Capacitance of a capacitor A, when a dielectric is placed between the capacitor then charged to potential difference 110V is?

Options

A. `2 xx 10^(9)F`
B. `2.5 xx 10^(9)F`
C. `3.6 xx 10^(6)F`
D. `3 xx 10^(6)F`

Answer

A

Explanation

Here, area of the plates of capacitor A, `A_(1)= 0.04m^(2)`
Area of the plates of capacitor B, `A_(2)= 0.02 m^(2)`
Separation between the plates of the two capacitors, `d_(1)= d_(2)= 8.85 xx 10^(-4) m`
Suppose C. and C.. is the capacitor of two parts of capacitor A. The total capacitance will be given by `C_(1)= C. + C.. =(epsi_(0)KA_(2))/(d) + (epsi(A_(1)-A_(2)))/(d)`
SInce `A_(1)= 2A_(2)` it follows that `C_(1)= (epsi_(0)KA_(2))/(d) + (epsi_(0)A_(2))/(d)= (epsi_(0)A_(2))/(d) (K+1)`
`=(8.85 xx 10^(-12) xx 0.02(9+1))/(8.8 xx 10^(-4))`
`=2 xx 10^(-9)F`