# Why we cannot use Gauss's Law to find the Electric Field of a finite-length charged wire?

One of my physics books has a nice example on how to use Gauss's Law to find the electric field of a long (infinite) charged wire. However, at the very end of the example, the author ends by saying Gauss's Law cannot be used to find the electric field of finite-length charged wire. I could not understand why not.

Please this is not a homework problem or nothing like that. This is just a question I have and I have not being able to solve it in my mind yet.

-

Gauss' law is applicable for a finite wire. But, it's useless in this case.

In the infinite example, you assume some things due to symmetry, namely:

It's pretty obvious why these things can be assumed--moving up and down the wire should not change $\vec E$, so we take it constant. Also, there should be no direction bias, so $\vec E$ has no component along the wire.

In this way, $\oint \vec E.\mathrm dS\to\oint|\vec E|\mathrm dS_{curved}$ (since it's perpendicular), and then $\oint|\vec E|\mathrm dS_{curved}\to |\vec E| \oint\mathrm dS_{curved}$ (since it's constant at a given radius)

Once E is outside the integral, we can easily integrate the $\mathrm dS_{curved}$ term.

On the other hand, for a finite wire, we have:

The translational symmetry along the wire is lost (while the symmetry under rotations and the symmetry under reflection through the origin are preserved), and the form of $\vec E$ is not predictable{*}.

So we can't remove the dot product, and we can't take $\vec E$ out of the integral. Since the integral is a closed integral, it is sort of like a definite integral in that we can't just differentiate the equation to get an answer. So we can't solve the Gauss' law integral, so we're stumped. You have to use Coulomb's law and find it by taking elements.

*$\vec E=\frac{k\lambda}{R}\left((sin\alpha+sin\beta)\hat e_r +(cos\alpha-cos\beta)\hat k\right)$. See? Not predictable.

-
Thanks for the answer. But I have a quick question. I understand that if the Electric field is tangent to the Gaussian surface on the ends, then the electric flux will be zero. However, I don't understand why you said that the electric field cannot have any component along the wire? I don't see why not? By the way, my book agreed with you. – leocod Apr 23 '12 at 4:26
@leocod: If you don't want to wait for another answer and are satisfied with this one, you can "accept" it by clicking the green tick next to it :) – Manishearth Apr 23 '12 at 4:28
@leocod: However, consider the situation where you have a pretty dang long wire of length $l$. Now you can approximate the situation as an infinite wire except at distances $\Delta l \not\ll l$ from the ends. – dmckee Apr 23 '12 at 4:37
@leocod: Aah, you edited it. The "no component along the wire" is for infinite wires. See, the wire should not have an "up" or "down". An ant walking up the infinite wire should feel the same as an ant walking down it. But, if the electric field had a $\hat k$ component, then the ant walking up could say "I am walking along $\vec E$", and the ant walking down would say "I am walking against $\vec E$". This goes against the symmetry of the situation. – Manishearth Apr 23 '12 at 4:40