A capacitor of C1 = 10 F is connected to a battery, and a charge of 100 C builds up on the capacitor. Then, the battery is disconnected, the capacitor is discharged, and a second capacitor of capacity C2 = 20 F is connected to it in parallel. Then, the two capacitors are connected to the same battery. What is the charge on the plates (in C) of the second capacitor at that time? Q=CV C1+C2=C t=RC q=Q0e^(-t/T) So far the only thing I have come up with to do is find the voltage of the battery by doing 100=10V. That would mean the battery is 10V, but I do not know if this is right or what to do next.