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Here is a picture of the given circuit. Initially the switch is connected to the left side, and the question is what are the charge and potential difference across each capacitor ( capacitance is in $\mu F$ ).

I want to make sure my understanding of this problem is clear. Here is my analysis. Before the switch is closed, all the potential from the $100$ V battery is stored into $C1$. After the switch is flipped to the right, $C1$ discharges some of its potential to $C2$ and $C3$, determined by their capacitances.

All three capacitors are connected in series, so $$C_{tot} = \Big( \frac{1}{C1}+\frac{1}{C2}+\frac{1}{C3}\Big)^{-1}$$

Giving way to explicitly calculate the voltage across $C2$ and $C3$. In this case the voltage is given $V=(\frac{C_{tot}}{C})\cdot \Delta V$. Where $\Delta V$ is the "total potential" = $100V$. In series, the charge across $C2$ and $C3$ is equal. However, the potential across $C1$ cannot be calculated this way, nor is the charge equal to the charge across $C2$ and $C3$. Instead, the potential across $C1$ is given via Kirchhoff's law : $V1 -V2 -V3 = 0$.

The role of $C1$ is certainly a bit different from the other two in this circuit, but I can't accurately explain why it behaves differently as it relates to the voltage and charge, even though it is connected in series with the other two. I think it is because $C1$ is the "source of voltage" in the right circuit. How exactly does $C1$ relate to the other two capacitors?

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It is understandable that you assumed that the capacitors are in series. But notice that once $C1$ is charged, the top plate (assume for simplicity that these are parallel plate capacitors) of $C1$ is positively charged. And once the switch is flipped, the top plates of capacitors $C2$ and $C3$ are positively charged.

Now, because the positively charged plates of $C1$ and $C2$ are directly connected (and so are the negatively charged plates of $C1$ and $C3$), it is actually the case that $C1$ is parallelly connected to the combination of $C2$ and $C3$. Also, notice that it is indeed the case that $C2$ and $C3$ are connected in series because the negatively charged plate of $C2$ is connected to the positively charged plate of $C3$.

From here on, I urge you to follow through on your ideas to see if you can solve the problem.

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When you use the equivalent capacitance formulae for series and parallel capacitors you have at the back of your mind the following ideas.

Capacitors in parallel all have the same potential difference across them.
Capacitors in series all have the same charge on them.

So in this case the two capacitors on the right have the same charge on them so they are in series and those two capacitors which are in series have a potential difference across the both of them which is equal to the potential difference across the capacitor on the left so they are in parallel.

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