Find final charge and final potential difference?

Question

A 2.46 µF capacitor (#1) is charged to 901 V and a 7.19 µF capacitor (#2) is charged to 579 V. The capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. Find final charge and final potential difference on both.

When I tried doing this I got 4.962189e-4 as on of the charges ..but it's not right.. so I dunno what I'm doing wrong.. >_<

Answer 1

q=cv

for the first cap q=(2.46E-6)(901)=2.2E-3 C
for the second cap q=7.19E-6*579=4.2E-3 C

When the capacitors are connected in parrallel the charges sum. So the total charge is 6.4E-3. This makes sense since the charges have to go somewhere -- they're not canceling themselves out. The total capacitance is 9.65E-6 F and the equilibrium voltage is q/c=6.4E-3/9.65E-6=661 V which makes sense as this is between the two voltages that you started out with. Another way to think of this is "what voltage would you need to use to charge two parrallel connected caps so they are charged to those values.