1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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

    \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
    Last edited by a moderator: Apr 24, 2017
  4. Nov 27, 2009 #3


    User Avatar
    Homework Helper
    Gold Member

    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook