A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric ,materials having dielectric constants `K_(1) , K_(2), K_(3)` and `K_(4)` 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

Options

A. `(2)/(K) =(3)/(K_(1)+K_(2)+K_(3))+(1)/(K_(4))`
B. `(1)/(K) =(1)/(K_(1))+(1)/(K_(2))+(1)/(K_(3))+(3)/(2K_(4))`
C. `K = K_(1)+K_(2)K_(3)+3K_(4)`
D. `K = (2)/(3)[K_(1)+K_(2)+K_(3)]+2K_(4)`

Answer

A, B, C, D

Explanation

Equivalent figure

`C_(1)=(2in_(0)AK_(1))/(3d),C_(2)=(2in_(0)AK_(2))/(3d),C_(3)=(2in_(0)AK_(3))/(3d)`
`C_(4) = (2in_(0)AK_(4))/(d)` and `C = (in_(0)AK)/(d)`
Equivalent capacitance of parallel connection of
`C_(1), C_(2), C_(3)`
`C . = C_(1)+C_(2)+C_(3)`
`=(2 in_(0)AK_(1))/(3d)+(2in_(0)AK_(2))/(3d)+(2in_(0)AK_(3))/(3d)`
`= (2in_(0)A)/(3d)[ K_(1)+K_(2)+K_(3)]`
Now if equivalent capacitance of series connection of C and `C_(4)` then
`(1)/(C)= (1)/(C) +(1)/(C_(4))`
`(d)/(in_(0)AK)=(3d)/(2in_(0)A(K_(1)+K_(2)+K_(3)))+(d)/(2in_(0)AK_(4))`
`:. (2)/(K) = (3)/(K_(1)+K_(2)+K_(3))+(1)/(K_(4))`
`[ because ` Dividing each term term by `(d)/(in_(0)A)`]