Capacitor Circuit (Capacitor as source)

[mtbhrd]

Homework Statement

http://img141.imageshack.us/img141/6173/capcircuitji2.jpg

Basically, I have to get charges and voltages as these switches are opened and closed on this capacitor circuit.

First, Switch A closed, B open. (C₁ is charged.)

Then, switch A opens, and B closes.

V_i=5v
C₁=4.6x10^-6 F
C₂=2.3x10^-6 F

Homework Equations

Q = VC

The Attempt at a Solution

Okay, so the first part is easy. Charge on the the first capacitor Q = VC, [(5)*(4.6x10^-6)] = 2.3X10^-5 C = Q_1i

Now, the second part. Is charge conserved? So that
Q_1i=Q_1f+Q_2f

Expanding that with Q=VC,
C₁V_1i = C₁V_1f + C₂V_2f

Can we say that V_1f = V_2f? So V_f would be 3.35v? Then charges could be found on each capacitor... Is this the correct way to go about this?

Or, Is the voltage the same, 5v across all the capacitors? (Being in parallel.)

 

[LowlyPion]

First C1 will have the charge determined by the 5v and the capacitance.

As you note the charge is conserved when A is opened. Then the effective capacitance is changed by closing B.

Charge the same, but capacitance larger means the voltage is lower

V = Q/(larger C) means lower V.

 

[mtbhrd]

You mention a large effective capacitance..As in combining the two as one effective capacitor in parallel (C1 + C2)?

Using Q = VC , taking the original Q_1i and using (C1 + C2) I get the same 3.35v as I got before.

 

[LowlyPion]

I didn't run the numbers. That looks about right.

I was just trying to provide you some guidance.

 

[mtbhrd]

Right, thanks.

So, with that voltage, charge on each cap (Q=VC) is 3.35v * C1 and then 3.35v * C2

I get Q_1f=1.55x10^-5 C and Q_2f=7.6x10^-6 C

However, when I run conservation of energy, U=.5QV (U_i = U_1f + U_2f) it doesn't work...

EDIT: Well, more simply put: Is it voltage or charge that is the same for both capacitors in this second ciruit situation where switch A is open, and b is closed, allowing C1 to charge the circuit.