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**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 *d***E**, 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 *d***E**‘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 *d***E****horizontals** 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 *d***E****verticals** 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 *d***E** 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 *d***E** 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 4*π* *ε**0* 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 *r*2 will be equal to big *R*2 plus *z*2.

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 *R*2 plus *z*2. Since *r*2 is equal to *R*2 plus *z*2, then *r* will be the square root of *R*2 plus *z*2.

Now moving on, electric field is going to be equal to integral of *dE*, and that is *dq* over 4*π* *ε**0* little *r*2, and little *r*2 is big *R*2 plus *z*2 and times cosine of *θ*, which is *z* over square root of *R*2 plus *z*2.

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 2*π* 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 2*π* *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 2*π* times *dΦ*.

Now we can go back and write down our integral in explicit form. For *dq*, we will have *Q* over 2*π* *dΦ*. Therefore this part is basically *dq*, and in the denominator we have 4*π ε**0*. *R*2 plus *z*2, 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 *R*2 plus *z*2. 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, 2*π*, 4*π* *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 2*π*, which will go to the denominator, times 4*π ε**0*, and then we have *z* divided by *R*2 plus *z*2 times square root of *R*2 plus *z*2 will give us *R*2 plus *z*2 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 2*π* radians. Therefore the boundaries of the integration will go from 0 to 2*π* radians.

Integral of *dΦ* is going to give us *Φ*, which we will evaluate this at 0 and 2*π*, and if we substitute 2*π*, this is going to give us just 2*π*. If we substitute 0, it will just give us 0. Therefore, we will end up with *Qz* over 2*π* times 4*π ε**0* times *R*2 plus *z*2 to the power 3 over 2, and from the integration we will end up with 2*π*. This 2*π* and 2*π* in the denominator will cancel, so our final expression for the electric field will turn out to be *Qz* over 4*π ε**0* times *R*2 plus *z*2 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 *z*2 outside of this power bracket. And if we do that we will have *Qz* over 4*π ε**0*. *z*2 is going to come out from the power bracket of 3 over 2 as *z*3 since *z*2 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 *z*3.

Inside of the bracket, since *R*2 doesn’t have *z*2 multiplier, we’re going to divide that by *z*2 and plus 1, once we take the *z*2 outside of this bracket. Now we have the ratio that we were looking for. Since *R* over *z* is much smaller than 1, *R*2 over *z*2 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 *z*3 will cancel. We’re going to end up with *z*2 in the denominator. Therefore this expression will be approximately equal to *Q* over 4*π ε**0* *z*2.

**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.**