In the figure three are three metallic large plates. The middle plates carries a total charge q. Plates 1 & 2 (which are uncharged) are connected by a wire. Find the charge induced on each surface of 1 & 2. Given `l_(2) = 2l_(1)`

Answer

`q_(1("outer"))=(q)/(2),q_(1("inner"))=(-2q)/(3)`
`q_(2("outer"))=(q)/(2),q_(2("inner"))=(-q)/(3)`

Explanation

Charge on (1) + (2) is zero
`rArr q_(3)-q_(1)+q_(1)-q+q_(3)=0`
`rArr q_(3)=1//2`
(1) & (2) have same potential
`rArr (q_(1)l)/(A epsilon_(0))-((q-q_(1)))/(A epsilon_(0))2 l=0`
`rArr -q_(1)+(q-q_(1))2=0`
`rArr -3 q_(1)+(q-q_(1))2=0`
`rArr -3q+2q =0`
`q_(1)=-(2q)/(3)`
Alternative Solution :
Let the charge on ourter surface of plates (1) be x and on the inner (interfacing) surface of plate )1_ be y, then charge distribution on other plates is shown in figure (by using charge conservation and E = 0 inside the metallic plates)
For electric field to be zero inside the plate 1

`=(x)/(2A epsilon_(0))=((q-x))/(2A epsilon_(0))`
`rArr x=(q)/(2) & q-x =(q)/(2)`
Since `V_(A) =V_(C) rArr V_(A) - V_(B) = V_(C) - V_(B)`
`rArr E_(1) l_(1) = E_(2) l_(2)`
`rArr = (y)/(A epsilon_(0)) l_(1) ((-q-y))/(A epsilon_(0)) 2 l_(1)`
`y l_(1)=2ql_(1)-2yl_(1)rArr3yl_(1)=-2yl_(1)y =(-2q)/(3)`
So, `-q-y=-q+(2q)/(3)=(-q)/(3)`
So, charge on outer surface of plate 1 is
`x = (-q)/(2)`
Charge on inner (interfacing) surface of plate
1 is `y = (-2q)/(3)`
Charge on outer surface of plate 2 is
`q - x = (q)/(2)`
Charge on inner (interfacing) surface of plate
2 is `(-q-y)=(-q)/(3)`
`q_(1("outer"))=(1)/(2),q_(1("inner"))=(-2q)/(3)`
`q_(2("outer"))=(q)/(2),q_(2("inner"))=(-q)/(3)`.