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When $r \rightarrow \infty$, $E \rightarrow 0$ for a point charge or set of charges or a finite charge distribution. While this seems obvious, I cannot find a reason why this is true when inspecting Maxwell's equations and the Lorentz force law. I thought however that all of electrodynamics was contained in Maxwell's equations and the Lorentz force law. Why then, does $E \rightarrow \infty$ when $r \rightarrow 0$.

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    $\begingroup$ 1. It's $E\to 0$. 2. If you didn't say $E\to 0$ as $r\to\infty$, you'd have $F = qE \neq 0$ at infinity. Does that make sense to you? $\endgroup$ – ACuriousMind Mar 13 '15 at 15:04
  • $\begingroup$ I fixed the question. What is inherently wrong with having $F =/= 0$ at $r \rightarrow \infty$. Couldn't $F$ just be really small, but nonzero. $\endgroup$ – user71646 Mar 13 '15 at 15:07
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You are correct that the vanishing of the field does not follow from Lorentz' force law and Maxwell's equations alone. An additional physical argument is needed:

If you didn't have $E\to 0$ as $r\to\infty$, you would have non-vanishing force $F = qE \neq 0$ at infinity. That is physically non-sensical because it would mean that a charge influences charges that are infinitely far away measurably. It is the physical boundary condition that we must impose on the physical solutions to Maxwell's equations in order to preserve locality, otherwise we would be living in a universe where every charge pulls or pushes on every other charge in a non-neglegible manner, no matter where these charges are.

It is rather evident that that doesn't describe our universe.

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  • $\begingroup$ So $E \rightarrow 0$ as $r \rightarrow \infty$ is a universal physical boundary condition (since expiremtnally we don't deal with infinite charge distrubutions unless we are approximating). Since this is a universal boundary condition, is this a law? $\endgroup$ – user71646 Mar 13 '15 at 15:18
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    $\begingroup$ Coulomb's Law can be derived from Gauss's Law. Is anything else (boundary conditions) needed? $\endgroup$ – garyp Mar 13 '15 at 15:18
  • $\begingroup$ @garyp: You can equivalently base electromagnetism on Coulomb's law + SR (i.e. the transformation rules of electric and magnetic fields as well as the four-current), since you then can always go to the rest frame and use Coulomb's law, then transform back. $\endgroup$ – ACuriousMind Mar 13 '15 at 15:20
  • $\begingroup$ @Donald: Well, it's not really a law because you often solve for different boundary conditions, e.g. you have a capacitor known to be at a certain voltage, but it is something you have to say to derive things like Coulomb's law from Maxwell+Lorentz. $\endgroup$ – ACuriousMind Mar 13 '15 at 15:22
  • $\begingroup$ @garyp: Ah, I see what you mean. Have a look at Wikipedia's comment on that, you need there the "boundary condition" that the field is spherically symmetric. You always need some kind of physically sensible requirement, be it the vanishing at infinity, or be it spherical symmetry, or be it something else. $\endgroup$ – ACuriousMind Mar 13 '15 at 15:32
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Since you speak about point charge, let us check the Gauss law (to start with something well known)?

$\nabla \cdot E = \frac{\rho}{\epsilon_0}$

We know that divergence has something to do with flow across some closed surface (sphere surface is the best) around the $\rho$ - the same in integral form is:

$\oint\limits_{\delta\Omega} E \cdot dS = \frac{1}{\epsilon_0}\iiint_\Omega \rho dV$

$\Omega$ is volume around your charge (sphere for us), $\delta \Omega$ is the surface of the $\Omega$. I think I can use delta function for the point charge $\rho=\delta(x,y,z)$. This way we reduce the right side of the above to $\frac{1}{\epsilon_0}$, we can certainly have it a charge of one electron $e$, or some $q$, but always constant.

$\oint\limits_{\delta\Omega} E \cdot dS = const$

means, that whatever sphere $\Omega$ (diameter $R$) we use, the integral of $E$ over its surface is constant.

The last step : $\oint\limits_{\delta\Omega} 1 \cdot dS$ is the surface and goes like $R^2$ with $R \rightarrow \infty$. To keep the previous integral constant, you need $E \sim \frac{1}{R^2}$. The answer is $E \sim \frac{1}{R^2} \rightarrow 0$ for $R\rightarrow \infty$

Was that correct? I wonder why \oiint doesnot work...

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  • $\begingroup$ Your assuming that the electric field doesn't rotate as $r$ changes. $\endgroup$ – user71646 Mar 13 '15 at 16:13
  • $\begingroup$ Should it with point charge? $\endgroup$ – jaromrax Mar 13 '15 at 16:26
  • $\begingroup$ OP is taking the position that from Maxwell's equations and Lorentz force law electrodynamics follows. Nowhere in those equations is the direction of $E$ specified. $\endgroup$ – user70720 Mar 13 '15 at 16:40
  • $\begingroup$ @Donald though your question is answered, I don't see why this approach would be inadequate if the E-field is rotating? $\endgroup$ – Gonenc Mar 13 '15 at 17:21
  • $\begingroup$ I think this argument is fine, as long as symmetry is also invoked to argue what direction the E-field doesn't point. $\endgroup$ – BMS Mar 13 '15 at 17:59
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using V.E=p/e0 in integral form that is gauss law you get your first answer E going to zero as r go to infinity.

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  • $\begingroup$ I don't think this is right, because E could rotate such that E gets more and more perpendicular to the area vector as $r \rightarrow \infty$ $\endgroup$ – user71646 Mar 13 '15 at 15:18
  • $\begingroup$ what do u mean by E becoming more and more perpendicular to area vector as r→∞ $\endgroup$ – AMAN SANGAL Mar 13 '15 at 15:25
  • $\begingroup$ at infinite distance for a finite amount of charge with finite extension in space we can consider E to be in radial direction at every point of gaussian surface. i.e. we have E*A = Qenclosed. as this area is infinte so that gives E=0 $\endgroup$ – AMAN SANGAL Mar 13 '15 at 15:28
  • $\begingroup$ Why can we consider $E$ to be radial? $\endgroup$ – user71646 Mar 13 '15 at 15:36
  • $\begingroup$ because at infinite distance the charge distribution can be suitably considered symmetric at the gaussian surface considered $\endgroup$ – AMAN SANGAL Mar 13 '15 at 15:49

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