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How should one interpret ##\vec{f}=0## in an ideal battery?

  1. Feb 1, 2015 #1
    In a circuit there are two forces that act on the charges to keep the current uniform through out,
    ##\vec{f}=\vec{E}+\vec{f_s}##, where ##\vec{E}## is the electrostatic field and ##\vec{f_s}## is the electric field produced by chemical reactions. Inside an ideal battery, ##\vec{E}## and ##\vec{f_s}##, oppose each other such that ##\vec{f}##, is zero. What exactly would be the correct way of thinking of ##\vec{f}=0##? I mean obviously the charges are moving, but it looks like in the battery they are not.
     
  2. jcsd
  3. Feb 1, 2015 #2

    Dale

    Staff: Mentor

    I don't think that this is a correct analysis, but f=0 would imply that the charge is not accelerating rather than that it is not moving.
     
  4. Feb 1, 2015 #3
    I got it from Griffiths' Introduction to Electrodynamics. I know that would imply the charges are still moving, but how do I know that current is being produced?
     
  5. Feb 1, 2015 #4

    Dale

    Staff: Mentor

    Current = charges moving. They are the same thing. If you know that the charges are moving then you automatically know that there is current.
     
  6. Feb 1, 2015 #5
    But if the net force on the charges are zero, shouldn't the charges move about randomly cancelling out any current?
     
  7. Feb 1, 2015 #6

    Dale

    Staff: Mentor

    No. If the net force is zero then they continue in a straight line at a constant speed.
     
  8. Feb 1, 2015 #7
    So how do I know they started in a straight line?
     
  9. Feb 1, 2015 #8

    Svein

    User Avatar
    Science Advisor

    Yes, but the moment you close a circuit on the outside of the battery, the current reduces the charge and thereby the electric field. The net field will then be different from 0 and start pushing charges through the battery.
     
  10. Feb 1, 2015 #9
    Ohh okay. You see in the textbook I was reading, it made no distinction as to whether or not the circuit was open.
     
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