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

Answer

B

Explanation

`C_@ = 9 pF`
Dielectric 1 –
`k_1=3, t=d//3`
Dielectric 2 –
k = 6 , `t=(2d)/3`
Capacitor are connected in series

`1/C= 1/(C_(1) ) +1/(C_(2) )`
`=1/((Aepsilon_0k_1)/(d_1))+1/((Aepsilon_0k_2)/(d_2))implies 1/C =d_1/(Aepsilon_0k_1) +d_2/(Aepsilon_0k_2) `
`=d/(3Aepsilon_0xx3)+(2d)/(3Aepsilon_0xx6) implies1/C =(2d+2d)/(18 Aepsilon_0)`
`=4/18 d/(Aepsilon_0)implies C =18/4xx9`
`C = 40.5pF`