A parallel plate capacitor has a capacity `80 xx 10^(-6)` F, when airs present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 V by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

Options

A. ` 4.56 xx 10^(-3)` C
B. ` 25.3 xx 10^(-3)` C
C. ` 120 xx 10^(-3)` C
D. ` 125 xx 10^(-3)` C

Answer

(a)

Explanation

Charge through the wire,
`Delta q = Delta CV = (C_(2) - C_(1))V =(K - 1) sigms V`
` = (20 - 1)( 80 xx 10^(-6))(30) = 4.56 xx 10^(-12)`
= `4.56 xx 10^(-3)` C