Charge on Capacitors in Series After Switch Flipped

[eridanus]

A capacitor, a resistor, and another capacitor is connected in series. The first capacitor, C_1, has an initial charge of Q_0, and C_2 is initially uncharged. The switch is flipped at t=0, what is the charge on each capacitor as a function of time?

So I thought
$$Q_1 = Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}}$$

$$V_1 = Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}}/C_{1}$$

$$Q_2 = C_{2}/C_{1}(Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}})(1-e^{-t(C_{1}+C_{2})/RC_{1}C_{2}})$$
but this is apparently wrong

where am I messing up?
thanks.

 

[Astronuc]

Is the configuration ---C1----R---C2 ---- ?

as opposed to

V_in-----C1---+---C2--- V_out
|
R
|
----------+---------​

The charge diminishes on C1, but increases on C2, so the solution cannot be exp(-At), which would go to zero at t = infinity. The solution must go to some constant, and the charge would be partitioned according to the relative capacitances.

 

[eridanus]

this is the diagram given
http://www.columbia.edu/~kqc2101/circuit.gif

so for Q_2 what would be the voltage drop V that charges C_2 in the equation
$$Q=CV(1-e^{-t/\tau})$$
?