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A question on Electromagnetic theory

  1. May 11, 2014 #1
    Hi,
    In EM theory to derive the 'wave equation for electric field in a conducting medium' we make ρfree to be zero, but we still use σ to be nonzero.
    My question is why we are doing this ?
    σ denotes the conductivity of the material and a material can not conduct if it has no free carriers.
    So a non zero σ always implies a non zero free carrier. And ρfree is defined to be the charge density which does not take part in polarization, like the free carriers in the metal.
    Thank You
     
  2. jcsd
  3. May 11, 2014 #2
    i'm still learning e&m myself. but to my knowledge, within conductors, the free charge carriers end up on the outside of the conductor making ρinside=0. but ρinside is essentially ρfree since ρ refers to a volume charge density.

    i may very well be mistaken, but based on the context, i'm going to assume that σ should not be the conductivity; i think the σ in whatever you were looking at was referring to the surface charge density. if it is a conductor we are dealing with, then i don't think σfree should be zero since, if i'm not mistaken, conduction happens to be surface phenomenon simply due to the fact that free charges end up on a conductor's surface to minimize the energy of the charge configuration.
     
  4. May 11, 2014 #3

    clem

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    In a perfect conductor, all free charges are on the surface. With finite conductivity, signa, there is a current inside the material with j=sigma E.
     
  5. May 11, 2014 #4
    σ is the conductivity of the material. It is related to the current density by j=σE.
    Where j is the conduction current density inside the material, and E is the applied electric field.
     
  6. May 11, 2014 #5
    So why we are assuming ρfree to be zero for conductors?
     
  7. May 11, 2014 #6
    well adapt the equation to the problem's dimension: use K (surface current density) instead of J (volume current density)
     
  8. May 11, 2014 #7
    It is correct as far as I know. I think you are taking about the equation K=σv, where σ is the surface charge density.
     
  9. May 11, 2014 #8

    ChrisVer

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    Because there is no free charge density in the conductors... all the free charges move on the surface.
    Otherwise, if you had free charge densities, you'd need to have electric field within the conductor, which is not the case.
     
  10. May 11, 2014 #9

    phyzguy

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    It is possible to have a finite (even very large) conductivity but have rho = 0. Since you have two charge carriers, all that is needed is that the charge densities of the positive and negative charge carriers are equal. In a metal, for example, if the charge density of the electrons is equal to the background charge density of the positive ions, then rho = 0. Similarly, a plasma can have very high conductivity, but zero charge density. The ideal equations of magnetohydrodynamics basically assume that the charge density of the plasma is zero and the conductivity of the plasma is infinite. The two charge carriers move in opposite directions to support the current flow.
     
  11. May 11, 2014 #10
    Are this rho and ρfree equal?
    Do the stationary ions contribute in ρfree?
     
  12. May 11, 2014 #11

    ChrisVer

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    Do you know what is [itex]\rho_{free}[/itex]?
     
  13. May 11, 2014 #12
    Yaa, ρfree is the free charge carrier. They do not contribute in the polarization of a material. They contribute to the free current density.
     
  14. May 11, 2014 #13

    ChrisVer

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    So if you look at Griffith's introduction to electrodynamics, there is a good reasoning why you put [itex]\rho_{free}=0[/itex]
    The main idea is that the continuity equation for a conductor is written as:
    [itex]\dot{\rho}_{free}= -\frac{\sigma}{\epsilon} \rho_{free}[/itex]
    The solution of this is:
    [itex] \rho_{free}(t)= e^{-(\frac{\sigma}{\epsilon})t} \rho(0)[/itex]
    So if you add in the conductor some initial free charge [itex]\rho(0)[/itex], it dissipates in a characteristic time [itex]\tau=\frac{\sigma}{\epsilon} [/itex]
    For a conductor the [itex]\sigma[/itex] is large or taken infinite... thus the free charge disappears very fast (it goes on the edges/surface of the conductor). So there is no mistake to take that it's zero (it becomes zero instanteously) or you can just wait for it to become zero...
     
  15. May 11, 2014 #14
    But what about the free electrons present in the conductor will they not contribute to the ρfree. So as long as free electrons are present how can we make ρfree = 0
     
  16. May 11, 2014 #15

    ChrisVer

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    They all go on the surface....
     
  17. May 11, 2014 #16

    ChrisVer

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    Also about your ions question, the deal is that ions don't move... But you can see the absence of an electron as a moving positive charge towards the other way (hole)...
     
  18. May 11, 2014 #17
    Yaa, but they don't leave the conductor. So as a whole the conductor should have ρfree, may be on the surface of the conductor, isn't it ?
     
  19. May 11, 2014 #18

    ChrisVer

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    and how are you supposed to define a volume density [itex]\rho[/itex] for something that only exists on a surface?
    Inside the conductor, there are no free charge densities, otherwise they would create an electric field. Then because of that electric field, the charges would again move so that they would immediately cancel it.
    Then Gauss's law will again give you 0 free charge densities
    Conductors as a whole, yes, they can have a charge...
     
  20. May 11, 2014 #19
    Great, many many thanks to you friend. I have got it.
     
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