Example 4- Electric field of a charged infinitely long rod
Now, we’re going to calculate the electric field of an infinitely long, straight rod, some certain distance away from the rod, a field of an infinite, straight rod with charge density, λ coulombs per meter.
In this case, we have a very long, straight, uniformly charged rod. Let’s assume that the charge is positive and the rod is going plus infinity at this end and minus infinity on the other end, and we’re interested with the electric field that it generates big R distance away from the rod. Since this is an infinite rod, we can place our coordinate system over here as y and choose its origin at this location.
Like in the previous examples, we’re going to choose an incremental segment along the rod, very, very small, and call the amount of charge associated with this segment as incremental charge of dq. Relative to our coordinate system, it will be y distance away from the origin and its length is going to be incremental distance of dy. Just a small increment to its y position.
This element is so small so that we will treat the amount of charge associated with this incremental element like a point charge. Therefore, a positive point charge sitting over here with a magnitude of dq is going to generate an electric field at the point of interest such that it will be pointing radially outward direction. Let’s call that electric field as incremental electric field of dE. Of course, if the charge distribution were negative, then we would have ended up with an electric field pointing radially inward just in opposite direction to this one.
Well, since the distribution is symmetric along this axis, which is basically, in a way like a bisector of this distribution of this rod, we can always find another dq below the origin, a symmetric one. This one is also the same distance away from the origin. It will have the same thickness of dy and that too will generate its own electric field at the point of interest in radially outward direction. Since both of these two charges have the same magnitude and they are the same distance away, little r, to the point of interest, and let’s denote that point of interest with P, the magnitude of the electric fields that they will generate will be equal to one another.
Now, if we introduce a coordinate system such that our point of interest is located at the origin, let’s say x and y-coordinate system, we can easily see that for every dq above the origin, we will have a symmetric dq below the origin. And when we add these electric field vectors vectorially, first we will resolve them into their components relative to the coordinate system that we introduce. Simply by considering these two by taking their projections along x, we will have their x components. We can easily see that the x components will align in the same direction along x axis, but the y components will align in opposite directions along y.
Since these electric field vectors, dE‘s are equal to one another, their components will also have the same magnitude. Therefore along y-axis, we will end up with vectors with equal magnitudes and opposite directions. When we add them vectorially, they will cancel.
So from our vector diagram, we can conclude that the total electric field, which is the vector sum of all these incremental electric field vectors generated by the incremental charges along the distribution, will add along x-axis. The resultant electric field will be in positive x direction because the y components will cancel due to the symmetry. First, let’s make a note of that by saying that dEy‘s, the y components, will cancel due to the symmetry. So the resulting electric field is going to be the vector sum of all the x components and the summation over here is nothing but integration.
So as a first step here, we need to express the x component in explicit form with respect to this coordinate system. To do that, we will take the advantage of the right triangles which are forming when we resolve the electric field vectors into their components. To do that, we need to define an angle and of course we can either use this triangle over here or the other triangle over here. In other words, we can either define this angle, the angle that dE is making with the x-axis, or we can define the angle that it makes with the y-axis. Both of them will eventually give us the same answer. It’s our choice.
If we use this angle over here, then in this triangle, the x component is the adjacent side in this electric field triangle relative to the angle θ, and the trigonometric function associated with the adjacent side is cosine. Therefore, dEx is going to be equal to dE times cosine of θ.
Here, by using the plane geometry, if this angle is θ, this angle will also be θ and obviously also this angle. Using now these triangles, we can express the cosine of θ. These triangles are forming from the distances. We can express cosine of θ as the ratio of adjacent side, which is big R to the hypotenuse, and that is little r.
By applying Pythagorean theorem to any one of these triangles, little r2 is going to be equal to y2 plus big R2. In doing so, we express the distance r in terms of the big R, which is the distance we are given, and in terms of the variable y and that is associated with the position of the incremental charge that we choose along this distribution. Furthermore, after expressing cosine of θ, we can calculate the electric field magnitude generated by any one of these dq‘s by using Coulomb’s law and that is Coulomb constant, 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. The distance between dq and the point of interest is r, so we’ll have square of r.
Okay. One more thing that we need to express is this dq incremental charge. Now, when we look at the statement of the problem, we see that, for such a charge distribution, we are given the charge density λ coulombs per meter. When we are dealing with infinite distributions, depending upon the type of the distribution, we have to be given the charge density because the total charge, the net charge, along an infinite distribution will be infinite.
