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Electric field arising from a uniform charge distribution of infinite extent 
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#1
Nov1606, 02:32 PM

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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, wellknown solution, but I have not encountered it before. Can anyone provide a reference or some insight? 


#2
Nov1606, 06:36 PM

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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. Claude. 


#3
Nov1606, 08:59 PM

P: 3

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?



#4
Nov1706, 12:44 AM

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Electric field arising from a uniform charge distribution of infinite extent
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.
Daniel. 


#5
Nov1706, 10:51 AM

P: 3

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 actionatadistance) 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??? 


#6
Nov1706, 01:37 PM

P: 223

[tex] \vec{E} = V_0 (x \hat{x} + y \hat{y}) [/tex] satisfies Gauss' law quite nicely, so it turns out that there are an infinite number of solutions for the given charge distribution. 


#7
Nov1706, 11:51 PM

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#8
Nov1906, 06:51 PM

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Claude. 


#9
Nov1906, 07:09 PM

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"Generalization of the electrostatic potential function for an infinite charge distribution" Palma et al. American Journal of Physics, 71 (8) 813815. 


#10
Nov1906, 07:30 PM

P: 546

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). 


#11
Nov2106, 06:08 PM

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P: 1,477

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. Claude. 


#12
Nov2306, 12:48 PM

P: 546




#13
Nov2306, 06:45 PM

Sci Advisor
P: 1,477

Claude. 


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