[SOLVED] capacitor problem 1. The problem statement, all variables and given/known data A 40 pF capacitor is charged to 3 kV and then removed from the battery and connected in parallel to an uncharged 70 pF capacitor. What is the new charge on the second capacitor? Answer in units of nC. 2. Relevant equations pF= 1 F x 10^-12 kV= 1 V x 10^3 nC= 1 C x 10^-9 Q = CV 3. The attempt at a solution I thought that because they're connected in parallel that means the voltage is the same across them so I used Q=CV to find charge. For the second capacitor the capacitance is 70 x 10^-12 C and I used the voltage of 3 x 10^3 V. Multiplying them together I got 210 x 10^-9 or 210 nC. I submitted this online to our answer service and it was wrong, and I don't know what I did wrong.