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Poisson and Helmholtz equations

  1. Dec 23, 2008 #1
    Hello guru's,

    I've been trying to figure out a way to incorporate an electric field source in the Helmholtz equation, and have been accumulating lots of question marks in my head. So in case of no static charge,

    [tex]\nabla^{2} E - \mu (\epsilon\frac{d^{2}}{dt^{2}} + \sigma\frac{d}{dt}) E = 0
    [/tex]

    In the quasistatic case, I know many people use Poisson's,

    [tex]\nabla \sigma \nabla (V - V^{p}) = 0[/tex]

    with a voltage source [tex]V^{p}[/tex], or

    [tex]\nabla \sigma (E - E^{p}) = 0[/tex]

    Unfortunately my electric field is in hundreds of MHz range. So (1) can I use the following equation with source [tex]E^{p}[/tex], since the second term seems to be the "admittivity times electric field" term?

    [tex]\nabla^{2} E - \mu (\epsilon\frac{d^{2}}{dt{2}} + \sigma\frac{d}{dt}) (E - E^{p}) = 0
    [/tex]

    (2) I wonder why in the low frequency limit the Helmholtz equation doesn't reduce to the Poisson's in the inhomogeneous media, since the [tex]\sigma[/tex] is in the spatial gradient in the Poisson equation. In my case the media is inhomogeneous. I did see some papers on scattering that used the Helmholtz equation in the inhomogeneous media, so that makes me wonder all the more.

    Any comments would be greatly appreciated!
     
  2. jcsd
  3. Dec 23, 2008 #2
    Sorry the Poisson's equation forms

    [tex]\nabla \circ (\sigma \nabla (V - V^{p})) = 0[/tex]

    [tex]\nabla \circ (\sigma (E - E^{p})) = 0[/tex]
     
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