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Homework Help: Charge Flown Through Battery In Charging Capacitor

  1. Jul 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Image URL:
    http://img13.imageshack.us/img13/1935/capacitor.png [Broken]
    Before the switch was closed, Capacitor 1 had a charge of magnitude CE/3 on its plates and Capacitor 2 had a charge of magnitude CE/6. The third capacitor was initially neutral.
    Find the charge flown through battery when switch 'S' is made to close where E is the emf of the battery.

    3. The attempt at a solution
    Firstly, I solve this problem using the conventional method. I assume that a charge 3q flows out from the positive end of the battery. It divides itself in the ration 2:1 between capacitors 1 and 2, in order to keep the potential across them equal, since they are connected in parallel. Therefore the net final charges on the capacitors are:
    1== (CE/3)+2q
    2== (CE/6)+q
    3== 3q

    From potential balancing in the circuit when steady state is achieved, I get:
    therefore charge flown through battery=3q=2CE/5

    Till now, I had no problem. But after this method, an idea struck me:
    I calculated the initial total +VE charge=(CE/3)+(CE/6)=CE/2

    Now, I calculated the final total +VE charge.
    C(equivalent for the circuit)=3C/5
    The net positive charge on the equivalent single capacitor would have been 3CE/5.

    This extra positive charge must have come from the charge flown through the battery.

    Charge flown through battery=3CE/5 - CE/2= CE/10 (which is different from the one calculated previously)

    This method worked right until now and I thought it was another version of charge conservation. But this problem really shocked me literally. Please explain the anomaly with the second method.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 3, 2009 #2


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    Your second approach is correct.
  4. Jul 3, 2009 #3
    I wonder if you could please explain to me the reason behind it. If method 1 is wrong, then there must be something that I did wrong in it. What is that?
    PS. Sir, please answer me only if you are sure about it.
    Last edited: Jul 3, 2009
  5. Jul 3, 2009 #4
    This statement is wrong. Think about the potentials of each capacitor more carefully.

    Your second approach is actually the typical, compact way to solve capacitor questions. Do you mean you don't understand why it's correct?
  6. Jul 3, 2009 #5
    Actually, I dont understand, why the first method is wrong? Potential difference across 1 is E/3
    across 2 is also E/3 and across 3 is 0 initially. So why wont the charge divide itself in ratio 2:1 ???

    Yes, please explain to me the 2nd method via the method of charge flow from the battery. I am simply confused.
  7. Jul 4, 2009 #6
    This is for all the members at PF. I still need help on why the first method went wrong. queenofbabes and rl.bhat told me the first method was wrong. But I stillcant figure out, why? Please help me!!!
  8. Jul 4, 2009 #7
    Well, think about the amount of charge you start with in the problem. Is the assertion in your first method still correct?

    I would stay put and type a more detailed explanation, but unfortunately I'm busy too =( But if u can wait another 12 hours or so, I should be able to come up with something.
  9. Jul 4, 2009 #8


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    1== (CE/3)+2q
    2== (CE/6)+q
    3== 3q

    This assumption is wrong.
    If the capacity 3 has 3q charge, then the parallel combination of 1 and 2, which is in series with 3, must have 3q charge including the initial charges.
  10. Jul 5, 2009 #9
    I would still ask, why? What makes capacitor 3 to have a total charge which is the sum of the charges present on 1 and 2?
    See, if I take this assumption:
    1== (CE/3)+2q
    2== (CE/6)+q
    3== 3q
    to be true,


    1== 0.6CE
    2== 0.3 CE
    3== 0.4 CE

    0.6CE+ 0.3 CE is not equal to 0.4 CE, yet there seems to be no problem in the circuit, if I analyze it.
    Potential drop across the parallel combination of 1 & 2=0.6 E
    Potential drop across capacitor 3= 0.4 E
    Net drop= E
    which is the emf of the battery.
    Which law of physics did my assumptions in method 1 violate?
  11. Jul 5, 2009 #10
    Sure, I am ready to wait another 12 hours. All I want is my concepts to be crystal clear. I'll be waiting for ya.
  12. Jul 5, 2009 #11


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    In series combination of capacitors, magnitudes of the charges in the all the plates must be the same.
    If two capacitors are connected in series to a battery, plate one of the first capacitor is connected to the battery, plate 2 of the first capacitor is connected to the plate one of the second capacitor. Plate two of the capacitor is connected to the battery. According to the conservation of charges, charge on the plate one of the first capacitor must be equal and opposite to the charge on the plate 2 of the second capacitor.
  13. Jul 5, 2009 #12


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    Hi Ritwik! thanks for your PM of last night. :smile:
    I'll just add this to what rl.bhat :smile: has just said about conservation of charge:

    imagine the switch is closed, but there is no emf …

    what happens to the charge then? :wink:
  14. Jul 5, 2009 #13
    I think you are assuming the initial charges are somehow fixed (even if so the assumption is still wrong). You don't see why capacitor 3 must have the same amount of charge as the sum of 1 and 2? I'll try illustrating it in my attachment.

    now coupled with the hints from other posters, hopefully you can now understand what's going on.

    Attached Files:

  15. Jul 5, 2009 #14
    OK, I get what you want to say. But I guess that it would be correct only if the capacitor in series were uncharged initially.

