Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mathematical derivation of the Faraday cage from the Maxwell Equations

  1. May 6, 2006 #1

    mma

    User Avatar

    Hi,

    We know that in a space region free from electric charges and surrounded by a conducting surface, the electric field must be zero (this is the Faraday cage).

    I suspect that this statement can be derived directly from the Maxwell equations, but I don't find this derivation anywhere. Could somebody show me that?
     
  2. jcsd
  3. May 6, 2006 #2

    mma

    User Avatar

    Is this question so simple, that it had to be moved to Introductory Physics? :(

    Then why does nobody answer me?
     
  4. May 6, 2006 #3

    mma

    User Avatar

    I'm afraid that this isn't trivial. I've found only this guidance about it here:

    So "it can be shown". But, how?
     
  5. May 6, 2006 #4

    Astronuc

    User Avatar

    Staff: Mentor

    Well trivially, it's Gauss's law. Inside the hollow conductor there is not charge, so the enclosed charge is zero, so the electric field is zero everywhere.

    Now more directly, consider the most trivial case of the center of a hollow sphere, with 'uniform' charge on the surface. For each charge, there is an equal charge diametrically opposed, and therefore at the center the electrical fields (vectors) are equal and opposite, so they cancel.

    Now, consider any point, off-center. One cannot apply the opposite point charge, but rather one must consider opposing surfaces, dA, which would have charges [itex]\sigma_1\,dA_1[/itex] and [itex]\sigma_2\,dA_2[/itex]. Now think if two cones with vertices touching (and having same solid angles) and colinear (parallel) axes, with heights [itex]r_1[/itex] and [itex]r_2[/itex]. The E from one is just [itex]\sigma_1\,dA_1[/itex]/[itex]r_1^2[/itex] and the other is [itex]\sigma_2\,dA_2[/itex]/[itex]r_2^2[/itex], but realize that [itex]dA_i[/itex] is proportional to [itex]r_i^2\,d\Omega_i[/itex], where [itex]d\Omega[/itex] is the solid angle enveloped by cones and subtended by [itex]dA_i[/itex].

    So Ei is proportional to 1/[itex]r_i^2[/itex], and [itex]dA_i[/itex] is proportional to [itex]r_i^2[/itex], and the term cancel which then leaves equal charges ([itex]\sigma\,d\Omega[/itex]) opposing each other, and therefore the electric fields cancel, i.e. [itex]\vec{E}\,=\,0[/itex].

    Voila!
     
    Last edited: May 6, 2006
  6. May 6, 2006 #5

    mma

    User Avatar

    Not too good. This "proof" doesn't use the presence of the conductive surface. So, it states that the electric field in every chargeless volume vanishes, what is obviously false.


    The other solution is applicable only in this very special case :(

    But what is the general solution?
     
    Last edited: May 6, 2006
  7. May 6, 2006 #6

    Astronuc

    User Avatar

    Staff: Mentor

    No, the other solution that I provided ends up being a surface integral. If one makes [itex]d\Omega[/itex] very small and integrates the E field over all [itex]\Omega[/itex], one discovers that the entire integral vanishes, for any arbitrary surface enclosing a volume. All the example shows is that there is an equal charge on one side of the volume opposing the other charge, and the net E field is zero.

    For non-spherical surfaces, and particularly nonsymmetric surfaces/volumes, the integrals can be more complicated.
     
  8. May 7, 2006 #7

    mma

    User Avatar

    OK, your solution is good for any shape.
    But only for
    • uniform charge distribution on the surface
    • zero external electric field
    The point is that Faraday cage precludes external electric field to enter into its interior.
    Evidently, the exterior electric field affects the distribution of the free charges on the metal surface such way that the algebraic sum of the electric field caused by these free charges and the external electric field is zero. I am curious of the proof of this statement.
     
  9. May 7, 2006 #8

    mma

    User Avatar

    For we know that the electrostatic potential on a closed conducting surface is always constant, and the electric field is the gradient of the potential, my question sounds mathematically:

    Let's proof that if the potential on a closed surface is constant, and if there is no singularity of the potential inside the region bounded by this surface, then the potential in this whole inner region is constant too.
     
  10. May 7, 2006 #9

    mma

    User Avatar

    I'm afraid that this isn't true in general. A simple counterexample the V(r)=r potential, and a sphere as the conducting surface.

    So, we need some extra conditions about this potential, of course on the bases on the Maxwell's equations.

    Any idea?
     
