A parallel plate capacitor with square plates is filed with four dielectrics of dielectric constants `K_(1),K_(2),K_(3),K_(4),` arranged as shown in the figure. The effective dielectric constant K will be :

Options

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

Explanation

`C_(1)=(K_(1)in_(0)L.L//2)/(d//2)=K_(1)in_(0)L^(2)` , `C_(2)=K_(2)in_(0)L^(2)` , `C_(3)=K_(3)in_(0)L^(2)" "C_(4)=K_(4)in_(0)L^(2)`
`C_(1)C_(2)` are in series, `C_(3)C_(4)` are in series
`C_(eq)=(C_(1)C_(2))/(C_(1)+C_(2))+(C_(3)C_(4))/(C_(4)C+C_(3))=(K_(1)K_(2))/(K_(1)+K_(2))+(K_(3)K_(4))/(K_(3)+K_(4)), C_(eq).(K_(eq)in_(0)L^(2))/(d)` So `K_(eq)=(K_(1)K_(2))/(K_(1)+K_(2))+(K_(3)K_(4))/(K_(3)+K_(4))`.