    Lets shift our discussion to a 2 capacitor system in diagram given below.
    http://img200.imageshack.us/img200/6681/capacitor.jpg [Broken]
    Capacitor 1 has a charge of 0.6 CE initially.
    Capacitor 2 has a charge of 0.4 CE initially.
    I connect their series combination to a battery of emf 'E'. I would say that no charge would flow from the battery.
    And according to you, some charge must flow from battery since this is not the steady state wherein plate 1 of capacitor 1 has an equal and opposite charge as plate 2 of capacitor 2, right?
    And my question is what would make the charges flow? There is no potential difference anywhere to support charge flow. I have marked potential regions in my diagram V1,V2 and V3(initially).

    V1-V2=0.6E ..................... via capacitor 1
    V2-V3=0.4E ..................... via capacitor 2
    V1-V3=E ..................... via the battery

    adding equation one and two:
    V1-V3=E ........................ which seems perfectly in agreement with the battery potential difference. So will the charges yet flow???

    First of all, thanks a lot for responding to my request. Well, I am in a conceptual dilemma. Had there been no emf, the capacitors would bear the same magnitude of charges on their plates. But this is not my problem dear tim. Just see the example I gave above. Do you think the charges would flow out from the battery? If yes, then why? There seems to be no potential difference as such.
    Last edited by a moderator: May 4, 2017
  16. Jul 5, 2009 #15
    Imagine both capacitors in your example were uncharged to start with. Connect the batteries, and positive charges flow to the left plate of C1. Now, for every positive charge entering the left plate, a negative charge leaves the right plate. This negative charge MUST go somewhere, and it will go to the left plate of C2. And this SAME amount of negative charge will leave the right plate of C2 and return to the battery, satisfying the conservation of charge.

    Do you see it now? The charges displaced by the right plate of C1 can only end up on the C2 If not you'll have a net charge on the system which is not what your started with. Now do you understand why capacitors in series MUST have the same charge?

    So in your example, we know effective capacitance is 0.5C, so overall charge is 0.5CE. Now we have to be careful: what we mean here is that the left plate of C1 has 0.5CE of charge, as does the right plate of C2. After all we're now treating the system as one capacitor. Of course, correspondingly there is 0.5CE of charge on the right plate of C1 and left plate of C2, but it does NOT affect our analysis here. It does NOT contribute to the "overall charge" we just calculated.

    Thus what happens is that 0.1CE of charge flows from right plate of C1 to left plate of C2. 0.1CE of charge flows from left plate of C1 to battery, and 0.1 CE of charge flows from battery to right plate of C2.

    Now let's see what went wrong with your argument. You are quite right to say that the potential differences add up to EMF of battery, but the problem is that this is NOT a steady state! Cuz the right plate of C1 is at a higher potential than the left plate of C2, so there is an emf driving charges around.

    Now, are you able to answer tim's question, and subsequently your original question, correctly?
  17. Jul 5, 2009 #16


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    When you connect a capacitor ( charged or uncharged ) to a source of EMF, the electrons from one plate will flow to other plate until the potential difference across capacitor is equal to the applied EMF. In the given problem, the potential difference across the system is E/3. It is connected to a source of EMF E. The electrons from left plates of C1 and C2 will flow to the right plate of C3 until the PD across the system is equal to the applied EMF.
    Here the battery does not supply the charges. It acts like a pump, which transfer the electrons from one plate to other plate.
  18. Jul 7, 2009 #17
    Yup! No objections till now.

    It doesnt necessarily follow that the capacitors in series always have same charge.
    I would have believed all this, if and only if I had no doubts regarding this particular statement below.
    Can you prove that when the inner surface of plate 1 of capacitor 1 has a charge +0.6 CE,inner surface of plate 2 of capacitor 1 has a charge -0.6CE, inner surface of plate 1 of capacitor 2 has a charge +0.4 CE & inner surface of plate 2 of capacitor 2 has a charge -0.4CE , a potential difference exists between the plates????
    I cannot think of any such way. Please just prove to me that there is a potential difference between the plate 2 of capacitor 1 and plate 1 of capacitor 2, I will be able to understand the whole of it then.
  19. Jul 7, 2009 #18
    Let's try another angle of looking at things. You can always find the effective capacitance of a network of capacitors, which means it is equivalent with replacing all the capacitors with a single "effective capacitor". This single capacitor must have the same charge on each plate right? In our example, these two plates are the left plate of C1 and right plate of C2. This effectively means that C1 and C2 have the same charge.
  20. Jul 8, 2009 #19
    Yes. Let's follow this through and we will find why Ritwik's first answer is right:

    Take sign convention as positive = clockwise round the circuit.

    With no battery, a charge q flows in the negative direction
    until (CE/2 - q)/(3C/2) = q/C
    giving q = CE/5.

    Now insert the battery, and a charge of 3CE/5 flows in the positive direction.

    Net charge flow = 2CE/5.

    The initial charges on the capacitiors are different, and so
    are the final ones.

    The error in your second method lies in assuming that capacitor 3
    has the same charge (EC/2) as 1 and 2 combined. The question clearly
    states that capacitor 3 was neutral (uncharged) initially.

    Last edited: Jul 8, 2009
  21. Jul 10, 2009 #20
    Yup! I agree with david completely. If you still think you are right, please prove that there is a potential difference between the inner plates, in the state I mentioned,which you say was not the stady state!
  22. Jul 10, 2009 #21
    Feel free to PM me anytime!


    (King of adults)

    BTW I'm not sure what miniscule-tim was trying to get at when
    he said pretend the battery wasn't there.
    I would say pretend there was a battery of E/3.
    Then you close the switch and nothing happens.
    Now insert a battery with emf 2E/3, and your correspondents
    can cheerfully say """combined capacitance""" (they love formulae)
    = 3C/5.
    Last edited: Jul 10, 2009
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