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Electric field arising from a uniform charge distribution of infinite extent

  1. Nov 16, 2006 #1


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    Consider a uniform charge distribution occupying all of (flat) spacetime,

    [tex]\rho(t,x,y,z) = \text{constant} \;\;\;\;\; ,\; (t,x,y,z) \in R^{1,3}[/tex]

    Because this charge distribution is translationally invariant, it seems reasonable to expect that the electric field arising from the charge distribution is zero,

    [tex]E(t,x,y,z) = 0 \;\;\;\;\;\;\;\;\; ,\; (t,x,y,z) \in R^{1,3}[/tex]

    But then the electric field does not appear to satisfy Poisson's equation,

    [tex]\nabla\cdot E = \rho/\epsilon_0[/tex]

    Presumably this problem has a simple, well-known solution, but I have not encountered it before. Can anyone provide a reference or some insight?
    Last edited: Nov 16, 2006
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  3. Nov 16, 2006 #2

    Claude Bile

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    The solution is simple, E = 0 everywhere as symmetry demands.

    The crux of the paradox is the boundary conditions of the problem. It is implicitly assumed in Poisson's equation that the charge density tends to zero as one approaches infinity. Your scenario of course, does not have the charge density tending to zero, hence Poisson's equation is actually rendered invalid in this case.

    Poissons equation is rendered invalid not just for this equation, but for an infinite line charge, infinite sheet charge, in fact, any charge distribution where the boundary condition of charge density tends to zero as we approach infinity.

    Why does Poisson's equation possess this condition? Because it is derived from how we define work and potential, which requires the condition that potential and thus charge density is zero at infinity.

  4. Nov 16, 2006 #3


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    Followup question

    Claude, thankyou for casting some light on the matter! You mentioned that Poisson's equation is not applicable to charge distributions of infinite extent. Are Maxwell's equations defined for such systems? Also, do you know of a book which discusses this?
  5. Nov 17, 2006 #4


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    Maxwell's equations in integral form are valid for any distribution of (electric) charge, it's just that you have to take into account the boundary conditions on th charge & current which occur when you pass from the (always) correct integral form to the differential form. The point i'm trying to make is that you don't have to take the differential equations for granted.

  6. Nov 17, 2006 #5


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    Daniel, thanks for the clarification; I can see how, for an infinite charge distribution, Gauss's Law applies but is not equivalent to Poisson's equation.

    I find it surprising that the equations in integral form are more fundamental than the equations in differential form. Contrast this with particle physics, where the need to keep all interactions local (no action-at-a-distance) leads to the equations of motion being expressed in differential form.

    I wonder whether it would be possible to obtain a differential formulation by dividing the charge distribution into two patches and introducing a potential for each patch? Poisson's equation could be applied to a single patch because each patch has a boundary. This is somewhat reminiscent of the need to use two patches, each with its own vector potential, to describe the field arising from a magnetic monopole. The two vector potentials are related by a gauge transformation; could something similar apply to the two scalar potentials? Perhaps there is a topological charge hidden in this problem???
    Last edited: Nov 17, 2006
  7. Nov 17, 2006 #6
    Actually, you can come up with a large number of electric fields which satisfy the equation in general, so the problem is worse than just "E = 0". You can check to see that
    \vec{E} = V_0 (x \hat{x} + y \hat{y})
    satisfies Gauss' law quite nicely, so it turns out that there are an infinite number of solutions for the given charge distribution.
  8. Nov 17, 2006 #7
    How does Poisson's equation assume the potential vanishes at infinity? Mathematically, you can get there from the integral form of Gauss' Law by applying the divergence theorem and recognizing that any curl-less vector field can be represented as the gradient of a scalar field. All of this is done locally without reference to boundary conditions. So, where does the problem come from?
  9. Nov 19, 2006 #8

    Claude Bile

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    The definition of V demands that the potential be zero at infinity. Once you include V in your calculations you therefore are implicity assuming some boundary conditions. Gauss Law and Poisson's equation therefore does make reference to boundary conditions, it is just buried in how quantities are defined.

    Last edited: Nov 19, 2006
  10. Nov 19, 2006 #9

    Claude Bile

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    I don't know of any books, my readings on this subject has been limited to journal articles. This journal article may contain some of the discussion you are after.

    "Generalization of the electrostatic potential function for an infinite charge distribution" Palma et al. American Journal of Physics, 71 (8) 813-815.
  11. Nov 19, 2006 #10
    The definition of V in no way assumes that. V is simply defined to be that scalar field such that [tex]\vec{E} = -\nabla V[/tex]. This makes no assumptions about behavior at infinity. All it assumes is that [tex]\nabla \times \vec{E} = 0[/tex]. For that matter, adding an arbitrary constant or scalar function of time to V will not change [tex]\vec{E}[/tex]; so, if you give me a V that vanishes at infinity, I can give you an equally valid one that doesn't.

    It's certainly true that in solving Poisson's equation (or Gauss' Law, or what have you) that you need to impose boundary conditions to get a unique solution. However, no boundary conditions are actually assumed in the definition of V (or [tex]\vec{E}[/tex] for that matter).
  12. Nov 21, 2006 #11

    Claude Bile

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    At some point you need to set a reference point for V. If we use differential laws, we MUST use the reference that V approaches zero at infinity, because this reference is set as we move from the integral form to the differential form (via the definition of an absolute potential). If we stick with the integral form, we can use our own reference point.

    EDIT: I should include that the definition for absolute potential is the work required to move a +1 C charge from infinity to a given point. The value for absolute potential at infinity must therefore approach zero.

    Last edited: Nov 21, 2006
  13. Nov 23, 2006 #12
    This is simply not correct. To solve the differential equation for the potential, we need to set boundary conditions. However, the differential equation is valid independent of the boundary conditions. Furthermore, we have near complete freedom to set our boundary conditions. If it were necessary that the potential be 0 at infinity, we wouldn't have full gauge invariance.

    This definition is implicit in the integral equation for the potential; but it is not quite fundamental. I could choose that energy is not 0 at infinity without changing any physical consequences.
  14. Nov 23, 2006 #13

    Claude Bile

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    The problem isn't with the boundary conditions of Poisson's equation etc. it's a problem with the boundary conditions of the quantity V. All this hoo-ha can be avoided by choosing the reference point for V at the origin for example, and everything works out fine. Does this not indicate that it is the boundary conditions of V that results in the apparent paradox present in Poisson's equation for infinite charge distributions?

    Yes, the definition of an absolute potential in this fashion does introduce a constraint, the very constraint in fact, that results in the paradox.

    Absolutely, but while physical consequences don't change, mathematical ones do. As I have pointed out, it is our choice of our reference for V that results in the apparent paradox, if we choose a different reference, the paradox is avoided.

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