Why can infinite planes be approximated as Gaussian surfaces? A little background: I'm an undergraduate studying Electrodynamics, currently in Chapter 8 of Griffiths.
A question I came across (8.4 part a for those curious) asks for a calculation of the force exerted by one point particle on another point particle of equal charge. This is meant to be done through means of:
$\oint_S \bar{T} \cdot d \vec{a}$
$\bar{T}$ being the Maxwell stress tensor.
I'd expect that you'd have to create a closed surface around one of the point charges, but this question explicitly wants the surface integral to be done over the plane. Easy enough, but my question is why would this be a viable closed surface?
The explanation I've been given so far from the lecturing professor (and wikipedia) is that the plane is an approximation of a closed surface. It seems that as the "bubble" (see the below cross-section illustration) extends to infinite size, its function becomes negligible and what ultimately matters is the infinite plane.

This explanation makes intuitive sense, but I feel is a little hand-wavey. Why is this the case? Does it work for all systems of particles (continuous and not continuous), or just for a point particle. My intuition would tell me that if you zoomed in on the very edge of a sphere, it would start to look like a plane (but again this explanation is not very mathematical).
Any insight would be greatly appreciated.
 A: The key calculation is given in example 8.2, the problem that reads:

Determine the net force on the "northern" hemisphere of a uniformly charged solid sphere of radius $R$ and charge $Q$.

Part of the solution there is computing the surface integral of $\overleftrightarrow{T}\mathrm{d}\vec a$ over the hemispherical "bowl", yielding
$$F_{\text{bowl}}=\frac1{4\pi\epsilon_0}\frac{Q^2}{8R^2}.\tag{8.23*}$$
I omitted the intermediate calculation here, hence the asterisk, otherwise it's indeed equation $(8.23)$ from [1].
In our case of two charges we have to make an observation that, as $R\to\infty$, the field (and thus Maxwell's stress tensor) becomes closer to that of a single charge $2q$ (assuming equal signs). This can be rigorously described in terms of multipole expansion, whose leading term dominates at infinity.
Now, your doubt can be cleared by looking at the expression: we have $R$ in denominator, which makes this term vanish at $R\to\infty$, which is exactly the limit where our flat surface between the two charges becomes an infinite plane. Now only the contribution of the plane remains.
References

*

*Griffiths, David J. Introduction to Electrodynamics. Fourth edition. Pearson, 2013.

A: Imagine the infinite plane orthogonally intersected by a cylindrical Gaussian pillbox of area $A$. Force lines are normal to the infinite plane, therefore the totality of force lines exiting the cylinder are through the ends of the cylinder.  If the charge density of the infinite plane is $\sigma$, and the integral only needs to be evaluated over the two ends, then
$$g(2A) = 4\pi GM = 4\pi G\sigma A$$
hence
$$g = 2\pi G\sigma$$
This is a constant, independent of the length of the cylinder.
