A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is d. The space between the plates is now filled with two dielectrics. One of the dielecrics has dielectric constant `K_(1) =3` and thickness d/3 while the other one has dielectric constant `K_(2)=6` and thickness `2d//3`. Capacitance of the capacitor is now

Options

A. `1.8pF`
B. `45pF`
C. `40.5pF`
D. `20.25pF`

Answer

C

Explanation

As both capacitors are connected in series,
`therefore C=(C_(1)C_(2))/(C_(1)+C_(2))` . .. (i)
where `C_(1)=(K_(1)epsi_(0)A)/(d//3)` . . (ii) and `C_(2)=(K_(2)epsi_(0)A)/(2d//3)` . . (iii)
, It is given that, `(epsi_(0)A)/(d)=9pF`.
On substation equations (ii) and (iii) in equation (i) we get the result `C=40.5pF`.