Hello everyone. This problem we are asked to find a show the um well basically we have to find the charge first and then the radius of a uniformly charged ring whose electric field were given some information about. So if you want to know you know how you work at this church distribution, then you can easily look that up online. But basically if you have a uniformly charged ring, then the electric field along the center axis of the line that's going that goes through the center of the of the ring is going to be a function of the distance from the center of the ring. And it is given by this formula over here. So basically there's going to be the usher feel is going to be pointing along this access because all the other components that are pointing away, so perpendicular to this axis are canceling by symmetry. So this is a highly symmetric electric field. So it's only pointing in the X. Direction. And it's given by K times Q over times X over X squared plus a squared to the three halves. So A. Is the radius of the ring and X is the distance away from the center of the ring. Now we're kind of interested in two limits. Um if we look at the graphs that we were given, then we see that e times X squared um kind of evens out after a while and it becomes a constant value. So first we're interested in finding out why that might happen. So what we want to do is we want to look at the limit where X is much much larger than the radius of the rig. So looking at that, we can actually make an expansion of the denominator in the electric field. So we're going to say that this function of a square plus X squared to the minus three has is equal to x squared to the minus three halves, times one plus a squared over X squared to the minus three halves. And the limit that X is much greater than a. This quantity of a over X or S and especially a squared over X squared becomes very very very small. So this is a very tiny number. Now that warrants a taylor expansion. So we then do the taylor expansion of dysfunction and we see that you know, X squared to the minus 3/2 is x squared square rooted to the minus three. So it's just X to the minus three. And then multiply that by the taylor expansion of one plus a squared over X squared to the minus three halves. And we don't have to go very far. We just have to go to the first term essentially. Um Usually when you're doing something like this, you want to find the first non vanishing term here. The first nomination term is the first term, which is just one. Um But I've written about the second term just so that we see how the expansion proceeds. And in the L. F says we contain the lss contain all the other terms that are higher order and a squared over X where such as eight squared over X squared, all squared, so 8 to 4 over extra for and so on. Um So essentially this numerator or denominator function becomes one over execute uh times one minus 3/2, A squared over expert. Um So then plugging that back into the expression of the electric field, we find that the anguish of the electric field is given by K times Q times X times one over X cubed, times one minus to b over two A squared X squared. And this is approximately equal to k times Q times X over X cubed where one of the excess cancels. So we have K time ski over experts. So very far away from this church distribution. Um the electric field that looks like that of a point charge, which is generally true for almost basically every non infinite charge distribution. So if we look at that, then notice that if X is very large, right? This is the approximate form. So if I multiply by X squared on both sides, then I get the E times X squared is approximately constant. Right? It's approximately equal to K times cute the charge of the of the ring. So we can use this information and the graph that were given to find that the charge on the ring is going to be the magnitude of e times X squared divided by K. So Kay schools constantly just nine times 10 to 9. So putting everything together, we find 45/9 times 10 to 9 the units are nicely canceling. So we've got that, uh, units of 45 comes in units meter square columns, whereas the units of cake comes in euros, meters square, Coolum squared. And so one of the powers of the columns cancel and we're left with Q being equal to 5.0 times 10 to the -9 cool arms. So this is the first limit and this is the first part of the problem and the second part of the problem, whereas to find the opposite limit. So what happens when X gets very, very small and we're basically just getting closer and closer to the center of the ring wanting to find what the electric field there is. So what we're going to do here were saying that X is much, much smaller than the radius of the ring. So we're going to say that X squared plus a squared to the minus 3/2 is now approximately. Or we can, you know, factories a from this or a square from this. In that case we're going to head that this is a squared to the power of 3/2 times one plus X over a squared to the power of three were too. And again, in this limit X over is now the small quantity, so X squared over a squared is much much smaller than one. So this again warrants a taylor expansion and basically everything is the same as before. It's just X gets X goes in a place of a and it goes in the place of X from before. So the electric field works out to be if you carry out with the algebra, it turns out to be K times Q times X over a cube. Right? So what this tells us is that we can then divide by X on one of the one on both sides. To get that the ratio of E over X is also constant is equal to K. Types the charge of the ring divided by the radius of the ring cubed. So this allows us to re arrange for the radius of the of the cute of the for the radius of the ring. And we find that radius in the ring is the cube roots. So to the power of 1/3 of K times Q times X over E. So one over the ratio of X over eat. And then looking at the graph. Um and looking at the values there, we find that the radius has to be approximately 32 meters. So this this is using the fact that the over X. Is approximately 700 actually I am just realizing now, excuse me, I have made a mistake here. So this should be the cube roots. You write that out. So this will be so that we're on the same page. So this will be if you look at the graph, this is equal to 9.0 times 10 to the nine Times 5.0 times 10 to the -9 times one over 700. It is the cube root of this quantity is what the radius of the ring is. So this is going to be um 45 over 700. So it's equal to 0.40 m. So about 47 m. Okay, so 40 cm, it's approximately the radius of this rig.