Example 2- Electric Field of a charged ring along its axis
As another example of the applications of Coulomb’s law for the charge distributions, let’s consider a uniformly charged ring charge. Electric field of a uniformly charged ring with radius R along its axis z distance from its center.
We have a ring which is uniformly charged. In other words, the charge is distributed uniformly along the circumference of the ring. It has radius R, and we are interested with the electric field that it generates at a certain point on its axis which is z distance away from the center of the ring.
Let’s assume that the charge is positive and it has a value of Q coulombs. As in the case of all distribution problems, we choose an incremental charge element at an arbitrary location along the distribution. In this case, this is going to be a very small length or segment of the ring, and let’s call that, this arc length, as dS, and the amount of charge with this length is what we call incremental charge of dq.
Now the problem [inaudible 00:02:45] that, we will treat this dq like a point charge, so as if a point charge, a positive point charge sitting over here. We look at the electric field that it generates at the point of interest, and that is going to be pointing in radially outward direction with an incremental electric field of dE, since the charge over hear will be a positive charge.
By looking at the shape of the distribution, we can easily see that the distribution is symmetric along its axis. Therefore, we can always find another dq right across from this charge located at this point. That too will generate it own electric field, which is going to be also pointing in radially outward direction from that charge, something like this.
Both of these two charges will have the same magnitude of charge, and they are same distance away from the point of interest. Therefore the magnitude of the electric fields that they generate at this location will be the same. As a matter of fact, for every dq that we will choose along this ring charge, we’re going to have a symmetrical one across from it, and if you trace the electric fields that they generate at the location of the point of interest, we will see that they will be distributed along the surface of that cone, something like this. So, they’re going to be along the side surface of this cone, and for every every dq we will have a symmetrical one across from that.
If we introduce a proper coordinate system to be able to get the total electric field or the net electric field generated by all these dq‘s such that the point of interest is located at the origin of the coordinate system, by taking the projection from the tips of the electric field vectors, we can get their horizontal and vertical components with respect to this coordinate system. As you recall, the vector addition rule says that we can add or subtract the vectors directly if they lie along the same axis.
From the diagram we can easily see that the horizontal components of the electric field vectors will be aligning in opposite directions. Since dE‘s will have the same magnitude, their components also will have the same magnitude, and the horizontal components which have the same magnitudes and lying in opposite directions in this coordinate system, when we add them, they will cancel. Therefore only the vertical components of the electric field vectors will survive, so when add them vectorially, resultant vector is going to be pointing in outward direction along the vertical axis.
As you can see, without doing any calculation, simply by drawing a proper vector diagram we can conclude that the resultant vector is going to be in outward direction along the vertical axis. Not only in this problem but in all the problems that they involve vectorial quantities, our first step should always be drawing a proper vector diagram. Once that is established then we can introduce a proper coordinate system and take the advantage of the symmetry, if there is a symmetry in the problem, therefore simplify it and then just proceed to be able to calculate whatever we are trying to achieve in the problem.
So, as a first note then, we can say dEhorizontals cancel due to the symmetry. Therefore the total electric field will be sum of all the vertical components, and the summation over here is integration, integral of dEverticals will give us the total electric field. Therefore, we need to express the vertical component of the electric field, and to be able to do that we’re going to use the right triangles which are forming once we resolve the electric field vector into its components with respect to this coordinate system.
So we can use either this triangle over here or this triangle, and to be able to express the vertical component, so we need to define an angle. We can define either this angle or that angle. It’s our choice. Let’s just go ahead and try this angle and denote it as θ. It means that we’re going to be using this triangle over here, and in that triangle, the vertical side is the adjacent side with respect to angle θ. Therefore, since the trigonometric function associated with the adjacent side is cosine, hypotenuse dE times the cosine of this angle, cosine of θ, will give us the vertical component.
When we look at the expression inside of the integral, we will see that we can calculate dE from Coulomb’s law, and since this is something that we defined, θ, we have to express cosine of θ in terms of the given quantities. Coulomb’s law says that the magnitude of the electric field generated by the point charge of dq, this incremental charge that we’re treating like a point charge, is equal to Coulomb constant 1 over 4π ε0 times the magnitude of the charge divided by the square of the distance between the charge and the point of interest, and that is this little r.
Now, if we consider this big triangle over here, which is a triangle forming from the distances, we see that if this angle is θ, this angle will also be θ. And in this big triangle, and that is also a right triangle, little r is hypotenuse, and applying Pythagorean theorem, little r2 will be equal to big R2 plus z2.
And in the same triangle we can express cosine of θ, which is a ratio of adjacent side, and that is z, to hypotenuse, and that is little r. We can express the little r in terms of big R and z. Those are the given quantities. We will end up with z over square root of R2 plus z2. Since r2 is equal to R2 plus z2, then r will be the square root of R2 plus z2.
Now moving on, electric field is going to be equal to integral of dE, and that is dq over 4π ε0 little r2, and little r2 is big R2 plus z2 and times cosine of θ, which is z over square root of R2 plus z2.
Now, one more thing that we need to take care of in the integrand, and that is dq. We need to express dq in terms of the total charge of the distribution because we don’t know what dq is. dq is the amount of charge along the arc length of dS. Since our distribution is a line charge distribution, in other words, in this case the charge is distributed along the circumference of this ring, if we define linear charge density, which is total charge of the distribution divided by the total length of the distribution, and multiply that quantity by the length that we’re interested with, which is dS, we will get the amount of charge along that specific length.
