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D/dx of the electric potential

  1. Feb 28, 2005 #1
    By differentiating the electrostatic potential

    [tex]
    \Phi(\vec{r}) = \int\int_{\Omega} \frac{k_e dq(\vec{r'})}{|\vec{r}-\vec{r'}|}[/tex]

    with respect to x, y, and z, and asumming that [tex]\Omega[/tex] is independent of x,y, and z show the electric field, can be writen as

    [tex]\vec{E}=\frac{-\partial{\Phi}}{\partial{x}}\hat{\vect{e_x}}-\frac{\partial{\Phi}}{\partial{y}}\hat{\vect{e_y}}-\frac{\partial{\Phi}}{\partial{z}}\hat{\vect{e_z}}[/tex]

    the problem is how to do I do the diriv of the the dq(r') function? no idea...probley chain rule any hints also help
     
  2. jcsd
  3. Feb 28, 2005 #2

    StatusX

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    You're differentiating with respect to the unprimed coordinates so dq(r) is treated as a constant.
     
  4. Mar 1, 2005 #3

    dextercioby

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    There's no connection between "r" and "r' ".You can see that by taking a look at the derivation of that formula...Namely the variables of the Green function are naturally assumed to be independent...


    Daniel.
     
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