/(d/2)
C2=K2ε₀A/d
C2=K2C

C3=K3ε₀A/(d/2)
C3=2K3ε₀A/d
C3=2K3C

C1 and C2 are connected in Parallel. 
So 
C12=C1+C2
C12=K1C+K2C
C12=(K1+K2)C

Now C12 and C3 are connected in Series 

So 
Ceq=C12×C3/[C12+C3]
Ceq=(K1+K2)C×2K3C / [(K1+K2)C+2K3C]
Ceq=2K3(K1+K2)C² / C(K1+K2+2K3)
Ceq=2K3(K1+K2)C/(K1+K2+2K3)
KC=2K3(K1+K2)C/(K1+K2+2K3)
K=2K3(K1+K2)/(K1+K2+2K3)
OR
1/K=1/(K1+K2) + 1/2K3