Capacitor Charge Calculation: C1 and C2 in Series

[i_island0]

there are two parallel plate capacitors C1 and C2 connected in series across a battery of emf V with a switch S. The capacitor C1 has some initial charge q0 while C2 is uncharged.
I have connected the positive terminal of battery (of emf V) to the positive plate of capacitor C1.
Final charge on the two capacitors that i got are:
Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)
is my answer correct?

 

[rl.bhat]

In the series combination of the capacitors charge on each plate must be the same irrespective of the initial charge.

 

[i_island0]

sorry sir, but i m not able to understand your point. can you then tell me the charge on both capacitors as a function of time.

 

[dynamicsolo]

Final charge on the two capacitors that i got are:
Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)
is my answer correct?

I think these expressions need some clarification and examination. Is VC1 intended to represent the initial "voltage drop" across C1, that is,

VC1 = q0/C1 ?

If so, there is a problem: VC1 and q0 have different units (voltage vs. charge), so the difference (VC1 - q0) in your expressions would not be meaningful.

Perhaps it would be helpful to show how you arrived at these results. You will not be able to construct functions of time describing the charges, since this idealized problem has no resistances in the circuit: you will not be able to calculate current. In any case, I don't think your expression can be correct.