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Why Electric Fields are perpendicular

  1. Jul 8, 2009 #1
    1. The problem statement, all variables and given/known data
    If electric fields are perpendicular to the equipotential surface, what is the orientation of the field on the surface of a conductor (like metal)? Why is this so?


    2. Relevant equations

    E= ∆V/d

    3. The attempt at a solution

    I know that there is no charge within a conductor and it moves to the edges. I am confused with how this affects the electric field lines. I also know that electric field lines point out if a charge is positive and in if a charge is negative.

    I am just confused with electric fields when there is a conductor.

    Thanks :)
    Katie
     
  2. jcsd
  3. Jul 8, 2009 #2

    berkeman

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    Staff: Mentor

    If you use a DVM to measure the voltage difference between two points on a conductive surface, what voltage do you think you will measure (assuming little or no current is flowing in the conductor)? Does that help?
     
  4. Jul 8, 2009 #3
    You wont measure any voltage, does this mean that the electrical field would be zero?
     
  5. Jul 8, 2009 #4

    berkeman

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    Staff: Mentor

    Something like that. It means that the two points on the conductive surface have no "potential" difference between them. So in other words, you could call the surface of a conductor an __________ surface, right? And since the E field is always perpendicular to that kind of surface, what can you say?
     
  6. Jul 8, 2009 #5
    Oh an equipotential surface! so the electric field would be perpendicular to the surface!
     
  7. Jul 8, 2009 #6
    (sorry it took me so long I had to leave for physical therapy)
     
  8. Jul 8, 2009 #7

    berkeman

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    Staff: Mentor

    Bingo!
     
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