Example 4- Electric field of an infinite uniformly charged straight rod
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.
For the side surfaces, the electric field is perpendicular to the surface. Therefore, it is radially out. The area vector, an incremental area vector along the surface will also have its area vector perpendicular to that surface. So the electric field and the incremental surface area vector at that specific point will be in the same direction.
Similarly, if we go to a different region along the side surface, somewhere over here, for example, the electric field is there radially out and the corresponding area vector, surface area vector of an incremental surface in that region is also perpendicular to the surface. Therefore, they are going to be in the same direction. So the angle between them will be 0.
If we look at the top surface, there the electric field is again radially out. So for example, here it’s going to be pointing like this and the area vector now will be perpendicular to the incremental surface at that point and it will be perpendicular to that, pointing therefore in outward direction. This will be the case wherever we go along the top surface.
If we look at the bottom surface, we will have the similar type of situation. So here, for example, again, the electric field is again radially out. At the bottom surface, if we just look at this surface here, its area vector is perpendicular. Therefore, it’s going to be pointing in downward direction.
So, for the top surface and the bottom surface, the angle between electric field vector and the area vector will be 90 degrees, whereas as long as we are on the side surface, the angle between the electric field vector and the surface area vector will be 0 degrees.
Since our cylinder consists of the sum of these three open surfaces, the rectangular side surface, circular top, and circular bottom surface, we can express the whole closed surface integral as integral of E dot dA over the top surface, plus integral of E dot dA over the bottom surface, plus integral of E dot dA over the side surface. In other words, we take the integrals over these open surfaces — top, bottom, and side surface of the cylinder — and then if you add all these now together, then we will end up with the closed surface integral. On the right-hand side of the equation, it will be equal to q-enclosed over ε0.
Now, writing these expressions in explicit form, we will have E magnitude dA magnitude cosine of the angle between these two vectors. For the top surface, that angle is 90 degrees. Plus, for the bottom surface, E magnitude dA magnitude times cosine of the angle between those two vectors, and that too, 90 degrees for the bottom surface. For the side surface, we have E magnitude dA magnitude. Now, the angle between these two vectors is 0 degrees. Wherever we go along the side surface, the angle between E and dA is 0 degrees. And they will all be equal to, the sum of all these terms will be equal to q-enclosed over ε0.
Since cosine of 90 is 0, and as well as in this equation, the integrals over the top and bottom surfaces will not contribute to the flux at all. Whereas the only contribution is going to come from the side surface integration. Here, cosine of 0 is just 1. And as long as we stay along the side surface of this Gaussian cylinder, the magnitude of the electric field will be the same because we’re going to be same distance away, R distance away, from the source. Therefore, E is constant along the side surface. We can take it outside of the integral. What we are left, then, is E times integral of dA over the side surface of the cylinder, which is equal to q-enclosed over ε0.
Well, integral of dA over the side surface means, again, we’re adding all these incremental surfaces along the side surface of this cylinder. If you add all these dA‘s to one another throughout the side surface area of the cylinder, then eventually we will end up with the side surface area of this cylinder. The side surface is nothing but a rectangular surface.
Now, since we define the surface, we can give, therefore, dimensions related to its height or length because the only requirement over here is that its side surface will pass through the point of interest. Let’s say the height of the cylinder is equal to h, that we choose. If that is the case, then since this side is wrapping around the top and bottom surface, therefore the length of this side should be equal to the circumference of these circles, and that is 2π times big R. The height that we chose is just h, so the surface area of the side surface of the cylinder will be 2πR times h. Here then, we will have 2πR times h times electric field is equal to q-enclosed over ε0.
We handled the left-hand side of the Gauss’s law. Now, we’re going to look at the right-hand side, q-enclosed. q-enclosed, by definition, is the net charge inside of the region surrounded by Gaussian surface. If we look at over here, the Gaussian surface is surrounding this, of course, whole region as big as the volume of this cylinder. In that region, only this segment of the whole distribution is staying. In other words, if I just paint that region, it is only this much of length of the charged rod is inside the cylinder. The remaining parts are outside of the cylinder.
Since we’re dealing with an infinitely long charge distribution, we should be given the charge density. In other words, charge per unit length. So the given quantity over here is λ coulombs per meter. By knowing charge per unit length, we can express q-enclosed and that’s going to be equal to charge density, charge per unit length, times the length of the region that we’re interested with, length, this much of length, of the distribution. That length is equal to length of the cylinder that we choose, because only that much of the distribution will be staying inside of this hypothetical cylinder.
Therefore, q-enclosed is equal to λ times h. Now, if we go back to our expression, E times 2πRh is equal to λh over ε0. If you divide both sides by h, we can cancel them. And solving for the electric field we will end up with λ over 2πε0R. Again, if we introduce a unit vector in radial direction, we can express this in vector notation in this form.
If you compare this result with the result we obtained earlier by applying Coulomb’s law for the same example, you will see that we have the same result. You can also see that applying Gauss’s law is much easier and shorter than the Coulomb’s law for this specific case.