How Is the Charge on Capacitor C1 Calculated When Switched from Position A to B?

[Dell]

given the following:
http://lh3.ggpht.com/_H4Iz7SmBrbk/ShATOIWSXEI/AAAAAAAAA_4/M57_Bf03Z2M/s720/Untitled.jpg

and knowing:
C₁=400μF
C₂=500μF
ε=9V
R=20kΩ

the switch is closed to position A for a time of t=RC₁/2 and then moved to position B
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what is the charge of C₁ the moment the switch is moved from A to B??


as far as i can see none of my equations have anything to do with time, other than P=VI (since W=J/s) but i don't think that helps me very much,, how do i incorporate the element of time into this, i tried kirchhoff but that didnt really get me anywhere

q₁C₁+IR=-ε
since i don't know q₁ I or R i think this is not the right way to go,
i would think that the R needs to cancel out since the t is dependent on R, so i am looking for something that will give me t/R

i thought that since W=J/s, P=E/t = q²/2C * 2/RC₁ but that doesn't help either

 

[nickjer]

There is an equation for exponential charging of a capacitor in a RC circuit.

 

[Dell]

what is that equation

 

[nickjer]

You never solved for the differential equation of an RC circuit in your book or class?

Q = Q_0 \left( 1 - e^{-t/RC}\right)

That is the equation for a battery charging a capacitor in an RC circuit. Where Q_0 = C\epsilon.

 

[Dell]

nope, never seen that before, probably in the next few lessons, what is the root of this equations?

 

[nickjer]

Are you doing homework ahead of time?

Too much to explain but I found this on google:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/RCSeries.html

It explains how a capacitor charges and discharges in an RC circuit.