Why does the electric field vanish at infinity?

Question

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 $r \rightarrow 0$.

Answer 1

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.

Answer 2

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...

Answer 3

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.