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

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?