Question

find field due to infinitely long charged wire without gauss law

Answer 1

As another example of the applications of Gauss’s law, let’s consider now the electric field of an infinitely long, straight wire. Infinitely long, uniformly charged, straight rod with charge density λ per coulomb. Earlier, we did the same example by applying Coulomb’s law, and if you recall that example, we had to take a pretty complicated integral which required some trigonometry substitutions, and now we will do the same example by applying Gauss’s law.

We have a very long, straight, charged rod. Let’s assume, again, that it is charged positively and the charge is distributed uniformly along the length of this rod. It goes plus infinity at this end and minus infinity at the other end. We’re interested, again, some big R distance away from the distribution at this point, P, the field that it generates.

Now, if you recall the example that we solved by applying Coulomb’s law, we chose an incremental charge element at an arbitrary location and treated that charge element, dq, like a point charge, so as if a positive charge sitting over here which generated an electric field in a radially outward direction and an incremental electric field of dE. And we took the advantage of the symmetry of this rod relative to this point which we denoted as our origin. And the charge, the symmetric charge below this origin also generates its own electric field of dE at the point of interest in this direction.

We said that there will be always symmetric charges. For each incremental charge that we choose above the origin, we will find the symmetric one below the origin. When we add all these electric fields vectorially, we said that the horizontal components will add because they will lie along the same axis. But the vertical ones relative to this x–y coordinate system would cancel because they would be in opposite directions with equal magnitudes.

So we realized that all the net electric field will be the sum of the horizontal components, or the x components of these dE‘s generated by these incremental charges. Therefore, as long as we’re R distance away, we can say that the electric field is going to be pointing in positive x direction along this line. Well, we can generalize this for all the points surrounding this rod which are capital R distance away from the rod. In that case, we will see that the electric field generated by this rod is going to be in a radially outward direction.

So, as our Gaussian surface, let’s choose a cylinder such that its side surface passes through the point of interest right over here. In doing so, we’re going to end up with this hypothetical cylinder such that the rod is located along the axis of the cylinder. The surface of a cylinder represents a closed surface because it encloses a certain volume. And our charged rod is, of course, extending to infinity in both directions.

Well, Gauss’s law is integral of E dot dA over a closed surface s is equal to q-enclosed over ε0. The integral is taken over the whole surface of the cylinder. If we take our cylinder and cut it open, we will see that it will consist of a rectangular side surface and two circular surfaces for the top surface and the bottom surface. In other words, this rectangular surface wraps around these circular surfaces and forms the whole cylindrical surface.