In the statement of the Gauss Law if we happen to consider a surface at infinity what should we expect to get keeping in mind that electric field is 0 at infinity?

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  • $\begingroup$ What is "a surface at infinity"? $\endgroup$ – ACuriousMind Mar 22 at 12:09
  • $\begingroup$ Also having trouble understanding what you're asking. Do you mean something like a sphere of infinite radius? $\endgroup$ – JamalS Mar 22 at 12:36

Let's consider a few cases here.

There's two rough methods we can use here : either we consider the entire space directly, or we can take some limit process.

In the first case, this is simple enough. The boundary of $\mathbb{R}^3$ is the empty set $\varnothing$. Then we have

\begin{equation} \int_{\varnothing} \vec{E} \cdot d\vec{S} = \int_{\mathbb{R}^3} \rho(x) d^3x \end{equation}

The first term is zero, the second term is the total charge $Q / \varepsilon_0$. Therefore, $Q = 0$. Why is that?

The Gauss theorem is a specific form of Stoke's theorem, which is defined specifically for functions with a compact support, which means that in our case, $\vec{E}$ should be zero outside of a finite region of space. This cannot be true if $Q \neq 0$, because, for instance, by the shell theorem, any non-zero static will generate an EM field like the Coulomb field outside of a volume containing them, therefore it will never be zero outside of a compact region. Therefore, for any electric field where it makes sense to use Gauss's law on all of space, the total charge must be zero.

The second method is to simply consider sequences of volumes, ie we have a sequence $(\Omega_n)$ such that $\lim \Omega_n = \mathbb{R}^3$. What happens here will depend on exactly what sequence of shapes and charge distribution you're dealing with, using the usual theorems from analysis, but let's consider three cases here.

First, let's consider a point charge at the center. Then we have

\begin{eqnarray} \int_{S_{0, R}} \vec{E} \cdot d\vec{S} &=& \iint E_r R^2 \sin \theta d\theta d\varphi\\ &=& 4 \pi \frac{q}{4\pi \varepsilon_0 R^2} R^2\\ &=& \frac{q}{\varepsilon_0} \end{eqnarray}

Thi is independent of the radius, therefore taking the limit is fairly trivial here.

Now let's consider the case of a charge distribution that is not of a compact support. Let's pick for instance a (standard) normal distribution

\begin{eqnarray} \rho(r, \theta, \varphi) &=& \sqrt{2} 2 Q \frac{e^{-r^2}}{(2\pi)^{3/2}} \end{eqnarray}

The total charge is $Q$, and the charge in a ball is

\begin{eqnarray} Q_R &=& \int_S \int_{0}^R \sqrt{2} 2 Q \frac{e^{-r^2}}{(2\pi)^{3/2}} r^2 \sin \theta dr d\theta d\varphi\\ &=& \frac{8\sqrt{2} \pi Q}{(2\pi)^{3/2}} \int_{0}^R r^2 e^{-r^2} dr\\ &=& \frac{4\sqrt{2} Q}{(2\pi)^{1/2}} \frac{1}{4} (\sqrt{\pi} \textrm{erf}(R) - 2 e^{-R^2} R) \\ &=& \frac{Q}{(\pi)^{1/2}} (\sqrt{\pi} \textrm{erf}(R) - 2 e^{-R^2} R) \\ \end{eqnarray}

$\mathrm{erf}$ converges to $1$, the second term to $0$, so that our charge converges to $Q$. By symmetry, we get

\begin{eqnarray} \iint_{S_{0, R}} \vec{E} \cdot d\vec{S} &=& \iint_{S_{0, R}} E_r R^2 \sin \theta d\theta d\varphi\\ &=& 4\pi E_r R^2 \end{eqnarray}


\begin{eqnarray} E_r &=& \frac{Q}{4\pi r^2 (\pi)^{1/2}} (\sqrt{\pi} \textrm{erf}(r) - 2 e^{-r^2} r) \\ \end{eqnarray}

Everything converges nicely too, here, we have $E_r$ being asymptotically equivalent to Coulomb's law.

As a last example, take a constant charge distribution,

\begin{eqnarray} \rho(r, \theta, \varphi) &=& \rho \end{eqnarray}

By symmetry, the electric field should not point in any particular direction, and therefore should be zero, but for any volume,

\begin{eqnarray} \int_\Omega \rho(x) d^3 x = V(\Omega) \rho \end{eqnarray}

This is the old problem of the Newtonian cosmology for a universe with constant density. It has been attempted many different ways, with a variety of limiting process and approximation, but there are no good solutions for this (this is related to the fact that such a universe cannot have the appropriate boundary condition $\phi = 0$ at infinity).

Therefore, the verdict is that yes, you can use Gauss's theorem for infinite volumes, but be very careful, because it will only work for very specific cases. In particular, to have a well-defined Poisson equation (ie for the potential to be $\Delta \phi = \rho$), we should be able to define $\phi(\infty) = 0$.

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  • $\begingroup$ Thanks for the explanation. $\endgroup$ – Gourav Halder Mar 23 at 21:11

It wouldn't have practical meaning for infinite surfaces in infinity . Yes mathematically it would be valid, physically it wouldn't be valid.

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Yes it is valid in the limit to infinity. The field will tend to zero but the surface tends to infinity. The product if the two will remain equal to the enclosed charge.

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  • $\begingroup$ How will this be different for any other vector field ? $\endgroup$ – Gourav Halder Mar 22 at 10:26

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