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The relationship between electric potential and field.

  1. Jul 25, 2015 #1
    1. The problem statement, all variables and given/known data
    There are 12 electrons spread on the circumference of a circle with radius R evenly. What's the electric potential at the centre? Then the electrons are concentrated on the upper half of the circle, spread evenly. What's the electric potential at the centre now?



    2. Relevant equations


    3. The attempt at a solution
    Since electric potential is a scalar, the potential due to one electron is V=kq/R. By superposition principle, the potential due to all the electrons is simply 12kq/R=12V. Both situations have the same potentials.
    What confuses me is that the electric fields in both situations are indeed different. In the first case, the field is simply zero because they all cancel at that point, while it's nonzero in the second case. But electric field is the negative gradient of the electric potential, then how come the same potentials result in different fields?
     
  2. jcsd
  3. Jul 25, 2015 #2
    The field is related to the gradient. The same numerical value at a point may not have the same gradient.

    Gradient is like a two or three dimensional slope.

    Two functions can have the same value at a point but have different slopes.
     
  4. Jul 25, 2015 #3

    Orodruin

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    Because the field is proportional to the gradient of the potential and not the potential. Also in one dimension you can have functions which take the same value in a point but have different derivatives.
     
  5. Jul 25, 2015 #4

    Orodruin

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    To expand on that, the field depends on how the potential varies in a neighbourhood of a point, not on the value in the point only.
     
  6. Jul 25, 2015 #5
    But then how should I differentiate the two Vs(which have the exact same expression) such that they'll yield different electric fields?
     
  7. Jul 25, 2015 #6
    Think of potential like the height of the ground at a point. If the ground is flat, a ball won't roll. If the potential is flat, a charge won't move because there is no net electric force.

    But if the ground is sloped, the ball will roll. If the potential is sloped, a charge will accelerate, because there is a net electric force.
     
  8. Jul 25, 2015 #7

    Orodruin

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    Read my last post before this one. If you only know the function value in a single point you cannot take the derivative.
     
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