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Work Done on a System of Charges

  1. Dec 12, 2014 #1

    kmm

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    Something I have read in Griffith's Electrodynamics is confusing me. In deriving the Poynting vector, he says to suppose that we have some charge and a current configuration which at some time t produces fields E and B. In the next instant, dt, the charges move around. He then asks the question, "How much work, dW, is done by the electromagnetic forces acting on these charges in the interval dt?" From the Lorentz force law, the work done on a charge q is: [tex] dW= \mathbf{F} \cdot d \mathbf{l} = q( \mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} dt = q \mathbf{E} \cdot \mathbf{v} dt [/tex] But ## q= \rho d \tau ## and ## \rho \mathbf{v} = \mathbf{J}##. So the rate that work is done on all the charges in a volume V is: [tex] \frac{dW}{dt} = \int_{V} ( \mathbf{E} \cdot \mathbf{J}) d \tau [/tex] Great, but the problem with this to me is that this gives a nonzero result for work. But how can the total work done by a system on itself be nonzero? I'm sure it's something simple I'm missing, but I'm not sure.
     
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  3. Dec 12, 2014 #2

    mfb

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    Staff: Mentor

    This is the work done of the fields on the charges. You also have the kinetic energy and the energy of the fields in your total energy.
     
  4. Dec 12, 2014 #3

    Dale

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    You have to be careful about what you are defining as the system. Given that definition of work the system is the charges, but not the field. So the system is not doing work on itself. The fields are doing work on the charges.
     
  5. Dec 12, 2014 #4

    kmm

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    OK, except how would the total work done be expressed any differently if the question was, "what is the total work done by the charges?" I mean, whenever we ask what the total work is done by two charges, don't we always mean the field anyway? So I guess I don't understand the distinction of saying it's the work done by the field and not the charges when the force is always due to whatever the field is anyway. Can't I think of it this way: If I have two equally charged particles and place them near each other on the x axis, they will repel each other with one charge going to the right and one going to the left. Since work is ## \mathbf{F} \cdot d \mathbf{l} ## and the forces and displacements are opposite, I will get positive work done on each charge and therefore a nonzero value for the total work done. Perhaps this is what you were actually getting at?
     
    Last edited: Dec 12, 2014
  6. Dec 12, 2014 #5

    Dale

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    Work is the transfer of energy other than by heat. You have complete freedom to decide what "the system" is. Once you make that choice it determines what work is since it determines what counts as a transfer and what is just an internal rearrangement.

    If you choose that the system is just the charges then the system is acted on by the field and work is done as energy is transferred from the fields to the charges.

    If you choose that the system is the charges and the fields then no energy is transfered since the system is not acted on by anything external.

    The only time that confusion arises is if you are inconsistent and define the system as the charges only for determining work and as the charges plus the fields for determining if it there is anything external to the system. You can make either choice as long as you are consistent.
     
  7. Dec 12, 2014 #6

    kmm

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    I see, so my problem was that I thought of the system as just the charges at first, but then I was inconsistent by trying to include the field as part of the system at the end of the problem to give zero total work. Thanks for helping me with that.
     
  8. Dec 12, 2014 #7

    Dale

    Staff: Mentor

    No worries. It is hard to do with something as "nebulous" as fields. Glad I could help.
     
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