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0 Electric potential between point charges?

  1. Feb 28, 2013 #1
    Can someone explain how you can have a zero electric potential between opposite charges?
    a charge would move there. I thought electric potential was the measure of energy per charge. wouldn't a 0 electric potential imply that a test charge cannot have any potential energy at this point? how is this possible if it is moving.
     
  2. jcsd
  3. Feb 28, 2013 #2
    The electric potential due to a positive point charge is [itex]V_1=kq/r_1[/itex] where [itex]r_1[/itex] is the distance to the opposite charge. It's value will be positive due the charge being positive.

    The electric potential due to a negative point charge is [itex]V_2=kq/r_2[/itex] where [itex]r_2[/itex] is the distance to the opposite charge. It's value will be negative due the charge being negative.

    The potential (or potential difference) would be [itex]V_1-V_2[/itex] or [itex]V_2-V_1[/itex]. Since they are oppositely charged these expressions will linearly combine, either producing positive potential difference or a negative potential difference, but not zero.

    However, you can have a point in 3-space that the potential is zero due to their combination. For example,

    [tex]V=kq_1/r_1+kq_2/r_2[/tex]

    Set V=0, and you should be able to find a point where the potential is zero.
     
  4. Feb 28, 2013 #3

    rude man

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    If you had two equal and opposite charges and moved a test charge on a line from infinity perpendicularly to the mid-point between the two charges, the net force on the test charge would always be zero, so no work would be done by or on the test charge and its potential would by definition then be zero.

    Even if the charges were not equal and opposite, as long as they were of opposite polarity you could find a path from infinity to some point between them such that the net work done is zero. (I think!).
     
  5. Feb 28, 2013 #4
    So if you knew that the electric potential at some point in space was 0, what excatly does this tell you? it would obviously move at this point right?
     
  6. Mar 1, 2013 #5

    rude man

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    Yes, butthe force on the test charge would be perpendicular to the line of travel, all the way from infinity to its final position between the two opposite-polarity charges. What would be the work done?

    E = - grad(V). V can be zero but grad(V) finite! You can show this by moving the test charge a small amount dx towards one of the charges, recomputing V, then taking dV/dx.
     
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