A parallel plate capacitor of area `A`, plate separation `d` and capacitance `C` is filled with three different dielectric materials having dielectric constant `K_(1),K_(2)` and `K_(3)` as shown in fig. 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 :

Explanation

Let `C=(in_(0)A)/(d) ,C_(1)=K_(1) =(in (A)/(2))/((d)/(2))=K_(1)C`
`C_(2)=K_(2)(in_(0) (A)/(2))/((d)/(2))=K_(2)C`,
`C_(3)=(K_(3) in_(0) (A)/(2))/((d)/(2)),=K_(3)C, C_(3),=(K_(3) in_(0)(A)/(2))/((d)/(2)),=K_(3)C`

The equivalent circuit as shown

`C_(eff)=(C_(1)C_(3))/(C_(1)+C_(3))+(C_(2)C_(3))/(C_(2)+C_(3))`
` = (((K_(1) epsilon A//2)/(d//2))((K_(3) epsilon A//2)/(d//2)))/(((K_(1) epsilon A//2)/(d//2)+(K_(3) epsilonA//2)/(d//2)))+(((K_(2) epsilon A//2)/(d//2))((K_(2) epsilon A//2)/(d//2)))/(((K_(2) epsilon A//2)/(d//2)+(K_(2) epsilon A//2)/(d//2)))`
`C_(eff)=((K_(1)K_(3))/(K_(1)+K_(3))+(K_(2)K_(3))/(K_(2)+K_(3)))C`
`K_(eff) =((K_(1)K_(3))/(K_(1)+K_(3))+(K_(2)K_(3))/(K_(2)+K_(3)))`.