# Ampere's Law and Gauss's Law for EXACT CENTER of Finite Wire: Mathematical Justification [closed]

I have always seen it explained that:

Ampere's Law (in integral form) works whenever B is constant around a path, so that you can pull it out of the integral.

Similarly, if you can draw a surface over which E is constant, you can pull E out of the integral.

In the Gauss's Law case, it seems like I should (for a finite wire) be able to draw a surface of infinitesimal thickness and get the exact result. (The contributions through the sides of the cylinder cancel by symmetry).

In the Ampere's Law case, I know that it has to do with the current distribution not being a loop, but I don't know the mathematics behind it.

Could someone explain, in as mathematically rigorous way as possible why these things do not work?

Thanks!

• 1. What are "these things" you refer to in your question at the end, exactly? 2. Ampere's and Gauss' law always hold. They're just rather useless to compute the unknown fields from a known charge/current distribution if you don't have enough symmetry to get the field out of the integral. – ACuriousMind May 14 '15 at 0:33
• The way I understand the question you are asking if the fields are only an integration (necessarily numeric in the general case) away if you know the charge and current distribution? I always believed that to be the case, with the difficulty that the (numerical) integration is a non-trivial matter in itself. Where things get tricky is that we don't actually know the charge/current distribution for most realistic problems, and we need a full field solver in which both fields and charge/current distributions are variable. – CuriousOne May 14 '15 at 1:37
• I am asking why we cannot make our Gaussian surface an infinitely short cylinder, in the center, since E is entirely radial there (so there would be no flux through the sides of the cylinder there). Well, obviously we can make that a Gaussian surface, but I do not se how doing so would not get us the exact result.... (I can clarify again if this is still unclear) – Nate Wellington May 14 '15 at 13:21

Gauss's Law states $$\int\limits_{\partial V}\vec{E}\cdot\textrm{d}\vec{S}=\frac{1}{\epsilon_0}\int\limits_V\rho\textrm{d}V$$ Then by if the magnitude of the electric field is a constant along the surface $\partial V$, and theta is the angle between the unit normal to the surface and the electric field, you may calculate $$\int\limits_{\partial V}\|\vec{E}\|\cos\theta\textrm{d}S=\|\vec{E}\|\int\limits_{\partial V}\cos\theta\textrm{d}S=\frac{1}{\epsilon_0}\int\limits_V\rho\textrm{d}V\iff\|\vec{E}\|=\frac{\frac{1}{\epsilon_0}\int\limits_V\rho\textrm{d}V}{\int\limits_{\partial V}\cos\theta\textrm{d}S}$$ Non the less, for an finite wire you cannot achieve a reasonable surface for which the electromagnetic field is constant due to the wire having an edge which breaks the symetry. That's why you have to use gauss's law for infinitesimal particles (Which yields Coulomb's Law) and the principle of superposition (integrate over the wire).