1. Not finding help here? Sign up for a free 30min 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!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electrodynamical oblivion
Loading...