  11. May 7, 2006 #10
    If the Astronuc's demonstration didn't satisfy you, then try to solve the Laplacian equation
    [tex]\bigtriangledown^2 V=0[/tex]
    with
    [tex]V=V_0[/tex]
    on the boundary (using Femlab for example). This Eq. is obviously derived from Maxwell Eqs......

    The "counterexample" [tex]V(r)=r[/tex] you proposed is wrong because it does not satisfy the equations above.
     
    Last edited: May 7, 2006
  12. May 8, 2006 #11

    mma

    User Avatar

    OK, this seems better. Unfortunately, I am not an expert in PDE-s, and I haven't find the solution of the Laplace's equations for this special case (3-dimensions and constant as boundary condition). I am not curious of of simulations, but only of exact mathematical solution. Can anybody show me the solution of the Dirichlet-problem of the Laplace's equation in 3 dimensions with constant boundary condition?
    (of course I see that constant is a solution, but I don't know, why must it be unique)
     
    Last edited: May 8, 2006
  13. May 8, 2006 #12
    You can "feel" what's happen with the following 1-D example:
    [tex]\frac{d^2 V}{dx^2}=0[/tex] in [tex][a,b][/tex]
    and
    [tex]V(a)=V(b)=V_0[/tex].

    (the potential between two infinite metallic plates).
    From the first Eq. you have
    [tex]V(x)=mx+n[/tex]
    and after imposing the boundary conditions
    [tex]ma+n=V_0[/tex]
    [tex]mb+n=V_0[/tex]
    you get [tex]m=0[/tex] and [tex]n=V_0[/tex].
    So the only solution you get is [tex]V(x)=V_0[/tex].

    Hope it helps....
     
  14. May 8, 2006 #13

    mma

    User Avatar

    Yes, thanks. This is exactly what I thought after your first reply.
     
  15. May 8, 2006 #14

    mma

    User Avatar

    However, thinking in 1 or 2 dimensions instead of 3 can be very misleading.
    3-dimensional Euclidean space has completely different properties as 1 or 2-dimensional ones. E.g. the Banach-Tarski paradox is valid in 3 dimension, but not in 2 or 1 (this paradox states that you can cut a solid sphere into finitely many pieces and moving this parts, you can build 2 solid sphere with same size than the original one). Or, in physics, Zeeman's "Causality Implies the Lorentz Group" theorem is valid only in Minkoswski spaces having dimension more than 2.

    So we really need a valid uniqueness theorem for the solution of our Dirichlet problem.
     
  16. May 9, 2006 #15

    mma

    User Avatar

    I've found the proof of the uniqueness. This is the following.

    Suppose we have two different solutions of our Dirichlet-problem.
    For the Laplace equation is linear and the values of these functions on the boundary are equal, their difference is a solution of the Laplace equation with boundary condition 0. This implies that this difference function is 0, otherwise it would have a local minimum or maximum in the inner region, what contradicts to the fact that it is a solution of the Laplace equation. So the difference of our two solutions is zero, hence these two solutions are equal.
    q.e.d.
     
  17. May 10, 2006 #16

    mma

    User Avatar

    But we are not ready yet.

    Laplacian equation is valid only in electrostatics.

    How can we proof that Faraday cage shields electromagnetic waves as well?
     
  18. May 10, 2006 #17

    lightgrav

    User Avatar
    Homework Helper

    In the physical world, charge is a property that is intimately connected to ponderous objects. The sluggish response of the charge carriers in a conductor make it possible for microwaves to resonate in an empty cylinder.
     
  19. May 10, 2006 #18

    mma

    User Avatar

    Does this mean thet Faraday cage shields only static fields? How can then it prevent us e. g. from lightning?
     
  20. May 10, 2006 #19

    Astronuc

    User Avatar

    Staff: Mentor

    The current travels in the conductive material, not in the hollow (which is not conducting) and travels primarily in the outside surface of the conductor.
     
  21. May 10, 2006 #20

    mma

    User Avatar

    OK, but how follows this from the Maxwell-equations? Currents are generated by magnetic flux change. If magnetic flux can penetrate the conductive surface, then it's change can generate currents in a piece of conductor located inside the hollow.
    We proved yet only that electrostatic field can't enter into the hollow. The proof is based on the fact that the surface of the conductive shell is equipotent. But what is the situation, when the external field changes in time? Then the potential of the shell evidently changes in time and in a given time instance varies point by point on the shell.
    Can't it be that sudden changes of the electromagnetic field can penetrate the conducting surface? If yes, then how depends the transmission on the frequency? What is the mathematical description of this phenomenon?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Mathematical derivation of the Faraday cage from the Maxwell Equations
Loading...