A parallel plate capacitor is constructed using three different dielectric materials as shown in figure. The parallel plates across which a potential difference is applied are of area `A = 1 cm^(2)` and are separated by a distance d = 2mm. If `K_(1)=4, K_(2)=6 and K_(3)=2`, find capacitance across points A and B

Options

A. 1.56 pF
B. 2.56 pF
C. 2.65 pF
D. 3 pF

Answer

A

Explanation

Let `K_(1), K_(2) and K_(3)` be the dielectric for `C_(1), C_(2) and C_(3)` respectively.
From figure,
`C_(1)=(AK_(1)in_(0))/(2d),C_(2)=(AK_(2)in_(0))/dandC_(3)=(AK_(3)in_(0))/d`
Now," "`1/C_(s)=1/C_(2)+1/C_(3)=d/(Ain_(0)K_(2))+d/(Ain_(0)K_(3))`
`C_(s)=(Ain_(0))/d((2K_(2)K_(3))/(K_(2)+K_(3)))=(Ain_(0)K')/d`
where" "`K'=((2K_(2)K_(3))/(K_(2)+K_(3)))=(2xx6xx2)/(6+2)=3`
Now" "`C=C_(1)+C_(s)=(Ain_(0)K_(1))/(2d)+(Ain_(0)K^(1))/(2d)`
`=(Ain_(0))/d((K_(1)+K^(1))/2)`
by putting the values we have,
C = 1.56 pF