Calculating Charge on Second Capacitor Connected to Battery

[santina91]

A capacitor of C1 = 10 F is connected to a battery, and a charge of 100 C builds up on the capacitor. Then, the battery is disconnected, the capacitor is discharged, and a second capacitor of capacity C2 = 20 F is connected to it in parallel. Then, the two capacitors are connected to the same battery. What is the charge on the plates (in C) of the second capacitor at that time?



Q=CV
C1+C2=C
t=RC
q=Q0e^(-t/T)



So far the only thing I have come up with to do is find the voltage of the battery by doing 100=10V. That would mean the battery is 10V, but I do not know if this is right or what to do next.

 

[santina91]

note: it is all microcolumbs and microfarrhads

 

[LowlyPion]

note: it is all microcolumbs and microfarrhads

First of all - please. farads after Michael Faraday.

Second of all here's a μ to help you.

As to the capacitors I think all they want is for you to figure the battery voltage and then figure the charge on the second capacitor.

 

[skeptic2]

This problem is much easier than you're making it.
You have the formula Q = CV. You don't know what V is but you know it is constant.
C2 is twice as large as C1.
What happens to Q in the equation if you double the value of C and keep V constant?

 

[radenm]

just remember that U=CV
from the givens you found that the voltage of the battery was 10V
now all you have to do is use that voltage in the equation to find the energy of the second capacitor.