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Maxwell equation at the surface of a conductor - paradox?

  1. Jul 20, 2015 #1
    Assume that we have a conductor of any shape, say a ball of copper. At electrostatic equilibrium, it is well known that the potential inside this conductor is constant, for otherwise free charges would move from points of highest potential to points of lowest potential (this includes the surface of the conductor). This implies in particular that the electric field is null inside the conductor and on its boundary. Now, on the surface of the conductor, Maxwell equations give ##{\rm div} \vec E = \rho/\varepsilon##, so ##\rho = 0## everywhere on the surface. These equation are true even if an exterior forcing electrostatic field is applied to the conductor, which means that no charge move when applying an electrostatic field to a conductor; this is in total contradiction to what is teached about Faraday cages (say). So, what is wrong ?
     
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  3. Jul 20, 2015 #2
    ρ is not 0 on the surface. on a perfect conductor ALL of the charge is on the surface. the issue is that the electric field is NOT continuous on the boundary of a perfect conductor, but the potential is. when an external electric field is applied, charges in the surface DO move and the e-field inside a conductor is only 0 for static charge distributions. the charges re-arrange themselves so fast that it is barely noticeable unless the applied external field is not a static field. even then, the field inside a conductor is very very close to 0.
     
  4. Jul 20, 2015 #3
    Thank you for trying to answer. It is clear that you have not answered the question, but only stated commonly accepted facts (I observe nevertheless that your claim according to which the "e-field inside a conductor is only 0 for static charge distributions" contradicts the Faraday cage principle, which is valid in electrodynamic situations as well). In other words, I also know that the charges on the surface of a conductor should move under the influence of an electrostatic field, but as I wrote above, this contradicts Maxwell equation ##{\rm div}\vec E = \rho/\varepsilon## (since the electric field should be 0 on the surface of the conductor). This is the paradox and my question is : what is wrong ?
     
  5. Jul 20, 2015 #4
    The electric field is not 0 at the surface of the conductor. The charges on the surface of a perfect conductor do not move in an electrostatic field, classically. There is no paradox, you are making the paradox. The discontinuity of E gives you a nonzero Efield at the surface. It's not a difficult integral....
     
  6. Jul 20, 2015 #5
    I agree with cpsinkule, I don't see any paradox here.

    On the surface of an ideal conductor you will have a charge density which will cause an abrupt change in electric field so inside the charged shell you will have no field although outside there is electric field.

    Ideal conductors does not exist since for example charges will have to move infinitely fast in order to instantaneously neutralize changes in the external electric field but ideally they could create null field inside the conductor.
     
  7. Jul 20, 2015 #6

    Dale

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    @coquelicot you have to stop claiming that a conductor violates Maxwells equations. That is not what PF is about.

    Regarding your specific "paradox". Inside the conductor charges are free to move in any direction, so there cannot be an electrostatic E field in any direction. Because the E field is zero everywhere inside the conductor the divergence of the E field is zero and therefore the charge is also zero inside the conductor.

    At the surface of the conductor charges are free to move parallel to the surface, so there cannot be an electrostatic E field parallel to the surface.

    However, at the surface the charges are not free to move perpendicular to the surface. Therefore there can be an electrostatic E field perpendicular to the surface. If there is such an E field then the divergence of the E field at the surface is not zero and therefore the charge at the surface is not zero.
     
  8. Jul 21, 2015 #7
    I would like to point out that @coquelicot, which is one of my friends, has not claimed that a conductor violates Maxwell law (at least not in this thread), but on the contrary has stated that this is wrong and tried to undertand what seemed to him a paradox (presenting things as an apparent paradox is a common and well accepted way to ask questions) . Actually he is satisfied with the answer of @DaleSpam as it appears that he didn't take into account the orthogonal component of the electric field on the surface of the conductor. So, what was for him a paradox appears now to be a simple mistake, and this was exactely why he asked this question.
     
    Last edited by a moderator: Jul 21, 2015
  9. Jul 21, 2015 #8

    PeterDonis

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    The "Faraday cage principle" does not say that the e-field inside a conductor must always be zero, even in dynamic situations. If that were true, copper wires could not conduct electric currents, since doing that involves a nonzero (time varying) electric field inside the conducting wire.
     
  10. Jul 21, 2015 #9
    Your remark is interesting, and illustrates the fact that sufficient litterature about Faraday cages is lacking. I'm not sure that an electrically fed wire can be assimilated to a Faraday cage: it is more a circuit fed with a power supply. Anyway, I think it is well known that a Faraday cage isolates (more or less perfectly, according to the waves frequency and the quality of the cage) its interior from the exterior influence of EM fields, in dynamic situations or not. "Exterior influence" is the topic of this thread, and not circuits with "inner feeding" (I'm not claiming these terms make sense but trying to express what I feel).
    On the other hand, if the issue you pointed out is "perfect isolation", I would say equally that the claim according to which "the e-field inside a conductor is 0 in the electrostatic situation" is wrong as well: As coquelicot tried to explain in a previous (erased) thread (without claiming this is a physical truth), this depends upon the free charges that can be contributed by the conductor at any of its parts. Take for example a very very bad conductor : if you apply to it a strong electrostatic field, it is very unlikely (in my opinion) that the field inside the (bad) conductor will cancel.
     
    Last edited by a moderator: Jul 26, 2015
  11. Jul 21, 2015 #10

    PeterDonis

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    Thread closed for moderation.

    Edit: reopened with a reminder to avoid speculation and discussion of moderation topics here.
     
