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Work done in moving a charge through an electric field

  1. May 17, 2013 #1
    In electrostatics, when you apply an external force to a charge in an electric field to move it from A to B against the field, why is the force said to be equal to the force the electric field is exerting on the charge? How come we're allowed to neglect the initial slightly greater force needed to accelerate the charge?

    This comes from my text supposing that an external force moves a charge through the field without changing it's kinetic energy..
     
  2. jcsd
  3. May 17, 2013 #2

    WannabeNewton

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    The key point is that the kinetic energy is not changing. This means velocity must remain constant which can only happen if the net acceleration is zero which means the net force is zero so the applied force must balance the electrostatic force. Obviously if there was no need to keep the velocity constant, you could theoretically apply a force much greater in magnitude than the electrostatic force.
     
  4. May 17, 2013 #3

    jtbell

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    In general, the net work done on the charge by the external force equals the sum of the change in the charge's potential energy and the change in its kinetic energy:

    Wext = ΔPE + ΔKE
     
  5. May 17, 2013 #4
    What about the initial force need to get the charge moving? Wouldn't it initially have to be slightly higher?
     
  6. May 17, 2013 #5
    it appears that you are referring to something mentioned in a text. unless you mention what text it is and what is the context it will be difficult to make headway.
     
  7. May 17, 2013 #6

    jtbell

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    Yes, it would. The extra work gives the charge the KE that it has while it's moving. When it reaches its destination, you have to make the external force slightly smaller, so the charge comes to rest and the KE becomes zero again. The external work during this deceleration phase is smaller, by exactly the same amount that it was larger at the beginning.
     
  8. May 17, 2013 #7

    Zag

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    Hi chipotleaway,

    Your concern seem to come from a very practical/experimental point of view. I believe that text-books usually resort to that approach in which the particle is always in equilibrium because it is more didactic to the reader. It is easier to imagine a particle being taken through a complicated and unnatural path when it is in equilibrium at every point of the path.

    However, the truth is that in principle there is no necessity for that approach. Unless the problem you stated is vital for some aspect of an experiment you are running, you could easily just forget about the "extra push" and think of the path integral as a merely theoretical tool that you use to define the electrical potential and, thus, describe the physical situation.

    jtbell wrote a nice way of seeing things on the post above. But if that issue still bothers you, then another way to get around the problem you posed would be to think that in the real situation you started the motion of the charge a little bit before the beginning of your mathematical path, so that you wouldn't have to take the "extra push" into accound when starting to perform the integral at A, and then you would go all the way with constant velocity past point B, after which you stop calculating.

    This is all about which way you consider the the best one to visualize it. In the end the only thing that matters in the integral is the Field you have, and not the mechanical force you are applying to the particle.
     
    Last edited: May 17, 2013
  9. May 17, 2013 #8

    jtbell

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    You might also compare this to the situation where there's no electric field and force, and you simply move an object from rest at point A to rest at point B, with only the single external force.

    In order to get the object moving, you have to push on it a bit in the direction you want it to go, doing positive work on it and giving it a bit of KE. After that, you can let the object "coast" with no external force until it approaches point B. Then you push on it a bit in the opposite direction to its motion, doing negative work on it, reducing its KE to zero, and bringing it to a stop at B.
     
    Last edited: May 18, 2013
  10. May 20, 2013 #9
    Understood, thank you!
     
  11. May 31, 2013 #10
    Sorry to revive this thread but I was thinking about this again today and I wonder if you remove the external force and let the field bring it to rest, is the external force still thought of as having done work?
     
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