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

Answer

c

Explanation

The two capacitors formed are in parallel, hence capacitance of the combination `C=(C_(1)C_(2))/(C_(1)+C_(2))` .. .(i)
Where ,`C_(1)=(K_(1)epsilon_(0)A)/((d)/(3)) " " (becauseC=(epsilon_(0)A)/(d))` …(ii)
`C_(2)=(K_(2)epsilon_(0)A)/((2d)/(3))` ..... (iii)
`C_(eq)=((3K_(1)epsilon_(0)A)/(d)xx(K_(2)epsilon_(0)A*3)/(2d))/((3K_(1)epsilon_(0)A)/(d)+(3K_(2)epsilon_(0)A)/(2d))`
`=(9K_(1)K_(2)epsilon_(0)^(2)A^(2))/(2d^(2))xx(d)/(epsilon_(a)Axx18)`
It is given

`(epsilon_(0)A)/(d)=9pF`
Using given values
`C_(eq)=40.5pF`