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Electrodynamical oblivion

  1. Nov 26, 2005 #1
    I was searching for some proof on macroscopical theory about lattice behaviour under lightning. I found a passage in which the average in time of optical power is given by average of electrical field square times constant dielectric ratio four pi! How could I explain this concept?
    Regards
     
  2. jcsd
  3. Nov 26, 2005 #2

    Gokul43201

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    I'm not sure I understand what you are talking about - it would help if you provided a reference or link to the passage.

    In general, the energy density of an electric field is half the number you stated.

    [tex]\frac{d{\cal E}}{dV} = \frac{1}{2} \epsilon E^2~~(SI~Units)[/tex]

    [tex]\frac{d{\cal E}}{dV} = \frac{1}{8\pi} \epsilon_r E^2~~(Gaussian~Units)[/tex]
     
  4. Nov 26, 2005 #3
    I've just forgot to say you I'm in gaussian units... but your 8 is my 4
     
  5. Nov 26, 2005 #4
    Mayhap... the reason is this: you told about an energy density at a given instant, while the time average of my waves (PLANE waves) has brought new root-square-two terms (2 terms because E is the base of a power 2) forward in my formula...
    Is the Poynting vector involved in anyway?
    Thank you
     
  6. Nov 26, 2005 #5

    Gokul43201

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    [tex]E_{RMS} = \frac{E_0}{\sqrt{2}} [/tex]

    That would only make things worse.

    [tex]\frac{d{\cal E}}{dV} = \frac{1}{8\pi} \epsilon_r E_{RMS}^2 = \frac{1}{16\pi} \epsilon_r E_0^2[/tex]
     
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