Circuit with capacitors

1. Apr 23, 2015

toothpaste666

1. The problem statement, all variables and given/known data

2. Relevant equations
Q = CV

3. The attempt at a solution
C1 charges to Q0 = C1V = (8 x 10^-6 F)(12 V) = 9.6 x10^-5 C
when the switch is flipped the total charge is the same (Q0) but it is spread over the two capacitors which are now in parallel. so Q1 + Q2 = Q0
so C1V1 + C2V2 = Q0
but the capacitors are in parallel so they have the same voltage
so V(C1+C2) = Q0
V = Q0/(C1+C2) = (9.6x10^-5)/(8x10^-6 + 12 x10^-6) = 4.8 V
which is the answer to part a)
for part b)
Q1 = C1V = (8x10^-6)(4.8) = 3.84 x 10 ^-5 C
Q2 = C2V = (12x10^-6)(4.8) = 5.76 x 10^-5 C

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2. Apr 23, 2015

Staff: Mentor

Your work looks good. Did you have a question?

3. Apr 23, 2015

toothpaste666

Thank you. Just wanted confirmation that I understand correctly. Particularly that I was correct about the two capacitors being in parallel after the switch was flipped. Sometimes it is hard for me to tell.

4. Apr 23, 2015

toothpaste666

i guess i just wanted to be sure that C1 and C2 were in parallel after the switch is flipped.

5. Apr 23, 2015

Staff: Mentor

A rule for judging whether components are in parallel is that they share exactly two nodes. That's why components in parallel share the same potential difference.

A rule for judging whether components are in series is that they exclusively share a node (no other connections to other components at that node).

When you have only two components and they are connected as in this problem they satisfy both series and parallel criteria and you are free to interpret the scenario either way.

6. Apr 23, 2015

toothpaste666

So I could have done this by saying they are in series?
but then Q1=Q2=Q0 right?

7. Apr 23, 2015

Staff: Mentor

You could do it by saying that they are in series, but that would be a more complicated route to take. The parallel paradigm is the simple interpretation.

The charges would be the same as Q0 if they were in series when the initial charge was instilled (since they would then share the same charging current). But this is not the case here. One capacitor was charged all alone while the other remained uncharged. When they are connected via the switch, the charged capacitor acts as a source and THEN a shared current takes place. So the change in charge is the same for each from that point on. Current stops flowing when the potential differences on the two capacitors equalize.