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

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1 Answer
Jun 6, 2018

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.
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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
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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#