A parallel plate capacitor with air between the plates has capacitance of `9pF`. The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectrics 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. `45muF`
B. `40.5pF`
C. `20.25pF`
D. `1.8pF`

Answer

B

Explanation

`C_(1)=(K_(1)Ain_(0))/(d//3)=(9Ain_(0))/(d)`
`C_(2)=(K_(2)Ain_(0))/(2d//3)=(9Ain_(d))/(d)`
`C_(1)` and `C_(2)` are in series (check polarity)
`C_(eq)=(9)/(2)(Ain_(0))/(d)=(9)/(2)xx9=(81)/(2)-40.5pF`