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Maxwell's equations and Coulomb's law

  1. Feb 23, 2009 #1
    In Feyman's lectures on physics, he said Maxwell's first 2 equations in electrostatics, namely curl E =0 and div E=rho/epsilon, is equivalent to Coulomb's law and superposition principle,
    But for a particular charge distribution, we can always use Coulomb's law and superposition principle to determine one unique field, and when it comes to Maxwell's equation-if we just want to satisfy the 2 equations-we can add any gradient of a harmonic field to the field we get using Coulomb's law and superposition principle,thus we can get infinitely many solutions.
    So how can it be that they are equivalent??
     
    Last edited by a moderator: Feb 23, 2009
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  3. Feb 23, 2009 #2

    clem

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    Adding a gradient won't satisfy the divE equation.
    There is a uniqueness theorem for the divE and curlE dquations.
     
    Last edited by a moderator: Feb 23, 2009
  4. Feb 23, 2009 #3
    but i'm adding the gradient of a harmonic field,say, h, then laplacian h=0,which means
    div(grad h)=0, then div(E+grad h)=div E+0=div E, which still satisfies the first equation.Am I correct?
     
    Last edited by a moderator: Feb 23, 2009
  5. Feb 23, 2009 #4
    "There is a uniqueness theorem for the divE and curlE equations. "
    could you give some more details about the theorem? Thanks.
     
    Last edited by a moderator: Feb 23, 2009
  6. Feb 23, 2009 #5

    atyy

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    Last edited by a moderator: Feb 23, 2009
  7. Feb 23, 2009 #6
    Any localized charge distribution has a field that tends to zero at infinity, while any harmonic function is not even bounded at infinity. One of the boundary conditions is violated, so the gradient of a harmonic function may not be added.
     
    Last edited by a moderator: Feb 23, 2009
  8. Feb 23, 2009 #7

    clem

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    The uniqueness theorem depends on appropriate boundary conditions atyy suggests.
    The proof is in most EM textbooks.
     
  9. Feb 23, 2009 #8
    But what's the boundary condition for a known charge distribution?
    I cannot see how Maxwell's equations require that "localized charge distribution has a field that tends to zero at infinity"
     
  10. Feb 23, 2009 #9

    atyy

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    Knowing the charge (or current) distribution isn't enough to determine the electric field in the general case. Because electromagnetic waves can travel far from their sources, you could reasonably specify an incoming light wave at the boundary. In electrostatics, we assume there are no light waves by definition of statics, and the boundary conditions are set by eg. whether you have a conductor etc present.
     
  11. Feb 23, 2009 #10
    Since the [itex]\nabla\times\vec{E}=0[/itex], we can ignore incoming and outgoing electromagnetic waves.

    Choose some point as the center of a Gaussian Sphere of radius r so that all of the charge is contained within the sphere. Then the first of Maxwell's equations says, roughly, that:

    [itex]\nabla \cdot\vec{E}=\frac{Q_{enc}}{\epsilon _0}\sim 4\pi r^2 |\vec{E}|[/itex]

    Since the sphere is contains all of the charge, [itex]Q_{enc}[/itex] is constant, so that:

    [itex]|\vec{E}|\sim \frac{Q_{enc}}{4\pi \epsilon _0r^2}\propto \frac{1}{r^2}[/itex]

    So that the electric field tends to 0 at infinity. Another way to see this is through the multipole expansion, but I won't go into that right now.
     
  12. Feb 24, 2009 #11
    but if we don't know the field is radical ,we cannot write E*4pi*r^2=Q/epsilon, right?
     
  13. Feb 24, 2009 #12
    for example, in empty space there's a sphere (radius R) with a constant charge density rho, in electrostatics,then the electric field should be uniquely determined, but what's the boundary condition in this case?
     
  14. Feb 24, 2009 #13

    Vanadium 50

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    That the field at infinity is zero.
     
  15. Feb 24, 2009 #14
    but how do you know that without Column's law?I mean, just by Maxwell's equations.
     
  16. Feb 24, 2009 #15

    clem

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    You are correct that Max alone does not uniquely specify the E field, but you haven't understood the posts. Max + Boundary Conditions do uniquely specify the E field.
     
  17. Feb 24, 2009 #16
    Yeah, I just cannot figure out what the boundary condition is in my example.
     
  18. Feb 24, 2009 #17
    "That the field at infinity is zero. "
    "but how do you know that without Column's law?I mean, just by Maxwell's equations. "
    I just cannot figure out what the boundary condition is in my example, I don' understand where the condition "That the field at infinity is zero. " comes from.
     
  19. Feb 24, 2009 #18

    atyy

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    In the case of a point charge, Gauss's law plus isotropy of space gives you Coulomb's law, from which you can see that the field goes to zero at infinity.

    In the case of a charge distribution with no symmetry, it's hard to use Gauss's law, so one has to use "common sense" or "physical sense" to make the boundary condition at infinity the same as that of a superposition of point charges.

    You can see that Maxwell's equations alone don't contain all the physics in classical electrostatics when a conductor is present. There we need "common/physical sense" to tell us that the conductor is equipotential. In some cases with a conductor, the boundary conditions plus uniqueness enables us to solve the problem with the "method of images" in which we solve the problem by replacing it with a different problem.

    So the boundary conditions are determined by additional physics you choose according to the situation at hand, just like initial conditions.
     
  20. Feb 24, 2009 #19
    But if we must take isotropy of space into account, does it mean divE= rho/epsilon and curlE=0 indeed say less than Column's law?
     
  21. Feb 24, 2009 #20

    atyy

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    Hmm, yes and no. Do you count the principle of superposition as part of Coulomb's law?
     
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