Initially the switch is open for a long time. Now the switch is closed at t=0. Find the charge on the rightmost capacitor as a function of time given that it was initially uncharged.

Answer

`q = (CV)/(2) (1-(1)/(2)e^(-t//RC)`

Explanation

Before closing the switch charge distribution will be-
`V_(AB) = (V)/(2R)' R = (V)/(2)`

Let switch in closed at `t = 0` and charge on
right capacitors is `q(q_(1) = (CV)/(4)` between each capacitor) So charge on capacitor

For loop (1)
`V -(l_(1)+2l_(2))R -l_(1) R = 0`
`2l_(2)R = V-2l_(1)R ....(1)`
For loop (2)
`l_(1) R - (q)/(C ) = 0 Þ l_(1) R = (q)/(C )...(2)`
From (1) and (2)
`2l_(2)R = V -(2q)/(C )...(3)`
we have `l_(2) = (dq)/(dt) ..(4)`
From (3) and (4)
`2R(dq)/(dt) = (CV-2q)/(C )`
`overset(q)underset(q_(l)=CV//4)Ò(dq)/(CV-2q)=overset(t)underset(0)Ò(dt)/(2RC)`
`overset(é)(ê)"ln"(CV_2q)/(-2)overset(ù^(q))(ú)_(CV//4) (t)/(2RC)`
`(CV-2q)/(CV-CV//2) = e^(-t//RC)`
`CV -2q = (CV)/(2) e^(-t//RC)`
`2q=CVoverset(æ)underset(è)(C1)-(1)/(2)e^(t//RC)overset(Ö)underset(Ø)(./div)`
`q=(CV)/(2)overset(æ)underset(è)(C1)-(1)/(2)e^(t//RC)overset(Ö)underset(Ø)(./div)`
An soon as switch is closed at the same momnet charge sharen between two capaitor which is initial charge on each capacitor.

`tau = (R )/(2) xx 2C = RC`
Time varying charge on capacitor
`q = (V)/(2) xx 2C (1-e^(-t//RC))) +2q_(0)e^(-t//RC) = CV(1-e^(-t//RC) +(CV)/(2) e^(-t//RC)`
`q = (CV) (1-(1)/(2)e^(-t//RC))`
time verying charge on each capacitor `=`
`(q)/(2) = (CV)/(2) (1-(1)/(2)e^(-t//RC))`