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

Answer

D

Explanation

`C = (epsilon_(0)A)/(d) = 9 xx 10^(-12)A`
With dielectric, `C = (epsilon_(0)KA)/(d)`
`C_(1) = (epsilon_(0)A.3)/(d//3) = 9C, C_(2) = (epsilon_(0)A.6)/(2d//3) = 9C`
`:. C_("total") = (C_(1)C_(2))/(C_(1) + C_(2))` as they are in series.
`C_("total") = (9C xx 9C)/(18 C) = (9)/(2) xx C = (9)/(2) xx 9 xx 10^(-12)F = 40.5 pF`.