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Is e field equal to the negative derivative of electric potential?

  1. Sep 14, 2011 #1
    Since V is the integral of -E. Shouldn't E be derivative of -V. I asked my TA that and he said something about them not having that kind of a relationship, can someone explain this?
     
  2. jcsd
  3. Sep 14, 2011 #2

    kuruman

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    You need to sharpen your thinking. The electric field is a vector, so to specify it you need to specify its three components. These are

    [itex]E_{x}=-\frac{\partial V}{\partial x}[/itex]; [itex]E_{y}=-\frac{\partial V}{\partial y}; [/itex][itex]E_{z}=-\frac{\partial V}{\partial z}[/itex]
     
  4. Sep 14, 2011 #3
    Doesn't that just mean E is the negative derivative of V tho?
     
  5. Sep 14, 2011 #4

    kuruman

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    It means that the x component of E is the negative derivative of V with respect to x
    and that the y component of E is the negative derivative of V with respect to y
    and that the z component of E is the negative derivative of V with respect to z.
     
  6. Sep 14, 2011 #5
    ah okay thanks
     
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