Circuit with capacitors and switch

1. The problem statement, all variables and given/known data
This is the circuit diagram I am working with:

There is actually a switch under C3 that I didn’t draw. Initially, the switch was set so that the red dashed line was part of the circuit and the green dashed line was not (so C1, C2, and C3 charged but C4 did not). Then after the capacitors were fully-charged, the switch was flipped to the pencil-drawn circuit there.

I have to determine the charge on C4 in this new configuration after the switch was flipped.

Here are the values given:
ε = 100 V
C1 = 3 μF
C2 = 6 μF
C3 = 4 μF
C4 = 5 μF

2. Relevant equations
[tex]C = \frac{Q}{V}[/tex]
All capacitors in series have the same charge on them, if the voltage is held fixed. (That is the condition, right?)
Charge on conductors will spread so that the surface of the conductor is an equipotential.

3. The attempt at a solution

If I did it correctly, then before the switch was flipped, C1 and C2 each had 200 μC of charge and C3 had 400 μC of charge. At first I just tried distributing that 400 μC across both C3 and C4 (200 μC each) but that doesn’t work. Looking at it later, I see that those two capacitors are still connected to the rest of the circuit, which includes two more capacitors as well as a battery. So it appears to me that a lot of complicated stuff is going on here—charges can still flow throughout the entire thing since it’s all still connected (unless, maybe, since capacitors aren’t actually touching, there are a few segments of conductor in this circuit and each one can have its own charge, but I don’t think that’s possible because the two sides of a capacitor must have equal charges), and two of the capacitors are still in a circuit with a battery.. But it really can’t be this bad, because this was a former exam question so they can’t make it too complicated.

What parts of the circuit should I really be looking at? Where can charges flow and where can’t they? Will all capacitors have the same charge? I don’t think they will.

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