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I Dipole above infnite conductor

  1. Jan 11, 2017 #1
    An electric dipole p with arbitrary direction and is at distance a from plane infinite conductor at z=0.
    Using the image of the dipole
    ##p=(2pcos\theta \hat{z}+psin\theta \hat{x}##
    ##p'=(2p'cos\theta \hat{z}-p'sin\theta \hat{x}##
    220px-Image_of_dipole_in_plane.svg.png

    Using the following:##V=\frac{\vec{p'}.\hat{r}}{4\pi\epsilon_0{r}^2}##, i get ## V=\frac{p'(2cos\theta-sin\theta)}{4\pi\epsilon_0{r}^2}## Which i can now write as ##\vec{E}## by symple taking the gradient in spherical coordinates. I get: ##\vec{E}=p'\frac{(2cos\theta-sin\theta)\hat{r}+(2sin\theta+cos\theta)\hat{\theta}}{4\pi\epsilon_0{r}^3}##
    Now using one of Maxwell's laws i can get the charge distribution: ##\rho=\epsilon_0\nabla.\vec{E}=\frac{psin\theta}{4\pi{r}^4}## Does it make any sense that this is the result for charge distribution?
    Thank you guys, i'm not really sure how to interpret this
     
  2. jcsd
  3. Jan 11, 2017 #2

    haruspex

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    Not to me. I don't see how it could be radially symmetric.
     
  4. Jan 11, 2017 #3

    haruspex

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    I do not understand where that 2 comes from.
     
  5. Jan 12, 2017 #4
    I'm sorry, i wanted to edit my post, but somehow i can't :/. That 2 obviously shouldn't be there, it was a mistake when i was writing those in latex. I was doing the math again and i think i got it. If a moderator could delete this thread that would be good. Thank you anyway haruspex.
     
  6. Jan 12, 2017 #5

    haruspex

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    Glad you sorted it out. Don't onow why you could not edit the post.... was there no "edit" button?
    Forum policy is not to delete threads unless they violate some rule.
     
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