1. Two capacitors, C1 = 60.0 nF and C2, were connected in parallel with a power supply with E = 12.0 V. The two capacitors were then disconnected from the power supply and from each other and were reconnected with the positive terminal of one connected to the negative terminal of the other capacitor. After reconnection, the voltage across C2 was 4.00 V. C2 is less than C1. Calculate C2 3. well originally it says they are in parallel so q1 = 60*12 q2 = C2*12 After reconnection q1+q2 = q1' + q2' q1' = q2' ? because they are in series? ... if i use this i get the wrong answer but if i use this q1+q2 = q1' + q2' 60*12 + c2*12 = 60*8 + c2*4 8 because still basically the remaining voltage should be across the other capictor (since im thinking they are in series) I solve for C2 = and i get the NEGATIVE RIGTH answer -30nF..while answer is +30nF Help !!!!!!!!