Example 1- Electric field of a charged rod along its Axis
As a first example for the application of Coulomb’s law to the charge distributions, let’s consider a finite length uniformly charged rod.
Let’s try to calculate the electric field of this uniformly charged rod. Let’s say, with length, L, and charge, Q, along it’s axis. Let’s say, a, distance from one end of the rod. We have a finite length rod with a length of, L, and this is positively charged and the charged is distributed uniformly along its length. We’re interested with the electric field that it generates a distance away from one end of the rod at this point, P, our point of interest.
When we look at the form of distribution, we see that it is linear charged distribution, charge is distributed along the length of the rod and to be able to calculate the electric field of charged distribution, first we choose an incremental charge amount within the distribution at an arbitrary location, and treat the amount of the charge associated with that segment as incremental charge and treat it like a point charge. Therefore the problem induces into a form that we have a point charge sitting over here, with a magnitude of dq. It is a positive charge and it will generate a electric field at the point of interest and radially outward direction. Therefore the electric field generated by this dq, at this location , will be pointing to the right and will have magnitude of incremental field of dE.
Let’s introduce a coordinate system to our problem. Since we’re dealing with a one dimensional case, we’re going to introduce an x-coordinate system. Let us assume the other end of rod is located at the origin. Relative to this coordinate system, then, our incremental charge will be located x distance away from the origin, and the thickness of this incremental charge will be dx, just a little increment from that position.
And of course this dx is very, very small, so that we can treat the amount of charge along the length of dx, dq, like a point charge. Now, once we determine the electric field generated by this incremental charge at this location, then we can go ahead and calculate the electric field due to the next incremental charge at the same location. Then we do the same thing for the next incremental charge and so on and so forth throughout the length of the whole distribution.
So we will have their associated electric field vectors pointing in positive x direction, they’re all positive, and therefore total electric field will be the vector sum of all these dE‘s, which are generated by all these dq‘s, which eventually makes the whole charge distribution. Of course, we’re not going to add the electric field associated with these billions of billions of incremental charges, which eventually makes the whole distribution.
The addition process is going to be done, over here, through integration. Therefore, the total electric field is going to be equal to sum all these dE’s, in other words, the electric field generated by the incremental charges, which eventually make the whole distribution.
Now, we know to get the magnitude of this incremental electric field, or the incremental electric field generated by its source, which is dq, from Coulomb’s law, that is equal to 1 over 4 π ε0 times the magnitude of the charge divided by the square of the distance between the charge and the point of interest. In other words, we’re talking about this distance, and that’s what we call, r, the distance between the charge, the source and the point of interest, r2.
We can express this distance in terms of given quantities. We know that the whole distance from the origin, up to the point of interest is L plus a, therefore this distance is L plus a. Since the distance from the end of the rod to the location of the incremental charge is x, then r is going to be equal to total distance, L plus a minus x. L plus a minus x will be equal to r. In doing so we will express distance, r, in terms of the given quantities of length of the rod and distance a. Of course x is the variable because that is going to change depending upon the location of this incremental charge, dq.
So, we can say that r is equal to L plus a minus x. One other thing that we should take care and that is expressing dq in terms of the total charge of the distribution. The total charge of the distribution is Q. And since the distribution is at linear charge distribution, then dq is going to be equal to linear charge density λ, times the length of the region that we are interested with. In other words, the amount of charge along the length of the dx.
Well, since λ is equal to total charge of the distribution divided by the total length of the distribution, that will be Q over L. Therefore, dq is going to be equal to in, explicit form, Q over L times dx.
Now we can express our dE in explicit form as 1 over 4 π ε0 times dq and that is Q over L dx divided by r 2, which is going to be L plus a minus x quantity squared. This will be the magnitude of the electric field generated by this incremental charge, dq. The next one will generate a similar type of electric field, which will be represented with the corresponding x value for that incremental charge. To be able to add all these incremental electric fields, we will take the integral and that will in turn give us total electric field generated by the whole distribution.
When we look at our integrand, we see that one over 4 π ε0, Q and L are all constant. We can take it outside of the integral. Our variable is x, and if you start from the first incremental charge just at the origin, therefore x is going to start from 0, and if you add all these incremental charges along the length of the distribution, then it’ll vary up to the length of the distribution, so the boundaries will go from x is equal to 0, to x is equal to L.
