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Why is the surface of a conductor equipotential?

  1. Aug 26, 2010 #1
    I am having trouble understanding this. For example, say I have this:

    [positive point charge] [A][neutral conducting sphere] [...infinity]

    I know the point charge will create a field radially outward. But now I'm thinking of a test charge being brought in from infinity. If the sphere was not there, point A would obviously have a higher potential than point B. But somehow the conducting sphere being there would allow A to have the same potential as B. I do not understand this.
  2. jcsd
  3. Aug 26, 2010 #2
    The free charge on the conducting sphere will re-distribute in the presence of the field from the point charge. The electron density will increase near the positive charge (side A) and decrease on the opposite side. The field of the charge distribution will add up to the field of the point charge.
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