A 2.7 uF capacitor is charged by a 12 V battery. It is disconnected from the battery and then connected to an uncharged 4 uF capacitor. Determine the total stored energy before the connection, after the two are connected, and the difference between the two energies. OK. So for the first question, obviously, you just find E by using 1/2 Q^2/C. Sounds good. I think the second part is where it gets tricky. So I have this system that now has a total capacitance of 6.7 uF. My instinct is to just say, well, I have a 6.7 uF capacitance system with a charge of whatever I found in the first part, and I can just solve E = 1/2 Q^2/C. There is no voltage, right? Because neither capacitor is connected to a battery. So then the difference is just the second minus the first. I am assuming the second part is wrong because it seems too easy.