So if you move one more step then, we will have Q over L will go to the denominator, 4 π ε0 L. Inside of the integral we’re going to have dx over L plus a minus x quantity squared. Integrated from 0 to L. To be able to take this integral, we’re going to make a simple change of variable transformation, and we will see that let L plus a minus x is equal to u. Then, if you take the derivative of both sides, since L is constant, that is going to give us 0, and a is constant again, the derivative of that will give us 0. Derivative of minus x will give us minus dx, and that will be equal to du.
By looking at our integrand over here, we have dx at the numerator and if multiply both numerator and denominator by minus 1, the ratio will not change. But in doing so, we’re going to end with du in the numerator, and we will end up with u2 at the denominator. Of course, once we change the variable, the boundaries will also change, but we’re not going to calculate the new boundaries because we will go back to the origin of variable of x. So the electric field will be equal to –Q over 4 π ε0 L integral of du over u2 integrated from u1 to u2.
Moving on, –Q over 4 π ε0 L, integral of du over u2 is going to give us -1 over u, which will be evaluated at u1 and u2. This minus and that minus will make plus. Therefore, the electric field will be equal to Q over 4 π ε0 L times 1 over — now let’s go back to the original variable — u was equal to L plus a, minus x and we will evaluate this at 0 and L.
Electric field will be equal to then, Q over 4 π ε0 L, open parenthesis, first we’re going to substitute L for the x and if you do that, plus L and minus L will cancel from the first boundary, therefore we will end up with 1 over a and minus. Now we will substitute 0 for x and if you do that we will have 1 over L plus a in the denominator.
If we have common denominator for the quantities in the parenthesis, electric field will be equal to Q over 4 π ε0 L (L plus a minus a over a times L plus a). a‘s will cancel in the numerator. We’re going to end up with L and we can cancel that out eventually, with the one in the denominator, leaving us, electric field is equal to Q over 4 π ε0 a times L plus a.
Since, when we add all the incremental electric field vectors to one another, and since they all point in positive x direction, we can express our results in vector form, multiplying the magnitude of the vector by the unit vector pointing in positive x direction. That is defined as î, and that is from the fact if your equal in rectangular coordinate system, in general, in x, y z coordinate system, unit vector along x direction is called î, along y direction is called, ĵ and along z direction is called k̂.
So, our result tells us that the total electric field generated by this rod charge distribution, a distance away from it’s end, is equal to, this quantity, in other words, with this magnitude and it is pointing in positive x direction.
Here, we can look at a special case and that is the case of a much, much greater than L. In other words if our point of interest is far, far way along the axis, such that the distance a is much greater than the length of the rod, then L over a is going to be much, much smaller than 1.
So we can take the advantage of this ratio to obtain an approximate equation. If you consider the magnitude, E, in this form of course, we don’t have the ratio of L over a, to compare with 1, but we can express our equation by taking the L inside of the parenthesis, outside of the bracket, actually, sorry, the a outside of the bracket. In doing so we have a is going to come out. We have another a over here. a times a will give us a2. And inside of the bracket, since L doesn’t have any a multiplier, we will have L over a, and since we moved a outside of the bracket, we will end up with 1 over here, in a parenthesis.
Now, we can apply this approximation over her in comparing to 1, since L over a is much, much smaller than 1. 1 plus a very small number or 1 minus a very small number, then 1 will not really affect the overall result, so neglecting this in comparing to 1, we’re going to end up with E is equal to q over 4 π ε0 a 2 for this special case.
When we look at this result, we see that’s its a familiar result. As a matter of fact, this is identical to the case that if we had a positive charge of Q, point charge, sitting over here, and if we try to figure out the electric field it generates a distance away, and that from Coulomb’s law, the magnitude of that electric field will be simply Q over 4 π ε0 a2. They’re identical expressions. That enables us to conclude that if we go far away from the rod along its axis, we will perceive that charge that distribution like a point charge. So for that case the distribution will behave like a point charge. Then we can make a small note over here that distribution behaves like a point charge for distance, a, much much greater than the length of the rod.