Will the flux through an arbitrary closed surface be finite or infinite when a plane charge intersects the Gaussian surface?

Question

Let's consider a closed Gaussian surface (in red).

The white line and the white shaded part lies inside the Gaussian surface and the black line and the portion above it lies outside the Gaussian surface. The red Gaussian surface is intersected by a surface plane charge at the green line.

From Gauss law:

Flux due to portion above black line is zero because it lies outside the Gaussian surface. Flux due to white shaded part is finite because it lies inside the Gaussian surface.

Now we just have to consider flux over red Gaussian surface due to black, green and white lines. Now here is where I am having difficulty:

How can we find electric field at green line due to singularity? Similarly how can we find electric field at points on black and white lines which are infinitely close to green line?

If the electric field is finite everywhere in the green line, we can ignore its contribution to total flux (since the green line has infinitesimal area). Thus the total flux would be unaffected and will remain finite. However if the electric field is $(\text{infinite})^2$ everywhere in the green line:

$$\vec{E} \cdot \vec{dS}= \text{finite,}$$

we cannot ignore the contribution of green line to the total flux. And the total flux will be infinite.

Note that I say $(\text{infinite})^2$ rather than $\text{infinite}$ because of inverse square nature of electric field. When $r \rightarrow 0, \vec{E} \rightarrow (\text{infinite})^2$

So will the flux through an arbitrary closed surface be finite or infinite when a plane charge intersects the Gaussian surface?

Answer 1

Gauss' Law is clear. The total flux is finite. The Gaussian surface has an inside and an outside which divides the charge density into inside and outside. The integral of the charge density inside is finite. The total charge on the surface is zero because the boundary has no thickness.

If your concern is calculating the flux integral, that's not a problem, either. The electric field of a charged sheet is known to be finite everywhere. If your concern is the contribution to the integral from points on the surface, again, the thickness of the surface is zero, so its contribution is zero. (I suppose it's "a set of measure zero".)

A more realistic model might replace the charged sheet with a thin uniformly charged slab having no charge density singularities, and let the thickness of the slab approach zero. That might make it easier to see that everything is finite, but on the other hand, to show that might involve other Gaussian surfaces, so depending on your point of view, you might argue that the can has been kicked down the road.