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Finding the charge on each capacitor

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

E = 12V
C1 = 8 [uF]
C2 = 6.571 [uF]
C3 = 6 [uF]

Find:
1) CT
2) Q1, Q2, Q3 on each of the capacitors.

3. The attempt at a solution

CT was easy. I wasn't sure how to go about finding Q1, Q2 and Q3.

The idea I had was to simply find the voltage across each of them, but that means I have to ohm's law. But I can't use ohm's law with capacitors. If these were resistors I could. But, I don't have a formula that relates voltage, current and capacitance.

I only have Q = VC

Any clues?

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 Recognitions: Homework Help Total charge extract from the cell is Q = C_T*E. Across C2 potential difference is V = E. so Q2= V*C2. In series combination, charge are the same in each capacitor. Hence Q1 = Q3 = Q - Q2.

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Gold Member
 Total charge extract from the cell is Q = C_T*E. Across C2 potential difference is V = E. so Q2= V*C2.
Really? I wasn't sure if E=V in C2, but come to think about it it makes sense since the flow is uninterrupted.

So far so good. Simple enough. :)

 In series combination, charge are the same in each capacitor. Hence Q1 = Q3 = Q - Q2.
A lot more simple than I would've first imagined. .

My thanks!

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Recognitions:
Homework Help

Finding the charge on each capacitor

Quote by Femme_physics
 Quote by rl.bhat In series combination, charge are the same in each capacitor. Hence Q1 = Q3 = Q - Q2.
A lot more simple than I would've first imagined. .
If you would like to more fully appreciate this concept, there is nice visual information here (last diagram on that page): charge on series capacitance; and a discussion on a thread by another PF member here: why do capacitors 'in series' each carry the same charge?