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Force between capacitor plates at constant voltage

  1. Mar 31, 2014 #1
    I have a slight issue with this derivation.

    Firstly, displacing one plate by dx requires work dW=-Fdx where F is the internal force between the plates resolved in the direction of x. I.e F=-dW/dx.

    If I do work on the capacitor W, this plus the energy provided by the battery to keep the capacitor at constant pd (QV) equals the stored energy in the capacitor (QV/2).

    W+QV=QV/2
    W=-QV/2 ***

    Therefore F=-dW/dx=0.5V(dQ/dx)=0.5V2(dC/dx) where C is the capacitance. For a parallel plate capacitor, C=EoA/x thus F=-EoAV2/2x2.

    Now the force is attractive, which is a good thing. However I'm not too happpy about the starred line - why is the work needed by an external force to separate the plates negative? That can't be right (especially as when doing this for constant Q, I have F<0 and W>0).

    Thankyou :)
     
    Last edited: Mar 31, 2014
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  3. Mar 31, 2014 #2

    TSny

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    The energy supplied by the battery when work W is done to increase the separation of the plates is given by (ΔQ)V, where ΔQ is the charge supplied to the capacitor by the battery to maintain constant V. The corresponding change in energy of the capacitor is (ΔQ)V/2. So, you have

    W+(ΔQ)V = (ΔQ)V/2
    W= -(ΔQ)V/2

    Now consider whether ΔQ is positive or negative when the plate separation is increased.
     
  4. Mar 31, 2014 #3
    Well the p.d is given by σx/ε0 where x is their separation. Increasing x increases V but the battery wants to keep this p.d constant. So it must reduce σ, which is the charge density on the positive plate here, i.e it needs to add negative charge to it, so ΔQ is negative :)

    I've come across a slightly different approach in a book now however which I'm not too sure about:
    For constant charge they obtain F=-dW/dx as I did above, then say dW=QdV/2=-V2dC/2 is the energy stored by the capacitor and get the same as me.

    Then for constant p.d, they argue dW=VdQ/2 is now the energy stored on the capacitor, the electromotance does work VdQ and so the energy of the whole system is -VdQ. Now the problem:
    They argue that this is equal to -dW now, and not dW as I say. They then go on to use F=dW/dx whilst I'd go for F=-dW/dx, which obviously means the two signs cancel and we get the same answer. Obviously though, their work increment is dW=VdQ which is nice and positive. Why would their dQ be positive though?
     
  5. Mar 31, 2014 #4

    TSny

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    I don't understand this statement. VdQ/2 is the change in energy of the capacitor if dQ is interpreted as the change in charge of the capacitor. Are they using dW to denote the work done by the external force to increase the separation of the plates by dx? If so, then it is not correct to say dW = VdQ/2. As you noted, dW is positive while dQ is negative when the plate separation is increased.

    OK, that looks right. dQ will be negative if the plate separation is increased. The electromotive work is then negative. So, positive energy is given back to the battery.

    I don't understand this. -VdQ is a change in energy rather than an energy of anything. And I'm not sure what the "whole system" is referring to. I think I would have to see the complete and exact wording of their argument.
     
  6. Mar 31, 2014 #5
    It's probably (very) sloppy wording on my part sorry (I missed out the word increment in front of the two things you picked up). I shall copy it word for word.

    Suppose now that the same capacitor is maintained by a source of electromotance at a constant p.d V and that one conductor undergoes displacement dx. The increment of PE dW is this time equal to VdQ/2 and not QdV/2. The source of electromotance has to transfer charge dQ to keep V constant, and in doing so does work VdQ and thus loses this amount of energy. The increment in the energy of the whole system (capacitor+source) is thus -VdQ/2. This equals -dW. As before the work done on the capacitor is -Fdx where F is the internal force, and so -dW=-Fdx, F=dW/dx. Therefore F=0.5V2(∂C/∂x) holding V constant (which leads to my result).
     
  7. Mar 31, 2014 #6

    TSny

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    OK, this can really give you a headache.

    One thing that I don't like is the double use of "dW". First they say, "the increment of PE dW is this time .....". So, I guess here they are using the symbol dW for the increment of PE. (Or is this a typo and dW does not appear here?)

    Later they say, "the increment in energy of the whole system (capacitor + source) is thus -VdQ/2. This equals -dW." So, now it appears that they are using the symbol -dW for the work done by the external force. This is because it's the work done by the external force that equals the increase in energy of the system.

    Note that -VdQ/2 is actually positive when separating the plates since dQ is negative. Thus -dW is positive. Hence, in their notation, dW is negative when separating the plates.

    Now for the part that can drive you nuts. From what we just said, we have

    -dW = work done by external force =|Fext|dx = |Fint|dx.

    They write this simply as -dW = -Fdx "where F is the internal force". Comparing with above, we see this is only correct if they are interpreting the symbol F as a negative number so that -F = |Fint|. So, now they arrive at F = dW/dx.

    Thus, F = dW/dx = -(-dW)/dx = - (-VdQ/2)/dx = +0.5V2(∂C/∂x).

    Note ∂C/∂x is negative. This corresponds with F being taken to be a negative number.

    Anyway, that's the only way I can make everything consistent.

    (I think I need a drink now.)
     
    Last edited: Mar 31, 2014
  8. Mar 31, 2014 #7
    Haha thanks for that, you deserve one! So in short, I should stick to my own method.
     
  9. Mar 31, 2014 #8
    Just another quick related question - for the constant voltage case, the force is F=-ε0AV2/2x2 between the plates. So the work done to change their separation from d1 to d2 is then (by integrating -Fdx) 0.5ε0AV2(1/d1-d2) i.e 0.5V2(C1-C2). This seems weird because I would expect it to be 0.5V2(C2-C1) based on work done equalling change in PE of capacitor. It works as expected for the constant charge case.
     
  10. Mar 31, 2014 #9

    TSny

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    The work done by the external force equals the change in energy of the capacitor plus the change in energy of the battery.

    Change in energy of capacitor = 0.5V2(C2-C1) = -0.5V2(C1-C2)

    Change in energy of battery = V2(C1-C2)


    Work done by external force = -0.5V2(C1-C2)+ V2(C1-C2) = 0.5V2(C1-C2)
     
  11. Mar 31, 2014 #10

    TSny

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    To add to my previous post:

    If you consider the work done by the internal attractive forces between the plates as the external force pulls the plates apart, then that would be equal in magnitude but opposite in sign of the work done by the externally applied forces. So,

    Work done by internal force of attraction = -0.5V2(C1-C2) = +0.5V2(C2-C1) = change in PE of capacitor.
     
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