Two capacitors with capacitances of 1.0 μC and 0.50 μF, respectively, are connected in series. The system is connected to a 100 V battery. What charge accumulates on the 1.0 μF capacitor?
For capacitors in series: 1/C(equiv) = 1/C1 + 1/C2
V = Q/C
The Attempt at a Solution
I used the second equation (V=Q/C) to find the charge with V = 100 and C = 1x10^(-6) but I did not get the right answer. However when I found the equivelant capacticance and used that in the second equation to find Q I got the right answer. Can anyone explain whey the second why to do this problem is the correct way?