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**[SOLVED] capacitor problem**

## Homework Statement

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.

## Homework Equations

pF= 1 F x 10^-12

kV= 1 V x 10^3

nC= 1 C x 10^-9

Q = CV

## 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.