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Electric field of uniform charge density

  1. Jul 8, 2009 #1
    Let's say we have an electric field,

    (ax, 0, 0). (a is a constant)

    The divergence of the field is a, so the charge density is a*epsilon.

    This implies a uniform charge density, but the field only points in one direction! Furthermore, it switches directions at the x-axis. But that's weird, since the location of the x-axis and therefore the change in direction is dependent on where you place your coordinate system, but the field of a uniform charge density shouldn't be dependent on something like that!

    Also, what if you made the field

    (ax, by, cz) (where a, b, c are constants)

    The charge density is still uniform, (a+b+c)*epsilon, but now, there's some kind of radial field, even though there's still a uniform charge density! I don't understand what's going on.
     
  2. jcsd
  3. Jul 10, 2009 #2

    gabbagabbahey

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    Hmmm... that is a little curious. Clearly, if you had a uniform charge density filling all of space, you would expect the field to be zero everywhere. Otherwise it would exert a force on a test charge placed at any given spot, tending to move it to a different spot, yet due to the symmetry of the source there should be no difference between one location and another.

    At first glance, Coulomb's Law seems to confirm this. Assuming a constant charge density [itex]\rho[/itex] and choosing (without loss of generality) the field point to be at the origin ([itex]\textbf{x}=0[/itex]) we have, using spherical coordinates for the integration:

    [tex]\textbf{E}=\frac{1}{4\pi\epsilon_0} \int \rho(\textbf{x}')\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3} d^3x'=\frac{\rho}{4\pi\epsilon_0} \int\frac{\textbf{x}'}{|\textbf{x}'|^3} d^3x'=\frac{\rho}{4\pi\epsilon_0} \int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{r\sin\theta\cos\phi\hat{x}+r\sin\thet\sin\phi\hat{y}+r\cos\phi\hat{z}}{r^3} r^2\sin\theta dr d\theta d\phi=0[/tex]

    (Doing the angular integrals first yields zero anyways.....how about if you do the radial one first?:wink:)

    However, Closer inspection reveals that the integral is only conditionally convergent, not absolutely convergent, and so depending on how you sum the fields of all the little pieces of charge (carry out the integral), you can get any value you like for the field. For instance, if you divide the charge distribution into concentric spherical shells and add up the fields due to these shells of various radii (from r=0 to r=infinity) you get very different answers depending on how far away you choose the field point to be from the center of these shells (in my above calculation, I essentially assumed that the field point was at the center of these shells, but any other choice should be equally valid). See for example, Riemann's Rearrangement theorem
     
    Last edited: Jul 10, 2009
  4. Jul 12, 2009 #3
    That's pretty cool! How did you recognize that it was conditionally convergent? By placing the charge elsewhere I suppose. Anyway, doing the radial integrals first yields infinities which are even messier... I think problems like this point to difficulties with the mathematics, not the physics.

    So, I emailed David Griffiths, whose textbook this question is from, and got this as a reply:

    "I am pleased that you find this one a puzzle! I agree with you.

    The point is that Maxwell's equations (in this case Gauss's law) are differential equations, which do not uniquely determine the field. You need suitable boundary conditions as well. Ordinarily we forget about this, because symmetry (in the case, for example, of an infinite plane of charge) or (for localized charge configurations) the natural physical requirement that the fields go to zero at infinity supplies the extra information without our even thinking about it. But in this case neither is available. So if you turn the question around, and ask "What is the field produced by a uniform charge density throughout space?" the answer is simply ambiguous (you have supplied a second solution, but there are, of course, infinitely many more).

    I hope this helps. The moral of the story is that the laws of electrostatics must be supplemented by suitable boundary conditions, before the fields are uniquely determined."

    So because a uniform density all the way to infinity has no boundary conditions, Gauss's Law doesn't help here.
     
  5. Jul 13, 2009 #4

    gabbagabbahey

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    Wow, it is a very pleasant surprise to see Griffiths responded so quickly to your email (as the author of one of the most widely used electrodynamics texts, he must get many many emails)! :smile: However, I disagree with his conclusion on this.

    I would say that a uniform charge distribution throughout all space clearly possesses both rotational and translational symmetry, and in accordance with Coulomb's Law, the electric field it produces will as-well. The only vector that possesses both these symmetries is, of course, the null vector. Hence, the electric field must be zero in this case.

    I would argue further that carrying out the Coulomb integration in any manner that yields a non-zero result, is simply due to an incorrect ordering of terms in an integral which is not absolutely convergent.

    Finally, I would argue that Gauss' Law is simply wrong in this case (a bold assertion I know!). The reasoning for this is that Gauss' Law is not empirically derived from experiments the way Coulomb's Law is, but rather it is derived mathematically from Coulomb's Law (making Coulomb's Law more fundamental than the two relevant Maxwell's equations)---and that there is an error in the derivation for cases where the Coulomb integral does not converge absolutely. Namely,

    [tex]\mathbf{\nabla}\cdot \int \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3}d^3 x' \neq \int \left(\mathbf{\nabla}\cdot\frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3}\right)d^3 x'[/tex]

    For, if I recall correctly (and I'll have to look this up when I get a chance!), one of the conditions for being allowed to bring the operator inside the integral like this is that the integral is absolutely convergent.

    So, in this case, it is not solving Maxwell's equations which is at issue since the two relevant equations are actually invalid in this case. Hence his argument about not having sufficient boundary conditions is besides the point.

    My conclusion is thus:

    (1) A uniform charge distribution throughout all space produces an electric field of zero.

    (2)Any proposed electrostatic field (such as [itex]\textbf{E}=a\hat{x}[/itex]), in order to be a valid electrostatic field, must imply (through Gauss' Law) a charge distribution which renders the Coulomb integral absolutely convergent. Hence [itex]\textbf{E}=a\hat{x}[/itex] is not a possible electrostatic field.

    If anyone finds any holes in this argument please point them out to me.
     
  6. Jul 13, 2009 #5

    clem

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    This is correct, but that is because you have applied the BC that E be spherically symmetric at infinity.
    The original question was not about Coulomb's law.
    You have a lot backward's here. In fact, the most accurate experimental verification of the exponent 2 in the static Coulomb law is by testing Gauss's law.
    Maxwell's equations are more fundamental than Coulomb's law.
    The uniqueness theorem shows that Coulomb's law follows from Max's eqs. only for the appropriate BC. The integrals you give are only valid for special BC on phi and E that you have implicitly assumed
    This might be why the integrals are not unique. When an integral is not uniformly convergent at infinity, the result depends on the BC at infinity. This is consistent with solving Maxwell's equations with BC.
    He is right and you are the invalid here.
    This is true if E is assumed spherically symmetric at infinity.
    These statements are wrong. I know it is hard for you to envisage this situation,
    but 150 years have shown that you must start with Maxwell and BC.
    In this case, you can't start with the charge distribution without BC.
    You can start with E and deduce the charge distribution and the BC.
    In the case presented with E_x=ax, the BC at infinity is that E_x=ax,
    and there is no ambiguity.
    I have tried to.
     
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