Find `:`
(i) Potential difference between points M and N when switch 'S' is open.
(ii) Potential of point N after long time when switch is closed.
(iii) How much does the charge after long time when S is closed ?

Explanation

(i) when S is open there is no current at steady state. Therefore
`V_(M) = 18 ` volt
`V _(N) = 0` volt
`implies V_(M) -V_(N) = 18` volt
(ii) When 'S' is closed, after long time steady state reaches and current through resistor
` I = ( 18)/( 6+3)A`
`V_(A)-V_(M) = 6 xx 2 ` volt
`V_(N) -V_(B) = 3 xx 2` volt
`:. V_(B) = 0` volt `:. V_(N) = 6 `volt

(iii) When switch S is open charge on each capacitor is due to same potential difference 18 volt.
`q_(1) = 18 xx 6 muC `
`= 108 muC `
`q_(2) = 18 xx 3 muC ` ,brgt `= 54 muC `
When S is closed at steady state charge on each capacitor becomes.

`q_(1)^(')= 12 xx 6 muC = 72 muC `
`q_(2)^(') = 6 xx 3 mu C = 18 mu C `
` Deltaq_(1)= ( 72 - 108 ) = - 36 muC `
` Deltaq(2) = ( 18 - 54) = - 36 muC`