D/dx of the electric potential

  • Thread starter Phymath
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  • #1
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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
 

Answers and Replies

  • #2
StatusX
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You're differentiating with respect to the unprimed coordinates so dq(r) is treated as a constant.
 
  • #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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