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Permittivity values in conductors

  1. Mar 3, 2012 #1
    Hello All,

    I am a masters student in electronics engg and reading technical electrodynamics. Please let me know when the electric current density becomes more than the displacement current (for a conductor ) why is the value of ε≈-jσ/ω

    Quick replies will be highly appreciated!!

    Thanks and Regards,

    ksnf3000
     
  2. jcsd
  3. Mar 3, 2012 #2
    High ksnf3000,

    electric current density :

    [itex]J=σE[/itex]

    displacement current density :

    [itex]J_{d}=\frac{\partial D}{\partial t}[/itex]

    for a steady stead sinusoidal field,

    [itex]J_{d}=jωD=jωεE[/itex]

    the ratio is [itex]\frac{ωε}{σ}[/itex]

    when [itex]ωε=σ[/itex], the magnitudes become the same. Considering typical values of the parameters, this happen at very high frequencies.

    I hope that helps.
     
  4. Mar 3, 2012 #3
    Thanks for your reply Hassan2.

    The negative sign indicates the loss the imaginary part suffers, right??
     
  5. Mar 3, 2012 #4
    that j is not supposed to be there. The parameters are real, so, the right hand side must be real too. When comparing their magnitudes, j and also -1 becomes 1.
     
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