i_island0 said:
Moreover, I am thinking that before the switch was closed, the right plate of capacitor C1 and left plate of capacitor C2 had a total charge of -qo. Those plates are electrically isolated from rest of the circuit. So, the final charge on them should be same, i.e. -qo.
If the capacitors start out electrically neutral and a charge of +q0 is then applied to the left plate of C1, as you have been showing, the following will happen very rapidly (on a time scale of perhaps nanoseconds):
the applied charge will spread itself as uniformly as possible over the left plate of C1;
the electric field generated by that plate will attract a charge -q0 to the right plate from the conductive materials making up the right plate of C1, the connecting wire, and the left plate of C2; the potential drop across C1 will arrive quickly at q0/C1 ;
the same electric field of C1 will repel a charge in the amount +q0 to the left plate of C2; this will happen because the connected plates and wire were initially electrically neutral;
the charge of the left plate of C2 will in turn produce an electric field which will attract a charge -q0 to the right plate of C2, causing the potential drop across C2 to rapidly reach q0/C2 .
It is not necessary for the switch to be closed, as this effect is not caused by a flow of charge carriers, but by the action of the electric field from the "excess" charge. (When you get to Maxwell's Laws, you'll run across something called "displacement current". Certain actions of electric fields across open space mimic those produced by a physical current.) What you now have resembles the condition of the circuit if it started out uncharged and was allowed to be charged by the battery up to the point where the charge on each capacitor was q0.
Earthing does not enter into things in this circuit. Nothing in your diagram shows any components connected to the Earth; instead, you are simply applying some charge to an isolated set of conductive materials "linked" by electric fields across gaps.
q = CV(1 - e^(-t/CR)) + qoe^(-t/CR) where, C = C1C2/(C1 + C2)
I get this also: what you have is the sum of a "transient term",
(qo) · e^(-t/CR) , which goes to zero "in the long term", and an "asymptotic term", CV(1 - e^(-t/CR)), which goes to CV "in the long term". Again, this shows that the charge on the two capacitors will end up the same; the
voltage across each one will differ:
V1 = C·V/C1 and V2 = C·V/C2 , with V1 + V2 = V .
The transient term shows that, within reason, the initial charge q0 has no effect on the equilibrium situation. This can also be seen by going back to the differential equation
q/C1 + q/C2 + (dq/dt)R - V = 0 ,
or solving for dq/dt,
dq/dt = (V/R) - (q/RC) , again with C being the "effective capacitance".
The equilbrium solution is found by setting dq/dt = 0, giving
(V/R) = (q/RC) , or q = CV (as expected).
In terms of the properties of this differential equation, if q < CV , then dq/dt > 0 , meaning that the charge will rise asymptotically from its initial value q0 to the equilibrium value CV. Similarly, if q > CV, then dq/dt < 0 , so the charge will fall asymptotically from q0 to CV. This will apply as long as q0 is not
so big that the "breakdown voltage" of the capacitors is approached or exceeded.
Where does the excess charge q0 > CV go, if it doesn't matter in the end? The battery acts as a sort of reservoir and takes up excess energy associated with it. In the case of the battery charging the circuit (q0 < CV), the battery "discharged" and is the source of the energy to separate charges in the circuit. If there is excess charge (q0 > CV), the effect would be to "charge" the battery, supplying energy to it (much as what happens when other circuits send current toward the positive terminal of a battery); the reduction of potential energy in the circuit would also reduce the amount of separated charge.
All of this is somewhat idealized and is neglecting a certain amount of dissipation in the wires and the battery itself. I don't quite have the conservation of charge issue resolved in the case where q0 > CV . It seems like it ought to be possible to better describe what happens; I can't say I've ever seen that situation discussed. It may be that this model doesn't adequately apply in that case.