Lets say I have a coaxial cable, the internal material cylinder has radius $a$ and uniform volumetric charge density $\rho$, the external shield cylinder has radius $b$ and uniform superficial charge density $\sigma$, and the total charge of the cable is zero. I want to find the electric field at any point is space. Using Gauss's law I found $$E(r)= \begin{cases} \frac{r \rho}{2\epsilon_0} \quad \mbox{ if $r\leq a$} \\ \frac{a^2\rho}{2\epsilon_0 r} \quad \mbox{if $a\leq r<b$} \\ 0 \quad \mbox{ . .if $b<r$} \end{cases}$$ but what happen at $r=b$? Should I understand "the charge enclosed by the gaussian surface" as the charge in the interior of the volume having as frontier that surface or it does incude the charge in the surface itself? In the first case it would be $E(b)=\frac{a^2\rho}{2\epsilon_0b}=\frac{-\sigma}{\epsilon_0}$ while in the second one it would be just zero. Which one is the right interpretation?
3 Answers
The best way to understand this problem is to work out the solution in the case where the coaxial cable has a thickness $s$, and a volumetric charge density $\rho$.
If you do this, you'll see the field being nice and continuous, and changing inside of the cable. Then, when you take the $s\rightarrow 0$ limit, it makes sense why you get a discontinuity.
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Where to include a sheet charge of infinitesimal thickness at exactly r=b is not well defined. You can consider it as a delta function ·(r-b).Then the surface integral as a function of r gives you a step function which is not defined at r=b. But you could assign the half height value of the step to the surface integral at this point. Actually this is a rather artificial problem because the real surface charge has a finite thickness and you do not encounter this purely mathematical problem.
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A surface charge induces a discontinuity in the electric field normal to the surface (see also this question). So at $r=b$, there are, so to speak, two field values, depending on which direction you approach the surface from. This is, in the end, unphysical and it comes from an unphysical assumption in our modeling: there are no surfaces that have no volume but contain charges in nature - in nature, the surface will always have a certain width - thus the charges are not all "localised on a boundary of zero thickness".
To specifically answer your question: you should enclose the charges on the surface if your surface for Gauß' law contains them - which depends on whether you are outside of the cable or inside of it in the end.
