Capacitor is charged then connected to a capacitor and battery

1. Apr 28, 2014

TheEvenfall

Hello PF,

1. The problem statement, all variables and given/known data
The capacitor C_1 is connected to a battery as shown. After being charged, it is connected to the capacitor C_2 and the second battery as shown. When all the charges come to rest, what will be the charge on capacitor C_2?
http://imgur.com/4ciJRrn

2. Relevant equations
C = Q/V

3. The attempt at a solution
The initial charge on C_1 is Q_1 = C_1 V_1
After it is connected to the other circuit and in steady state, I used the loop rule to get the equation (q_1/C_1) + (q_2/C_2) = V_2, where q_1 is the new charge on C_1 and q_2 is the charge on C_2.
Now the problem is relating q_1, q_2 and Q_1. According to my professor's solution, -Q_1 = -q_1 + q_2.
He got that equation by doing what's shown here: http://imgur.com/WvKJ6rQ, and I have no idea where he got that from, and this is the point that I'm stuck on and why I'm posting this.
Using that equation we can find q_2 easily, but I don't understand how he got it.

Thank you.

2. Apr 28, 2014

Staff: Mentor

The lower plate of C1 and left plate of C2 along with the wire connecting them form an isolated "island" for charge. Whatever charge is on that island must remain, and there's no way for new charge to appear there.

So label the plates for their charges before and after the switching occurs. If the total charge on the island can't change then you can equate the sum of the charges on the plates for the before and after conditions.

3. Apr 28, 2014

TheEvenfall

Got it, thanks a lot.