Circuit with three capacitors and a switch I have a homework problem where I have a battery connected to a capacitor and a switch that connects the capacitor to 2 others in series/disconnects the circuit from the battery. 
The question is:

S is initially closed to the left until c1 is completely charged. Once charged, it closes to the right and remains closed there until it reaches equilibrium. Calculate the difference in potential of c1.

I'm stumped because I don't know how the charges distribute between the three capacitors once the circuit is closed, and so I'm not able to calculate the final voltage. I probably didn't do too good of a job translating the text so if you don't understand ask. Thank you.
 A: Remember: Capacitors in parallel all have the same voltage across their plates, but can have different charges. 
So, when the switch is closed to the left, C1 is has a certain charge and voltage. Then, when the switch is closed to the right, that charge in C1 distributes in such a way so that the voltage across all three capacitors is equal.
You can intuitively justify it this way—imagine if one of the three capacitors had a lower voltage. We know that electrons will move from higher potential to lower potential (like a ball rolling from the top of a hill to the bottom); so if one of the three capacitors had a different voltage, charge would just redistribute itself (since all of the lower plates are connected to one another) until none of the capacitors had a lower voltage; that is, until they all had the same voltage.
You should be able to solve the problem using this, and the various rules for capacitors connected to one another.
A: The total charge on $C_1$ ends up distributed across the three capacitors, so that the final voltage on the three of them will be the same.
Now you know that $Q=CV$. When the switch is thrown, the total charge $Q$ is the same, but the capacitance $C$ becomes 5 times greater (10+20+20).
