# Integral Form of Gauss' Law at Center of Finite Wire

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1. May 14, 2015

### Nate Wellington

At the exact center of a finite wire (i.e. a distance, say $L/2$ from each end), why can I not apply Gauss's Law in integral form to find an EXACT solution for the electric field?

At the center of the wire, $E$ is entirely radial, so it seems like I should be able to draw an infinitesimally $\epsilon$ thick cylinder as my Gaussian surface, pull $E$ out of the integral (it should be a constant in the limit $\epsilon\to 0$), and get an exact expression for the electric field. Obviously, I do not.

Could someone explain in as mathematical a way as possible why this is the case?

BTW, I have similar questions regarding Ampere's Law. It seems like I should be able to apply that to finite wires as well. I know that the reason that doesn't work is that it only works for loops of wire, but again, just from the math, I cannot explain that either.

Thanks!

2. May 14, 2015

### eifphysics

I am not an expert on this, but I will give your my thought. First, I think you need to specify that by finite wire, it means a finite line charge distribution(usually uniform, as the integral is easier to work with), not a current-carrying wire. Because there is hardly any electric field around a current-carrying wire, even though there is magnetic field around it. Then by exact center, do you mean at the central point of the wire? If so, then there is no electric field defined here, because we treat charges as concrete point without structure when we considering the problem. Mathematically, electric field vertical distance z away from the wire is given by E = kλ/z·[b/(z2+b2)1/2+a/(z2+a2)1/2] . You can see that E is not defined when r = 0.

For the Gauss's law case, I am talking about a wire with a uniform charge density lambda, as you say. And I am talking about finding the field a distance $z$ radially away from the midpoint of the wire. This is defined and easily calculable $\int_{-l/2}^{l/2} \lambda z dx/(x^2+z^2)^{3/2}$