The space between parallel plate capacitors is filled with four dielectrics of equal dimensions but of dielectric constants `K_(1), K_(2), K_(3)andK_(4)` respectively. If K is the dielectric constant of a single dielectric that must be filled between capacitor plates to have the same capacitance between A and B. Then we must have

Options

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

Answer

A

Explanation

The given capacitor may be supposed to be formed of four component capacitors `C_(1), C_(2), C_(3) and C_(4)`. The capacitor `C_(1) and C_(2)` are in parallel and `C_(3) and C_(4)` are in parallel.
From figure,
`C_(1)=(AK_(1)in_(0))/(2d//2)=(Ain_(0)K_(1))/d`
`C_(2)=(Ain_(0)K_(2))/d`
`C_(3)=(Ain_(0)K_(3))/d`
`C_(4)=(Ain_(0)K_(4))/d`
Now,
`1/C=1/C_(P1)+1/C_(P2)`
`1/C=1/(C_(1)+C_(2))+1/(C_(3)+C_(4))`
`1/((AKin_(0))/d)=1/((Ain_(0))/(d)K_(1)+(Ain_(0))/(d)K_(2))+1/((Ain_(0))/(d)K_(3)+(Ain_(0))/(d)K_(4))`
`(1)/(K)=(1)/(K_1+K_2)+(1)/(K_3+K_4)`