Since λ is already given, then dq can be expressed as linear charge density, which is λ, times the length of the charge, or length of the region that we’re interested with, which is dy. Charge per unit length times the length that we’re interested with, will give us the amount of charge along that length.
Now we can express our integral in explicit form that the electric field is equal to integral of dEx and that is integral of dE times cosine of θ and dE is dq, which is λ dy, divided by 4 π ε0 r 2. If we just write down the explicit value of r2, that will be y2 plus big R2 times cosine of θ and cosine of θ is big R divided by little r. The little r will be square root of y2 plus R2.
So this term over here is little r. This is dq and this whole quantity over here is dE. The whole quantity over here is cosine of θ. Of course, dE times cosine of θ is the x component of the electric field. By taking the integral, we are adding all of these x components to be able to eventually get the total electric field generated by this distribution.
Now, the next thing that we will look at, the boundaries. Our variable is y and it is associated with the position of this incremental charge in this coordinate system relative to this origin. Depending upon the incremental charge that we consider, then that position will change from minus infinity to plus infinity. Therefore we can write down over here, by saying that y is going to go from minus infinity to plus infinity.
When we look at the integrand, we see that λ is constant, big R is constant, and 4 π ε0 is always constant. Therefore we can take these quantities outside of the integral since they are constant. Of course we cannot take this R2 or that one outside because they are in parentheses with the variable. We will leave them inside of the integral. Then, the integral will take this final form. λ R divided by 4 π ε0 and inside of the integral we will have dy over, if we take the product of these two quantities over here, we will have y2 plus R2 to the power 3 over half.
Now if we go back to our diagram one more time, we see that the electric field generated by the incremental charges, in the upper half of this infinite rod, their x components are adding to the x components of the incremental charges located below the origin along this infinite, straight charged rod distribution. So we can calculate our integral from 0 to infinity and then multiply it by 2 because the contribution from the lower half will be equal to the contribution from the upper half of the dq‘s associated with this charge distribution. So for our integral, we can say that y is going to go from 0 to infinity and then we will multiply the whole thing by 2.
All right. In doing so, we can cancel 2 and 4 in the denominator, so we will end up with 2 π ε0. Then our integral becomes λ R divided by 2 π ε0 times integral of dy over y2 plus R2 to the power 3 over 2. The boundaries of the integral will go from 0 to infinity.
When we look at the integrand, we will realize that we are not going to be able to make the usual u transformation to this integral because if we call the quantities inside the parentheses as u, y is the variable, R is the constant, and if we take the derivative of this, we will have 2 y dy for the du term. Since we don’t have any y term in the numerator, then we cannot apply this transformation to simplify the integral to an integratable form.
For these types of integrals, we’re going to apply a unique transformation and we will say that, let y is equal to R times tangent Φ. In other words, we will define a new variable of Φ such that it is related to our original variable y through this expression. Of course we can ask then how I knew to make this transformation and the only answer to that is just experience. When you do these types of integrals several times, then you will remember also. I am going to go through this integral in detail because we will indeed face this type of integrals throughout the semester and for the different cases. Therefore it is important to learn how to take this type of integrals.
If y is equal to R tangent Φ, then by taking the derivative of both sides, dy is going to be equal to, of course the R‘s derivative will give us 0, and then plus the derivative of second term times the first one will give us R times derivative of tangent is secant squared, and in explicit form it is going to give us 1 over cosine squared Φ dΦ. Therefore under this new transformation, dy is going to be equal to this quantity.
Using this new variable in the denominator, and that is y2 plus R2 to the power 3 over 2, for y2 we will have R2 tangent squared Φ plus R2 to the power 3 over 2. Since R2 is common in these two terms, we can first take R2 common quantity parentheses, and that’s going to give us R2 times tangent squared Φ plus 1 to the power 3 over 2.
Now we can take R2 outside of the 3 over 2 power bracket and as you remember when we do that, we simply multiply by the superscripts, therefore R2 is going to come out as R3 and inside of the bracket, let’s write down tangent squared using the trigonometric identity that tangent is equal to sine over cosine. Therefore the tangent squared will be sine squared over cosine squared Φ plus 1 to the power 3 over 2.
