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Capacitor charge question

  1. Mar 23, 2013 #1
    If two capacitors are connected in parallel to a battery and then the battery is removed, what happens to the charge on each capacitor? Do they move from one capacitor to another till they are distributed equally or do they stay in place?
     
  2. jcsd
  3. Mar 23, 2013 #2
    Nothing happens. Each capacitor holds the same charge it had before disconnecting the battery.
     
  4. Mar 23, 2013 #3

    Drakkith

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    The charges are already distributed equally. This is because the voltage on each capacitor is the same since they are in parallel.
     
  5. Mar 23, 2013 #4
    Drakkith, they're not identical capacitors.

    If the charge stays in place, why so? Is it because they're being held together by the attraction from across the plates?

    As a follow up question, what happens if I now insert a dielectric between one of the capacitors, say C1?
     
  6. Mar 23, 2013 #5
    What do you mean exactly by a 'dielectric'?
     
  7. Mar 23, 2013 #6

    Drakkith

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    Doesn't matter. If the capacitors have charged fully, then both will have identical voltage and no current will flow between them.

    I think it's better to focus on the basics before trying to get into complicated "what ifs".
     
  8. Mar 23, 2013 #7
    I find it strange that someone who seems to not know what happens when 2 capacitors are connected in parallel to a battery is then able to ask about the behaviour of inserting a dielectric between the plates of one of the capacitors.
    How much knowledge does the poster have?
    Knowing this would enable a meaningful response.
     
  9. Mar 23, 2013 #8

    davenn

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    it will already have a dielectric between the plates
    so please clarify what you are inferring :)

    Dave
     
  10. Mar 23, 2013 #9
    Sorry if I wasn't clear. Here's the situation I have in mind.

    Charge 2 non-identical capacitors in parallel, with no dielectric between them (unless you count air). Remove power source. Insert a dielectric between one of the capacitors. Now what happens to the charges and voltages? Do they change/ do they stay the same - that was my question? Essentially how does the induced field on the dielectric affect the setup.

    As an aside, going back to the first qs, wouldn't the system move to a lower energy state if the charges DID move and distributed themselves equally. If someone can offer insight from the energy/force perspective that would be very helpful in regard to this qs.
     
    Last edited: Mar 23, 2013
  11. Mar 23, 2013 #10

    davenn

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    you are still not overly clear

    between them ? the 2 capacitors or the individual plates of the capacitors ?

    2 capacitors in parallel, even if different values are just going to be the same as one capacitor with the combined value of those 2 separate ones


    Insert a dielectric between one of the capacitors ? again, more or different dielectric ??
    there is already a dielectric between the plates else it isnt a capacitor.
    As someone said earlier, the voltage potential across each cap is going to be the same
    They are in parallel! :)

    cheers
    Dave
     
  12. Mar 23, 2013 #11

    Drakkith

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    I think he's asking what happens after you charge a capacitor, remove the power supply, and then replace the dielectric with a different one.
     
  13. Mar 23, 2013 #12

    jim mcnamara

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    Let's try a different tack here.

    It sounds like you saw or heard something, then you took the ideas and words they used, made some assumptions and asked your question. I'll bet if you explained where the question came from rather than what you think the question should be, somebody here will be able to give you a good answer.
     
  14. Mar 24, 2013 #13

    sophiecentaur

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    Where is the problem here?
    When the battery is disconnected, the two capacitors will have charges Q=V/C, V will be the same and the Qs will depend upon their Cs. Energy (CV2/2) will be stored in each capacitor. If you insert or remove the dielectric in one of the capacitors, some WORK will have been done, changing the potential. This will change the energy stored in the capacitors and charge will flow until a new V is established across the two capacitors. No magic and no paradoxical behaviour.
     
  15. Mar 24, 2013 #14
    Let me try again. The setup looks like this.

    http://www.technologyuk.net/electronics/electrical_principles/images/capacitor_10.gif [Broken]

    Initially there's no dielectric between each capacitor. They are simple parallel plates with an air gap between them. Q1 = V*C1, Q2 = V*C2. Then the battery is taken out of the picture and the power supply is removed. And here come my questions.

    1) Since the system would be in a lower energy state if the charges now distributed themselves equally over the two capacitors, what prevents them from doing so?

    2) I now insert a dielectric between the plates of capacitor C1. An induced electric field is setup in the dielectric which will have some impact on V1 across the plates of capacitor C1 and therefore (and this is where I'm in need of help) also on V2 and on the respective charges Q1 and Q2 of each capacitor. My question is what happens to V2, Q1 and Q2.
     
    Last edited by a moderator: May 6, 2017
  16. Mar 24, 2013 #15

    sophiecentaur

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    The total charge stays the same. The capacitance of one capacitor changes (increases). The dielectric is actually 'pulled' into the gap and work is done against friction - (or the dielectric would oscillate in and out for ever!!). The final (lower) V will depend upon Q(original)/C(total, new). Some charge will, of course, flow from the other capacitor into the one with the increased capacitance.
    NB As with all these situations, you have to include some loss mechanism / resistance / friction or you get a nonsense scenario.
     
  17. Mar 24, 2013 #16
    .

    The system is in the lowest energy state (for a given amount of total charge )when the voltages across the capacitors are equal, so nothing happens after you disconnect the battery. The charges on the capacitors will be V_battery * C1 and V_battery * C2

    The potential difference across the capacitor is the amount of energy you need to add an amount of charge to the capacitor, If this is the same for both capacitors, you can't get energy out by moving charge from one capacitor to the other.


    .

    Easy to find this out.

    -The total charge can't change after you disconnect the battery.
    -The potential difference across both capacitors must be the same.
    - Q = C*V for both capacitors.

    That's all you need
     
  18. Mar 24, 2013 #17
    I understand now, I was looking at it wrong, should have considered the definition of potential more thoroughly.


    Doesn't V1 change? The induced field in the dielectric reduces the net electric field across C1 - which in turn would reduce the potential difference across the plates of C1. Or am I wrong in this? This is where I'm getting confused, in my mind both C1 and V1 seem to be changing on introduction of the dielectric.
     
  19. Mar 24, 2013 #18
    I'm glad you mentioned this. I had realised that the dielectric would be sucked in before while working on a different problem and was quite puzzled by that. So work is done by the capacitor C1 against friction and that's where energy is lost. I'm still a little confused about this. This would mean, from the energy/work perspective, the system loses energy by the field in C1 doing work and the remaining energy is balanced (distributed) between the 2 capacitors. Am I right?
     
  20. Mar 24, 2013 #19
    V1 does change, and then a current will flow from the capacitor with the largest potential difference until the voltages are the same again. With two smallish capacitors (They must be, if you can pull out the dielictric) and only the resistance of the wires, this will happen much faster than you can pull out the dielectric.

    If instead of pulling out the dielectric, you pull the plates apart, it's easier to see that you must do work against the force between the plates.
     
    Last edited: Mar 24, 2013
  21. Mar 24, 2013 #20

    sophiecentaur

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    Yes. As I said before, this sort of problem can only be solved by allowing for some energy to be lost during the change of conditions.It has to be lost either in circuit resistance or by friction *- else the stored energy in the circuit never dissipates and you get mechanical oscillation.
    The classic 'paradox' problem involves connecting a charged capacitor to an identical uncharged capacitor. The difference in energy has to be 'explained'. Same basic flaw in the way the problem is posed.
    * In the end the energy would be lost by EM radiation.
     
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