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Potential in a conductor within an external field

  1. Apr 27, 2008 #1

    Just got a doubt, which is probably silly but nonetheless cannot solve.
    Say you have a conductor placed in an external uniform electric field. We know that charges will be induced on the conductor and distribute on its surface as to nullify the field inside the conductor. Then, at equilibrium, the conductor surface is equipotential: but, because of the induced charges due to the external field, should not we have that one side of the conductor has, say, excess postive charge and the other end negative ones, hence a difference in potential on the surface?

  2. jcsd
  3. Apr 28, 2008 #2
    but if I am correct, then the difference in potential contradicts the fact that the conductor's surface must be equipotential, doesn't it?
  4. Apr 28, 2008 #3


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    The potential difference due to the charge distribution cancels the potential difference due to the external field. Net, actual effect: no potential difference.
  5. Apr 29, 2008 #4
    Reiterating what redbelly said. When you are looking at just the charge distribution on the conductor and saying there should be a potential difference there you are not looking at the net effect anymore you are then neglecting to include the external field that induced the charge seperation in the first place and the effect this external field has on the potential.

    The electric field is the negative gradient of potential. If you agree that for the conductor to be in equilibrium the net E field inside the conductor must be zero (if its not, it has not reached equilibrium yet) then the potential has to be the same constant everywhere in the conductor otherwise the gradient of the potential would not be zero and you would have a non zero E field and therefore not in equilibrium.
  6. Apr 30, 2008 #5
    Ok, guys, you persuaded me. I guess I was misled by the charge separation being induced by the external electric field, so naively I thought: excess charge present at the two ends of the conductor = potential gradient, but obviously it is not like that.

    Thanks a lot for your time, much appreciated.
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