1. The problem statement, all variables and given/known data A 4.6 mu or micro FF capacitor is charged to a potential difference of 15.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 4.6 mu or micro FF capacitor then drops to 7 V. What is the capacitance of the second capacitor? 2. Relevant equations Q = CV Capacitor in series 1/Ceq= 1/C1+1/C2 3. The attempt at a solution I found the charge of the capacitor in the first scenario when the potential difference is 15 V Q = CV = 4.6E-6 F *15 V = 7.35E-5 C Since the potential drops to 7 V, you know the capacitors are in a series so that means the charge remains the same on both capacitors but potential changes Capacitor in series 1/Ceq= 1/C1+1/C2 Q = Ceq * V 7.35E-5 C = (1/(4.6E-6 F) +1/C2)*7 V I solved for C2... C2 = -4.6E-6 C - which is negative so its obvious that its wrong, but could someone tell me where I went wrong that would be greatly appreciated. Thank you.