# Homework Help: Conceptual problem on capacitors

1. Jan 23, 2012

### tsw99

It is neither a homework nor coursework problem, but I really want to know the solution.
1. The problem statement, all variables and given/known data
3 charged capacitors are connected as shown in the diagram. What will be their final charges, potential difference when equilibrium is reached? Units are arbitrary.
http://imageshack.us/photo/my-images/19/cap1zu.jpg

2. Relevant equations
Q=CV obviously

3. The attempt at a solution
(1) Should I use the charge conservation?
E.g. The total charge on left plates of C1 and C2 is 1+4=5 units, on right plate of C2 and left plate of C3 is -4+9=5 units, on right plates of C1 and C3 is -1-9=-10 units
(2) p.d. across C1 = p.d. across C2 and C3?
Equivalent capacitance across C2 and C3 = 1.2 unit

Then I cannot continue...

How about the case if the polarity of one of the capacitors is reversed? I feel really, really confused.

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2. Jan 23, 2012

### technician

The way the circuit is drawn and the polarities shown mean, I would say, that C2 and C3 are in series and C1 is in parallel with that combination.
Do you know how to calculate combined capacitance?

3. Jan 23, 2012

### Staff: Mentor

One way to approach such a problem, where capacitors have some initial charge, is to replace each capacitor by an equivalent circuit consisting of an uncharged capacitor of the same value in series with a voltage source equal to the initial voltage. Take care to maintain the correct polarity for the combination. Thus in your case:

Then to see how much charge is going to move in the circuit (and since they are all connected in series the same amount of current should flow through each), sum the voltages to find the net EMF and find the net capacitance. The ΔQ for that net capacitance should then be applied to the original capacitors with their initial charges. Pay attention to the current direction and thus whether charge will be added to or subtracted from individual capacitors.

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Last edited: Jan 23, 2012
4. Jan 23, 2012

### technician

I would say that you are correct to recognise that V1 =V2 +V3. This means that V1 is in parallel with (V2+V3) in series. This also means that Q2 = Q3 and the total charge = Q1 + Q2. (or Q1+Q3)
Therefore the combined capacitance = (Q1+Q2)/V1.
I would say that C2 and C3 are in series (as drawn) so the charge on C2 = charge on C3 = charge in the series arm of the circuit.
I would say the combined capacitance is C = (C1C2 + C1C3 +C2C3)/(C2 + C3)

5. Jan 24, 2012

### tsw99

Are you sure C2 and C3 are in series? Because they don't have same charges.

6. Jan 24, 2012

### tsw99

Thanks for your suggested solution, but I completely do not understand how it works. Could you please elaborate on it?
Using your method, I figure out the net EMF is 4V, but I do not understand "how much charge is going to move in the circuit", there should be none right? Because in fact it is an open circuit.

7. Jan 24, 2012

### Staff: Mentor

I put the open switch in there to make the point that there are initial conditions on the capacitors --- they're given some charge prior to be connected together as a completed circuit. No charges will move or current flow until the switch is closed, after which the circuit is identical to the one you described.

So, if there's an EMF of 4V and all the capacitors are uncharged to begin with, a while after the switch closes what will the be the individual charges on each capacitor? One way to figure this out is to first determine the equivalent capacitance of the three capacitors combined (They are all three in series). Then find the charge that the voltage supply will put on that equivalent capacitor. Add (or subtract, accordingly) that amount of charge to the original charges on the three capacitors.