By having common denominator inside of the bracket, we will have sine squared Φ plus cosine squared Φ divided by cosine squared Φ to the power 3 over 2. Again as you recall, sine squared plus cosine squared is just 1, so we will have 1 over cosine squared Φ in power bracket of 3 over 2. If we take that outside of the power bracket, then we will end up with 1 over cosine cubed, because we just multiply the superscripts, or the powers together. 2 times 3 over 2 will give us simply 3. Therefore the result will be R3 over cosine cubed Φ.
So for dy, we will replace it in terms of this new variable with this term and the denominator will take this form. If we go back and look at our integral over here, and let me express that one more time here, and that is electric field is equal to λ R over 2 π ε0 integrated from 0 to infinity dy over R2 plus y2 to the power 3 over 2. That’s correct.
In this new transformation, we replace dy with R over cosine squared Φ dΦ. λ R over 2 π ε0 and instead of dy we will have R over cosine squared Φ dΦ divided by, for the numerator we ended up with R3 over cosine cubed Φ in terms of this new variable. Of course, again, changing the variable will change the boundaries of the integral also. Integral will go therefore, from Φ1 to Φ2. Again, we do not need to calculate the boundaries because after taking the integral, we will go back to the original variable of y.
Here, this R and that R will make R2, which will cancel with the R3 in the denominator, therefore we are left only with one R in the denominator and cosine squared Φ will cancel with the cosine cubed. We will have, therefore, just cosine Φ over here. Under this new transformation, electric field integral will induce into this simplified form, which will be equal to λ over 2 π ε0 times the integral, 1 over cosine Φ will go to the numerator as cosine Φ and since this R is constant, we can take it outside of the integral. Then we will be left only with cosine Φ dΦ inside of the integral, which will be integrated from Φ1 to Φ2.
As you can see, with this transformation, we simplify that relatively complex integral into a simple form which we can easily take it. The integral of cosine is just sine Φ. Therefore we will have λ over 2 π ε0 R times sine Φ evaluated at 0 and infinity.
Now, we are going to go back. Actually it’s not evaluated at 0 and infinity, but at Φ1 and Φ2. Now we will go back to our original variable, because we didn’t calculate Φ1 and Φ2. In order to go from here to the original variable, we will just take advantage of the definition of the new variable Φ. As you recall, it was related to the original variable y, as y was equal to R tangent Φ. If we leave tangent Φ alone on one side of the equation, then we will have that tangent Φ is equal to y over R.
Here, we are going to draw a right triangle, which will satisfy this relationship. If this angle is Φ, since tangent Φ is equal to y over R, therefore the opposite side relative to this angle should be y and the adjacent side should be equal to big R and the hypotenuse of this triangle, applying Pythagorean theorem, will be equal to square root of y2 plus R2. Since we are interested with sine Φ, then we can easily use this triangle to express sine Φ and that will be equal to a ratio of opposite side, and that is y, divided by the hypotenuse, which is square root of y2 plus R2. Therefore this triangle helps us to express the sine Φ in terms of our original variable y.
Then the electric field becomes λ over 2 π ε0 R and for sine Φ we will have y over square root of y2 plus R2. We know the associated boundaries for this variable and they are 0 and infinity. If we substitute infinity for y, we will have infinity here and infinity in the denominator. Infinity over infinity is undefined, therefore we cannot just go ahead and substitute the boundaries right away. We have to make another small manipulation to our expression.
We will do that by taking the y2 inside of the square root outside of the square root. That will give us λ over 2 π ε0 R y over, taking y2 outside of the square root will come out as y, and inside we will have 1 plus R2 over y2 for the second term because R2 doesn’t have any y2 multiplier.
These y‘s will cancel and we will evaluate now this expression at 0 and infinity. Electric field will then be equal to λ over 2 π ε0 R, open parentheses, first we will substitute infinity for y. A number divided by infinity will go to 0, so we will end up with only 1 inside of the square root, which will come out as 1. We have 1 in the numerator. 1 over 1 will give us just 1. Minus, now we will substitute 0 for y and if we substitute 0 over here, a number divided by 0 will go to infinity, so the whole square root will be infinity. 1 over infinity will go to 0, therefore the second boundary will give us just 0.
Then the answer for the electric field generated by this infinitely long, straight, uniformly charged rod, with charge density of λ coulombs per meter, will be equal to λ over 2 π ε0 R. Since this electric field was in positive x direction, in order to represent this in vector notation, we multiply this by the unit vector pointing in positive x direction.