In the following diagram that plates are conducting and are held parallel then amount of charge that flows through the switch after closing it is

Options

A. `Q`
B. `(5Q)/(2)`
C. `(7Q)/(2)`
D. `2Q`

Answer

D

Explanation

`q_(1)+q-Q-q+q_(2)=5Qq_(1)+q_(2)=6Q`
`(q)/(epsilon_(A))d+((q-2Q))/(epsilon_(0)A)2d+((Q+q))/(epsilon_(0)A)4d=0`
`q+2q-4Q+4Q+4q=0 rArr q=0`
`E_(A)=0rArr (q_(1))/(2epsilon_(0)A)-(q_(2))/(2epsilon_(0)A)=0rArr q_(1)=q_(2)`
` rArr q_(1)=q_(2)=3Q`
`rArr Delta` q that flows through switch is `2Q`.