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Getting the electric potential function using the electric field components

  1. Dec 21, 2011 #1
    hello , I want to get the electric potential function with displacement
    V(x,y,z)
    as

    Ex= -8 -24xy
    Ey =-12x^2+40y
    Ez= zero

    when I perform intergration on Ex and Ey seperately then sum them , the answer is wrong , why ?? I mean when I do this and then perform partial derivatives again , the Ex in not the same , is this integration wrong ?
     
  2. jcsd
  3. Dec 21, 2011 #2

    SammyS

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    Hello ahmeeeeeeeeee. Welcome to PF !

    It's wrong, because you shouldn't sum them.

    Show what you get & we can help .
     
  4. Dec 22, 2011 #3
    Why can't I sum them ??


    the original function of v (given) is

    V = 8X+12YX^2-20Y^2

    from which I got

    Ex and Ey and Ez

    when I integrate again and sum I get

    V =8X + 24YX^2-20Y^2

    so the difference is that it is 24 instead of 12



    Well , I didn't study partial derivatives if there were something special about the integrations , But I tried many functions and I noticed that when I get

    Vx ----- I just take from it the things without (xy) ( Just X)
    vy ----- I just take the "Y"s

    and then take just one of the (xy)s from Vx or from Vy

    this what I "noticed" when I tried some functions , is it true ?? and why ??
     
  5. Dec 22, 2011 #4
    I mean , every repeated (xy)or (xy^2) or (x^2y) or etc.. I just take one of them
     
  6. Dec 22, 2011 #5

    SammyS

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    There is no special Vx, or Vy, or Vz. Only V.

    Ex = -∂V/∂x , Ey = -∂V/∂y , Ez = -∂V/∂z

    When you integrate -Ex to find V, there is a "constant" of integration. It's not really a constant, just a constant with respect to x, e.i. when you take the (partial) derivative of this "constant" of integration with respect to x, you must get zero. Therefore, this "constant" of integration, is actually quite possibly a function of y and z. You figure out what this "constant" is, by taking the partial derivatives w.r.t. y & w.r.t. z, and comparing the result with Ey & Ez .
     
  7. Dec 22, 2011 #6
    You cannot typically solve multiple-dimensional differential equations by direct integration because of the way the dimensions couple. You have to cleverly write a trial solution, plug it in to see if it works, then refine your trial solution based on what goes wrong when you plug it in. A typical college course on differential equations is a course in the art of making good guesses on trial solutions.
     
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