Fig. 25-39 shows a 12.0 V battery and four uncharged capacitors of capacitances C1 = 1.00 µF, C2 = 2.00 µF, C3 = 3.00 µF, and C4 = 4.00 µF. If only switch S1 is closed, what is the charge on (a) capacitor 1, (b) capacitor 2, (c) capacitor 3, and (d) capacitor 4? (figure at http://imgur.com/xOR2AU4)
V is same for capacitors in parallel, P is same for capacitors in series
The Attempt at a Solution
a) C1 and C2 are parallel, as are C3 and C4.
Ct (capacitance of total) = 1/(C1+C2) + 1/(C3+C4) = 0.48
Q = (Ct*Vt) = 0.48*12 = 5.71 µC
Q1+Q2 = 5.71 (because both come from positive end of battery)
Q1/(1µF) = 2Q1/(2µF)
3Q1 = 5.71µC
Q1 = 1.9 µC
The correct answer is 9 µC, and I think I went wrong with the calculation of the initial capacitance, but I'm really not sure.