[SOLVED] capacitor problem

[soccerj17]

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.

[kamerling]

The voltage on capacitor #2 after you connect them is equal to the voltage on capacitor #1 after you connect them, but this is NOT equal to the voltage on capacitor #1 before you connect them, so you can't use the voltage of 3 x 10^3 V

Use the fact that the total charge on both capacitors doesn't change after the connection.

[soccerj17]

So total charge is the original capacitance (40 pF) times the 3 KV so 120 nC. Then I find voltage by V = Qtotal/Ceq which is C1 + C2 then i use that to find Q by using Q=CV?

[soccerj17]

Alright i got it!