# When we apply Gauss's law should we consider also the charge over the gaussian surface?

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?

• By superficial charge you mean there is a surface charge? – ClassicStyle Sep 29 '16 at 14:57
• @ClassicStyle sorry, yes, I guess the english way to say it would be surface charge density – la flaca Sep 29 '16 at 15:08

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.

• So, if there is charge distributed exactly over the gaussian surface (I'm taking a gaussian cylinder that coincides with the external cylinder shield of the cable, so I'm not outside nor inside, but exactly in the boundary) I should consider that charge as enclosed by the gaussian surface? – la flaca Sep 29 '16 at 15:19
• The choice is really yours in the end. As the field is discontinuous, there is no well defined field value on the boundary - but you can consider the charge as enclosed by the Gaussian surface for the electric field immediately outward on the surface, which will usually be what interests most people more. – Sanya Sep 29 '16 at 15:24

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.

Where to include a sheet charge 𝜎 of infinitesimal thickness at exactly r=b is not well defined. You can consider it as a delta function 𝜎·𝛿(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.

• Ok, so the problem is that I can't use Gauss's law fo find the electric field at $r=b$, because the function is not defined at that point? Should I think it as diverging at every point over the external shield cylinder? – la flaca Sep 29 '16 at 15:25
• @Eliana - No the field is not diverging over the external shield cylinder it is zero for r>b. To have a field value for r=b you could assume that at r=b half the surface charge σ is outside and half is inside the Gauss surface, so that you could assign half the value of the field (corresponding to the total interior charge) to r=b: E(b)=-(1/2)·σ/ϵ0. – freecharly Sep 29 '16 at 17:12
• @Eliana - In reality your surface charge σ has a small finite thickness and when you move your Gauss surface over the radius r=b you would get at perfectly smooth continuous function for E(r) there resembling the step function you get when σ is assumed to be infinitely thin. – freecharly Sep 29 '16 at 17:23