initially the switch is open for a long time. Now the swithc is closed at t=0. find the caharge on the rightomost capacitor as a function of time given that it was initially unchanged.

Explanation

when the switch is open,
(i) `I = (V)/(2R)`
Also `q_0 = CIR = (CV)/(2)`
(ii) when the switch is closed, `(V)/(2) - I(R )/(2) - (q_1)/(C ) = 0`
Also `(V)/(2) - I(R )/(2) - ((q_0 + q-q_1))/(C ) = 0`
`(V)/(2) - I(R )/(2) -(q_0)/(C ) - (q)/(C ) +(q_1)/(C )=0`
,
`(V)/(2) - I(R )/(2) - (q_0)/(C ) - (q)/(C) +(V)/(2) - (IR)/(2)=0`
or `V - IR - (q_0)/(C) -(q)/(C) = 0` or `IR = V- ((q_0 + q))/(C)`
or `R(dq)/(dt) = V - ((q_0 +q)/(C ))` or `int_0^q (Rdq)/(V-((q_0 +q)/(C ))) = int_0^tdt`
or `-RC[ln{V- ((q_0 +q)/(C))}]_0^q = t`
or `-RC[In (V -((q_0+q)/(C)))/(V- (q_0)/(C))] = t` or `(V - (q_0+q)/(C ))/(V-(q_0)/(C)) = e ^((t)/(RC))`
or `V - ((q_0 +q)/(C)) = (V- (q_0)/(C))e^((-t)/( RC))`
But potential differences across each capacitors are same. So
`(q_0 +q - q_1)/(C) = (q_1)/(C)` or `q_0 +q =2q_1`
or `q_1 = (q_0 +q)/(2)`
Putting the value of q, we get
`q_1 = (CV)/(2)[1-(1)/(2) e^(-t)/(RC)]`