Solve Capacitor Circuit: Is My Method Correct?

[1st1]

Capacitor Circuit: Is My Method Correct?

Homework Statement

Four capacitors, a battery and a switch are assembled in the circuit below. Initially, the switch is set to position A and C4 is uncharged.
At t = 0, the switch is moved to B.
Find Q4, the charge on C4 when the switch is on B.
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam1/fa09/fig20.gif

Homework Equations

Q= CV

The Attempt at a Solution

The initial charge on the C3 capacitor = C3 * V1

Now as the switch to moves to B.

The total charges remains the same, but the voltage changes.

Then

Q = ( C3 + C4 ) V2

V2 = Q / ( C3 + C4 )


next with V2 found

Q4 = C4 V2

*****************

Ok, so that is the only way I've been able to figure this out, however, arent the capacitors in series? In that case I would have to change the equivalent capacitance equation to 1/(1/C3+1/C4) which doesn't work out. Can someone please explain what's going on?
Help would be greatly appreciated. Thanks in advance!

 

[Stonebridge]

The question doesn't quite make sense at the moment.
Is the switch first connected to A, and then to B?

 

[1st1]

Yes, I'm sorry its a three part question and I forgot the main part of the question statement. Fixed now.

 

[1st1]

Anyone able to help? I've got the answer, I just want to know if I am doing it correctly and whether the capacitors are considered to be in parallel or in series and why.

 

[Stonebridge]

Your method seems correct.
In the first part, C1 and C2 are combined in series, and the result of that combination is combined with C3 in parallel. C4 has no effect.
When the switch is on B, some of the charge on C3 moves on to C4 until the pd is the same across both (=V). The total charge that was on C3 (=Q) is now distributed between C3 and C4. So C3xV + C4xV= Q
This gives you V
Knowing V will give you Q for C4.

 

[1st1]

Your method seems correct.
In the first part, C1 and C2 are combined in series, and the result of that combination is combined with C3 in parallel. C4 has no effect.
When the switch is on B, some of the charge on C3 moves on to C4 until the pd is the same across both (=V). The total charge that was on C3 (=Q) is now distributed between C3 and C4. So C3xV + C4xV= Q
This gives you V
Knowing V will give you Q for C4.

Thank you, the bold part is where I was confused. But now I see I was looking at it the wrong way. Thanks!