Question

Three large metal plates are arranged as shown. How much charge will flow through the key if it is closed ?

Answer 1

Let on the plate 1, given charge `Q` distributed evenly as `Q/2` and `Q/2` on the either side. Similarly, let on plate 2 charge `2Q` is distributed as `Q` and `Q` on either side.

The charges need to be redistributed as: Two surfaces facing each other must have equal and opposite charges. Now induction will take place due to the higher charge `Q` than `Q/2`.
Therefore, `-Q` will be induced on the right side of plate 1. This induced charge will make total charge on this side of plate 1 as `Q/2-Q=-Q/2`
This will lead to redistribution of charges on three electrically isolated plates before closing of the switch as shown in the figure below.

After closing of switch we see that plates 1 and 3 are at the same potential and plate 2 remains isolated. Let `C_1 and C_2` be capacitance of three plates as shown. Without loss of generality we can assume that charge of both faces of plate needs to redistributed. Therefore, charges on facing sides of plates 1 and 3 only move. Let the final charge distribution be as given in the figure below

Charge on isolated plate 2

`q_1+q_2=2Q` .......(1)

Potential between plates

`V_(21)=V_(23)`
`=>q_1/C_(1)=q_2/C_(2)` .....(2)

Now `C_1=2C_2`, (due to separation being twice between plates.) With this we get

`q_1=2q_2`

Inserting this in (1) and solving we get

`2q_2+q_2=2Q`
`q_2=(2Q)/3`
Also `q_1=(4Q)/3`

Charge moved through switch `K`, movement from surface having more charge to surface having less charge is

`-Q/2-(-q_1)`
`=>-Q/2+(4Q)/3`
`=>(5Q)/6`