Therefore, dq will be equal to λ times dS, and here, λ is equal to total charge of the distribution, which is q, divided by the total length of the distribution, and that is circumference of this ring charge. Therefore it is 2π times the radius of the distribution.
Here, let’s try to express this arc length also in explicit form, and to do that let’s look at the angle that this arc length subtends. As we can see in this exaggerated picture, this arc length of ds with radius r will subtend an angle of, angle of dΦ, and using the definition of radian, we can express ds is equal to radius times the angle that it subtends.
If I redraw that picture over here in an exaggerated way, we have the arc length, and it is subtending a certain angle, dΦ, as a radius of r, and the length of ds, and that length is equal to R dΦ. As a matter of fact, if you recall the definition of radian, if we leave the angle alone, that’ll be equal to arc length divided by the radius. If ds is equal to R, in other words if the arc length becomes equal to the radius of the arc, then it is called that dΦ is equal to 1 radian, so that is the definition of radian.
Now if we go back to our incremental charge dq, we can express that charge in explicit form as the linear charge density Q over 2π R times ds, that is R dΦ. You see that radius R will cancel in the numerator and denominator, leaving us incremental charge in terms of the total charge of the distribution as Q over 2π times dΦ.
Now we can go back and write down our integral in explicit form. For dq, we will have Q over 2π dΦ. Therefore this part is basically dq, and in the denominator we have 4π ε0. R2 plus z2, this whole term is equal to the magnitude of the electric field generated by dq times cosine of Φ, and cosine of Φ in explicit form was z over square root of R2 plus z2. Therefore this term over here is nothing but cosine of θ, and dE cosine θ was the vertical component of the electric field.
We’re going to add all the incremental electric fields through integration along this ring charge distribution to be able to get the magnitude of the resultant vector, which is going to be pointing in outward direction. When we look at our integrand, we see that the r variable is Φ, Q is the total charge of the distribution which is constant, 2π, 4π ε, these are constants, and as well as the radius of the ring charge distribution and z, and that is the location of our point of interest relative to the center of the distribution.
They’re all constant, so we can take it outside of the integral. Constant, leaving us an expression Q over 2π, which will go to the denominator, times 4π ε0, and then we have z divided by R2 plus z2 times square root of R2 plus z2 will give us R2 plus z2 to the power 3 over 2, and integral of dΦ.
If you go back and look at the diagram of our distribution, as we add all these incremental charges to one another along this distribution, the corresponding angle dΦ is going to vary starting from 0 and going all around and coming back to the point that we started with to 2π radians. Therefore the boundaries of the integration will go from 0 to 2π radians.
Integral of dΦ is going to give us Φ, which we will evaluate this at 0 and 2π, and if we substitute 2π, this is going to give us just 2π. If we substitute 0, it will just give us 0. Therefore, we will end up with Qz over 2π times 4π ε0 times R2 plus z2 to the power 3 over 2, and from the integration we will end up with 2π. This 2π and 2π in the denominator will cancel, so our final expression for the electric field will turn out to be Qz over 4π ε0 times R2 plus z2 to the power 3 over 2.
Since the net electric field is pointing in outward direction along the axis, and if we recall rectangular coordinate system of x, y, and z and the unit vectors associated with these directions as î, ĵ, and k̂ along z, we can express this in vector form multiplying the magnitude of the vector by the unit vector pointing in the proper direction, which is k̂, indicating that, our total electric field is going to be pointing in z direction, in outward z direction or in positive z direction. Or one can also write it over here by saying that pointing in outward direction.
Now, let’s try to obtain an approximate expression for a special case, and that is for the distance z along the axis, which is much, much greater that radius of the ring case, if this is the case, then R over z will be much much smaller than 1. So it means that we can neglect this ratio in the first crude approximation in comparing to 1.
Now if you consider the magnitude of this electric field expression, to be able to obtain this ration, let us take z outside of the power bracket, in other words take z2 outside of this power bracket. And if we do that we will have Qz over 4π ε0. z2 is going to come out from the power bracket of 3 over 2 as z3 since z2 to the power 3 over 2 is equal to — we simply multiply the powers in order to get this equality, and 2 times 3 over 2 will give us z3.
Inside of the bracket, since R2 doesn’t have z2 multiplier, we’re going to divide that by z2 and plus 1, once we take the z2 outside of this bracket. Now we have the ratio that we were looking for. Since R over z is much smaller than 1, R2 over z2 is going to be even more smaller than 1. Therefore we can neglect this in compared to 1 in the most crude approximation. This z and z3 will cancel. We’re going to end up with z2 in the denominator. Therefore this expression will be approximately equal to Q over 4π ε0 z2.
If we look at this result, it’s a familiar result. As a matter of fact, this is nothing but a point charge with a charge Q will generate an electric field z distance away from the charge. In other words, this ring charge is behaving like a point charge. If we go far away along its axis to a distance such that its distance to the center is much greater than the radius of the distribution, behaves like a point charge for z is much much greater than R, and this makes sense because if we go so far away along the axis from the distribution, we will perceive that ring charge like a point charge.