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Potential energy difference related qs

  1. Aug 10, 2014 #1
    in bringing a test charge (+q ) from a postion to another in an electric field of another charge +Q an external force , F is applied which is just as same as the electric force, F(electric)

    how is this possible????

    doesnt an unbalanced force be required to...bring about this process?

    note: there is repulsive force between the charges...

    can anyone explain to me about what is happening here??
     
  2. jcsd
  3. Aug 10, 2014 #2
    I think....

    The force is applied is slightly greater than the electric force but not exactly equal.

    But there is not much difference between this two.

    so we take applied force equal to electric force.
     
  4. Aug 10, 2014 #3

    sophiecentaur

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    Work was done to bring about the situation you describe - in the same way that work is down when lifting something against gravity. If you are worried about the 'extra' force, needed to get things going (moving) then, if the field is conservative, this extra force is only there at the start and then F can be reduced near the end of the ride. In a practical situation there would be a small amount of Kinetic Energy in whatever it is that's carrying the charge but it all goes into the Potential Energy at the end.
    The simple scenario would be to consider the process being carried out infinitely slowly but the speed doesn't actually matter.
     
  5. Aug 10, 2014 #4

    jtbell

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    If there are no other forces besides the electric force and the force that you exert in order to move or hold the charge:

    1. Imagine holding the charge stationary at its original position. The net force is zero.

    2. Start moving the charge towards its new position. You briefly exert a force that is slightly larger than before, so the net force is nonzero and the charge accelerates.

    3. After the charge has started to move, you reduce your force to its original value, so the net force is zero again, and the charge continues to move with constant velocity towards its final position. If the electric force changes along the way, you change your force also, to keep the net force equal to zero.

    4. When the charge is close to its final position, you briefly exert a bit of "extra" force in the opposite direction to bring the charge to rest at its final position.

    The "extra" forces that you exert in steps 2 and 4 do work that is opposite in sign (positive work in step 2 and negative work in step 4) so they cancel out. The total work that you do during the entire process is equal in magnitude and opposite in sign, to the work done by the electric field.
     
  6. Aug 10, 2014 #5

    sophiecentaur

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    The point about a conservative field, however, is that you can apply any forces you like on the way from A to B and achieve the displacement in any time you choose. The total work done will always be the same. More force at one time will always be made up for with less force at another time (or even some forces in the 'wrong directions'). All that's required is that the charge is stationary at each end of the process.
     
  7. Aug 10, 2014 #6

    jtbell

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    Yes indeed! I was just simplifying the situation to (hopefully) make it easier to grasp. You can make my "extra" forces at the beginning and end as small as you like, and as brief as you like, so long as they "match up" so the charge comes to rest at the end. The process will still work out, and the total work that you do will still equal the change in the charge's potential energy. It will merely take longer for the charge to make the trip.
     
  8. Aug 10, 2014 #7

    sophiecentaur

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    My additional point was that they can be as Large as you like, too and you still get the same answer.
     
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