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B Potential Difference. across capacitors

  1. Apr 26, 2016 #1
    When a fully charged capacitor is connected to an uncharged capacitor with different capacitance, would the p.d. across each of them be the same?

    I thought that only the charge stored on each would be the same so p.d. across each would depend on their capacitance? Does that mean that they will not have the same p.d.? If they do not have the same p.d., will the sum of their p.d. add up to be the voltage stored in the originally uncharged capacitor?
     
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  3. Apr 26, 2016 #2

    phinds

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    How could parallel capacitors possibly not have the same PD? For that matter how could ANY elements in parallel not have the same PD?

    Now, whether or not the PD is the same as that on the original capacitor is a different matter. If you add two caps in parallel, what happens to the total capacitance?
     
  4. Apr 26, 2016 #3
    I should have stated it more clearly. The fully charged capacitor is only connected with the uncharged capacitor, so shouldn't they be in series in this case? This why I am confused because technically the p.d. should be "split" between the two capacitors when they are connected in series, but my textbook says they have the same p.d.
     
  5. Apr 26, 2016 #4
    I suggest that you make a circuit diagram.
     
  6. Apr 26, 2016 #5

    phinds

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    +1 on that
     
  7. Apr 26, 2016 #6

    berkeman

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    Does your textbook have drawings or figures for this topic?
     
  8. Apr 26, 2016 #7
    If you are connecting just two components together, automatically, they are parallel. The potential across each is the same. The charge distributes proportional to the capacitances capacitances
     
  9. Apr 26, 2016 #8

    berkeman

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    That is not true. The OP has mentioned series connection, so we need to wait to see what the textbook is asking about.
     
  10. Apr 26, 2016 #9

    David Lewis

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    When the caps are connected together in series, the stored electric charge redistributes itself. The voltage on the charged cap will go down, and the voltage on the uncharged cap will go up, until the voltage across both caps will (after some amount of time) be the same.
     
  11. Apr 26, 2016 #10

    berkeman

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    Not unless the circuit is completed somehow. Let's let the OP respond with the actual question out of their textbook please.
     
  12. May 2, 2016 #11
    Hi, this is the exact question on my book. It doesn't come with a diagram however. But I think the meaning is quite clear.
     

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  13. May 3, 2016 #12

    davenn

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    totally unclear
    we still don't know how the capacitors are connected
     
  14. May 3, 2016 #13

    Merlin3189

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    I thought it is clear enough. They are connected in series and parallel! If you have only two components this is always the case.
    You may have thought that only one wire of each capacitor is connected, but that would be a fairly trivial & pointless exercise. You could join all 4 leads to the same point and there are other trivial possibilities, but the only non-trivial way for 2 elements is in series and parallel at the same time.
    And although your mind, like mine, probably immediately thought electrolytics when you saw microfarads, the question does not say that. So it does not matter which end connects to which.
    It does not matter whether you say parallel so the voltages are the same, or series so that the sum of the voltages round the loop is zero. If there are only two components, the voltages must be equal in magnitude.
    Similarly the amount of charge transferred comes out the same whether series or parallel.
    It's like skinning cats.
     
  15. May 3, 2016 #14

    phinds

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    Exactly.
     
  16. May 3, 2016 #15

    jbriggs444

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    If these two capacitors have no other connections then they are both in parallel and in series. However, there is a catch.

    Make that circuit drawing. Label the two ends on each capacitor. A and B on the charged capacitor and a and b on the uncharged capacitor. If you join A to a and B to b and consider the two capacitors as being in parallel then you have:

    pd(AB) = pd(ab).

    If you join A to a and B to b and consider the two capacitors as being in series then you have:

    pd(AB) + pd(ba) = 0.

    When you keep track carefully, you see that you have to flip conventions for the potential difference across one of the two capacitors. The result is that both equations say the same thing.
     
  17. May 3, 2016 #16

    Merlin3189

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    I'm not sure what you're saying about conventions here?
    I think your series equation is exactly what Kirchoff would say. I can't see where any change in convention is required.
    My labels are the reverse of yours, but I get P(BA) + P(ab)=0, ∴ P(BA) - P(ba) =0 [since P(xy)=-P(yx) ], ∴ P(BA) = P(ba) or P(AB)=P(ab)

    But what I'd like to know is, what happens to the energy? I used to think it was dissipated in the resistance (of the wires and the capacitors themselves), but I was once told that this does not account for enough. (I've been trying for a while this afternoon to calculate this myself, but my maths is failing me.) The explanation I was given was that some is dissipated in EM radiation. While I thought that possible at the time, I've come to doubt it, so I'd like to see a proof that it can all be dissipated as I2R. If so, any links?
    capacitors_2.png
     

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  18. May 3, 2016 #17

    jbriggs444

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    You have the same capacitor with with the same endpoints and the same potential difference. This is transcribed into equations once as +4V and once as -4V. [Edit, nothing wrong with that. The equations, are, of course, correct]. There must have been some reasoning or some standard behind that assignment. That is the sort of convention that I am referring to.

    If the resistance is very high, it does account for essentially everything. If the resistance is low then the circuit will resonate due to the finite inductance in the wires. It amounts to an RLC circuit. Ultimately, the energy in the resonance will damp due to both resistance and EM radiation. [A glance at the equations for radiated power from dipole and monopole antennae makes it less than obvious how to calculate the radiated power for a given current, frequency and wire length]
     
  19. May 3, 2016 #18

    cnh1995

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    I am not sure if this is what you are looking for:
    In a simple RC series circuit, the energy stored in a capacitor after charging is ½CV2. It can be shown mathematically that the energy lost in the resistor is also ½CV2 (I don't know where EM radiation comes into picture. Edit: Reading jbriggs444's reply gave me some idea:wink:).
    The expression will be E=∫i(t)2R dt from 0 to ∞, where i(t) is ioe-t/RC. Solving this, we get E=½CV2.
    Similarly, in case of two capacitors connected in series and parallel at the same time, I believe it is possible to show that the energy can all be dissipated as I2R.
     
    Last edited: May 3, 2016
  20. May 3, 2016 #19

    Merlin3189

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    Thanks for those comments, esp. cnh and the maths. No wonder I was struggling, as I'd got i(t) wrong and couldn't find the integral. I just need to check that this is still ok for C1RC2 circuit, but I'm already feeling reassured!
     
  21. May 3, 2016 #20

    David Lewis

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    When charge travels from one capacitor to another, if we neglect the resistance of the connecting wires and the dielectric loss, the change in energy is zero.
     
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