    Last edited by a moderator: Jul 26, 2015
  12. Jul 26, 2015 #11

    @DaleSpam. I was previously satisfied with your answer. Nevertheless, I have now one more question. Let ##x,y,z## be an orthogonal system of axes whose origin is at some point of the surface of the conductor, and such that ##x,y## is parallel to the surface and ##z## is orthogonal to the surface, pointing to the exterior. As you have explained, the electric field is 0 inside the conductor, and its ##x,y## components are nul on the surface of the conductor. So, ##\vec E = (0,0,E_z)## and ##{\rm div}\vec E = {\partial E_z\over \partial z}##. Now, either the electric field ##\vec E## is not null at the surface of the conductor, in which case ##\vec E## is not continuous and ##{\rm div} \vec E## makes no sense (at least according to the usual definition of the derivative), or ##\vec E=0## on the surface of the conductor. But in this case too, the left derivative of ##E_z## according to ##z## is equal to 0 (since the e-field is equal to 0 inside the conductor), while the right derivative may or may not be equal to ##\rho/\varepsilon##. This means that the orthogonal ##z## component of the e-field is not differentiable at the surface of the conductor, since the left and right derivative are distinct (right ?), and in both cases, apparently, Maxwell equation ##{\rm div} \vec E = \rho/\varepsilon## cannot be asserted, at least without introducing some exterior concept (right ? this is not a personal speculation, I am simply trying to understand). I guess that the solution to this problem may involve distributions, but I can't figure out how. Have you any insight ?
     
  13. Jul 26, 2015 #12

    Dale

    Staff: Mentor

    Are you familiar with the Dirac delta distribution?

    The "function" describing a surface charge is the Dirac delta, whose integral is the discontinuous Heaviside theta function. It has the necessary mathematical properties.

    ##\nabla \cdot (0,0,H(z))=\delta(z)##
     
    Last edited: Jul 26, 2015
  14. Jul 26, 2015 #13
    Yes, but what does it mean regarding the e-field and Maxwel equations ? does an equation like ##\delta(z) = \rho/\varepsilon## make sense ?
     
  15. Jul 26, 2015 #14

    Nugatory

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    It can. You're right that we have a problem with ##\frac{\partial{E_z}}{\partial{z}}## at the boundary because ##E_z## is discontinuous there, but that's not fatal because the identification of that partial derivative with the divergence only works where there are no discontinuities. Instead, we can work with ##E_z(z-\epsilon)-E_z(z+\epsilon)## as ##\epsilon## becomes arbitrarily small. This quantity is sensibly defined across the discontinuity.

    You will find similar concerns, and will have to deal with them in a similar way, when you apply Maxwell's equations to the electrical field of a point particle in a vacuum. There's no problem taking partial derivatives of the field components away from the charge and getting a zero divergence, but you can't sensibly evaluate the partials at the central point. We can, however, find the non-zero flux through the border of an infinitesimal volume surrounding the central point.
     
  16. Jul 26, 2015 #15

    Dale

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    It means that the E field is 0 inside the conductor and normal to the surface outside. It also means that there is charge on the surface, but not inside the conductor. Exactly what we want.

    Of course, it is a classical approximation, so it breaks down if you look at such a small scale that quantum effects become important. But at classical scales it is reasonable and valid.
     
  17. Jul 27, 2015 #16
    @Nugatory. I don't see how the symetric derivative can be of any help to give a sense to the divergence in this case. Can you be more precise and show me equations, so I'll be able to grasp it ?

    @DaleSpam. More or less, I would say you the same. Can you show me how you introduce the Dirac distribution inside the equations, and deduce from Maxwell laws your assertions ?
     
    Last edited: Jul 27, 2015
  18. Jul 27, 2015 #17

    Dale

    Staff: Mentor

    It is just an ansatz. Based on experience with the distributions I guess the result and then I check if my guess is correct. That is how a large portion of these problems are solved.
     
  19. Jul 27, 2015 #18
    @DaleSpam. Thank you for your answers. I don't want to make you vasting time, in particular because I'm not sure this problem is of interest for you. You seem to say that you have guessed the form of the solution, and checked that it works. This is this "checking" that would have been interesting for me. But as I've just said, I don't want to make you vasting time and it's ok if you want to give up.
     
  20. Jul 27, 2015 #19

    Dale

    Staff: Mentor

    Ah, I misunderstood what you were asking.

    The checking is relatively straightforward.

    ##\nabla \cdot (0,0,H(z)) = \frac{\partial}{\partial z} H(z) = \frac{d}{dz} H(z) = \delta(z)##
     
  21. Jul 28, 2015 #20
    This is not what I meant. You assume that the e-field is discontinuous, say Heaviside-shaped. Then its derivative along z is Dirac. So far so good. But where are Maxwell equations in all of this ? How do you find the magnitude of the field near the surface, or more precisely, its orthogonal slope in the outer direction ? (if the field were not discontinuous, Maxwell equations would give ##\rho/\varepsilon##). Saying that the e-field is Heaviside near the surface without showing that this helps somewhere in equations, amounts to nothing else than an affirmation that can not be confirmed (at least without doing experiments) ; even if this is true, this is more an experimental hypothesis than a theoretical deduction, or than a theoretical hypothesis supported by satisfying theoretical consequences. In particular, this does not prove, apparently, that "there are charges on the surface" as you seem to have asserted above : we reached previously this conclusion from another point of view, using the Lorenz force acting on free charges.
     
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