# How to use Gauss's law on the surface of a conductor?

I have seen that the electric field on the surface of a conductor of any shape is $$E=\frac{\sigma}{\varepsilon_0}\; .$$ I have already got that the field near the surface of a conductor but my concern is for on the surface of a conductor. While using Gauss's law here, I'm unable to draw a Gaussian surface as a Gaussian surface can't pass through a charge but a line of charge.

How can a Gaussian surface pass through a line of charge whereas it's almost the same case as a charge placed on a Gaussian surface?

Your question is basically asking, "What is the electric field at the location of a surface charge?" The answer is that there is no mathematically well-defined answer. If we imagine a infinite flat surface charge $$\sigma$$, then Gauss's Law tells us that the difference between the normal components of the electric field change by $$\sigma/\epsilon_0$$ between any point above the surface and any point below the surface. This means that if $$z$$ is the coordinate normal to the surface, then we have $$\lim_{z \to 0+} E_z - \lim_{z \to 0-} E_z = \frac{\sigma}{\epsilon_0} \neq 0,$$ and so the limit of $$E_z$$ as we approach $$z = 0$$ does not exist.

How can a Gaussian surface pass through a line of charge whereas it's almost the same case as a charge placed on a Gaussian surface?

The best way to think about any electrostatics problem with a "infinite charge density" (line charge, point charge, or surface charge) is as the limit of a well-behaved volume charge density $$\rho$$. For example, a surface charge $$\sigma$$ can be viewed as the limit of a "slab of charge" with thickness $$a$$ and density $$\rho = \sigma/a$$, as $$a \to 0$$ and $$\sigma$$ stays constant.

If we try to use Gauss's Law to solve the problem of a "slab" of charge, everything is perfectly well-behaved — the problem above doesn't arise. The electric field is well-defined everywhere, including inside the slab; and we can apply Gauss's Law to our heart's content without worrying about what the contribution to the flux is from a point where the electric field is ill-defined and/or infinite. It's only when you insist on setting $$a = 0$$, rather than just saying "$$a$$ is very small", that you run into problems of this type.

For a line of charge, you would chose the Gaussian surface as an infinite cylinder. The cylinder encloses the charge, but does not cut through it.

Then you have $$\int \int \vec E.d\vec S = Q/\epsilon_0$$ On the cylindrical surface enclosing the linear conductor, the surface element is $$2\pi r L$$ where L is the length of the wire. Writing Gauss law and using symmetry argument to deduce that the electric field $$\vec E$$ is radial and on a surface of fixed radius, is therefore constant so that $$\int \int \vec E.d\vec S$$ is simply $$E S$$ we obtain $$2\pi r E L = Q/\epsilon_0$$ Introducing the linear charge density $$\rho = Q/L$$ one obtains $$E = \frac{\rho}{2 \pi \epsilon_0 r}$$

• my concern is shall we take the charges of the distribution to be inside the gaussian surface or outside of it so that we can apply E.dA = q/ϵ0. – sachinrath123 Jul 6 '17 at 1:27
• Always inside. Remember, you chose the gaussian surface so that it encloses the charge distribution. – Thierry Kauffmann Jul 6 '17 at 5:42
• then how the book says a gaussian surface can pass through a charge distribution but cant pass through a point charge,passing does not mean over something. – sachinrath123 Jul 7 '17 at 2:15
• A Gaussian surface is a mathematical construction. It doesn't exist. So it can go through a charge distribution, but it cannot be finer than a point. It cannot cut through a point. Again, the Gaussian surface is just a trick to compute the electric field. You create that surface to fit your needs. It's not like a physical surface :) – Thierry Kauffmann Jul 10 '17 at 21:39