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Solving Poyntings Vector for a point charge

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data

    It is found that Poyntings vector gives

    P = ExH = (mu0q2a2sin2(theta)/6pi2cr2)r

    This apparently leads to

    Total Power = (mu0q2a2/6pi2c)[tex]\int[/tex](sin2(theta)/r2)(2pir2sin(theta)d[tex]\theta[/tex])

    What I am unsure of is where the

    (2pir2sin(theta)d[tex]\theta[/tex])

    appears from. Can anyone help?
     
  2. jcsd
  3. Nov 27, 2009 #2
    [strike]The power is over the whole of the volume, and not just one component of the volume.[/strike] When you are coming from Cartesian coordinates to http://en.wikipedia.org/wiki/Spherical_coordinate_system#Cartesian_coordinates", we find

    [tex]
    \int dV=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\int_{-\infty}^\infty dz=\int_0^\infty r^2dr\int_0^\pi \sin\theta \,d\theta\int_0^{2\pi} d\phi
    [/tex]
     
    Last edited by a moderator: Apr 24, 2017
  4. Nov 27, 2009 #3

    gabbagabbahey

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    Errr... not quite...Poynting's theorem tells us that the power crossing a surface [itex]\mathcal{S}[/itex] is given by

    [tex]P(\textbf{r})=\int_{\mathcal{S}}\textbf{S}\cdot d\textbf{a}[/tex]

    (using [itex]\textbf{S}[/itex] to represent the Poynting vector and [itex]d\textbf{a}[/itex] to represent the infinitesimal vector area element of the surface)

    To find the power radiated, one usually uses a spherical surface of radius [itex]r[/itex] (so that [itex]d\textbf{a}=r^2\sin\theta d\theta d\phi\mathbf{\hat{r}}[/tex])) and then takes the limit as [itex]r\to\infty[/itex] (since radiation escapes to infinity).
     
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