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Problem 1Ring 2Two rings of radius R = 4 cm each, are 60 cm apart and concentric with a common vertical axis. Ring 1 carries a uniformly distributed charge of + 5 n...

Question

Problem 1Ring 2Two rings of radius R = 4 cm each, are 60 cm apart and concentric with a common vertical axis. Ring 1 carries a uniformly distributed charge of + 5 nC , and Ring 2 carries a uniformly distributed charge of 2nC , Point A is on the axis, halfway between the two ringsa) Find E,, the electric field vector created by Ring 1 at point A,b) Find Ez, the electric field vector created by Ring 2 at point A Find Enet, the net electric field vector created by the two rings at point A d) Ifa ch

Problem 1 Ring 2 Two rings of radius R = 4 cm each, are 60 cm apart and concentric with a common vertical axis. Ring 1 carries a uniformly distributed charge of + 5 nC , and Ring 2 carries a uniformly distributed charge of 2nC , Point A is on the axis, halfway between the two rings a) Find E,, the electric field vector created by Ring 1 at point A, b) Find Ez, the electric field vector created by Ring 2 at point A Find Enet, the net electric field vector created by the two rings at point A d) Ifa charge qz ~InC is placed at point A, find the net electric force vector applied on qz = Ring



Answers

The electric field on the axis of a uniformly charged ring has magnitude $380 \mathrm{kN} / \mathrm{C}$ at a point $5.0 \mathrm{cm}$ from the ring center. The magnitude $15 \mathrm{cm}$ from the center is $160 \mathrm{kN} / \mathrm{C} ;$ in both cases the field points away from the ring. Find (a) the ring's radius and (b) its charge.

Okay, so we're gonna do problem. Kept her Chapter 22 Problem 41 here. And it says a flat ring with any radius are not outer radius floor are not is uniformly charged in terms of the total charge. Q. Determine the electric field on the access at points. Uh, 0.25 are not. And for part B 75 are not okay, So first weaken. Using the hint that it gives us, we can treat the source as a superposition of two discs with one disc being. But radius are not with a chew charge of, um, it's a Megan of Charge. So que lust of zero and then another disk out here with the Radius four are not with a positive charge such that I'm gonna call this Q one. Cute to such. That Q one plus two equals the total charge, too. So we can now approximate the electric field at any point as just the sum of thes two fields. So for talking about in terms of A when we're in the distance, 0.25 are not from the centre. We can approximate both disks as infinite planes because they're both extend much farther away than where we are. But one has a positive charge and the other has a negative charge. So the only way we can or so these we're gonna add together and be zero, it is kind of hard, tricky to think so, if we're imagining these is infinite planes, they should have the same charged density. Right? So we have one. I think of this one as your really big one made with the charges density of Positive Sigma. And in here is a negative Sigma such that here. Well, while they're overlapping, that's gonna be effectively zero charge because these counter act each other and since we can for looking at just 00.25 are not away, are too effective discs that were thinking about overlapping each other. So the charges exactly canceling each other out, leading us to the zero electric field. It's kind of confusing to think about but trying to go go through it a couple times if you're confused because just think of it. I was too discs. You can also think about it as if you had two point charges of positive and negative Q on top of each other. If you're anywhere around them. It just looks like there's no charge there because they can't switch on their own part. B wants us to do it when ours 75 part. Not so now. We're so far away from the ring. We can approximate the field as a point charge. So if we approximated as a point charge, the electric field is simply one over four Pi absalon, not times the total charge, which we said was Q over the distance. Squared shoes. Simple is that?

For this problem on the topic of pulses law, we're told that a solid conducting sphere is carrying a charge Q. And it has a radius A. It is inside a concentric hollow conducting sphere within a radius B and outer radius. See if the hollow sphere has has no charge. We want to drive expressions for the electric field magnitude in terms of our for the region's less than a. Between A and B. Between B and C. And outside. See, we want to graph the magnitude of the electric field as a function of our, from zero to to see we want to find the charge on the inner surface of the holocaust hollow sphere on the outer surface and then sketch the field lines of the system. Now we'll use a Gaussian surface that is a ceo of radius R. And that is concentric with the conducting spheres. So for our less than A, We know that these points are within the conducting material. And so the electric field there will be zero between A and B. For a less than are less than B. We have the electric field strength E. To be won over four pi absolutely not times Q over R squared. Since there is a charge positive Q. Inside A radius R. For B less than are less than C. These points are again within a conducting material. And so the electric field zero. And for our greater than C. We have the electric field E. To be one over four pi. Absolutely not times Q over R squared. Since the total charge again enclosed is positive Q. And so we have the electric field for each region. We'll do part C. Next. And since the calcium surface of radius are for be less than are less than C must enclose zero. Net charge be charged on the inner shell surface must be minus Q. Now, since the hollow sphere has known a charge and has minus charge minus Q. On its inner surface, be charged on its outer shell surface must be plus Q. Okay, now we want to sketch a graph of the versus our and sketch the field lines, which will look as follows. And here we have for part B paragraph uh electric field E. Is a function of our And for party, the field lines sketched from sketch within the radius to see.

Days within Chapter 22. Problem 39 here. So it says a non conducting sphere radius are not is uniformly charged with charged in city row E. It is surrounded by a concentric metal conducting spirit. Kal Shell of Inter Radius are one outer radius are, too, which carries a net charge of positive. Cute. We want to determine the resulting electric field in the different regions. Okay, so let's go ahead and draw this else that we have a sphere of radius are not. And this is non conducting, uniformly charged with roadie. Outside of this, we have another's miracle shell with small radius are one big radius are too, and it has a total charge. Cute, so Q equals cute and plus que out because it's conducting so we can only have charged at the at the surface is the inner and the outer surface. Okay, part eh? Asked us. We want to figure out the fueled for or less are not will immediately. Let's figure out we gotta figure out how much charges in closed. So we drawn calc and surface, and we know that from the calcium sphere, the electric field has given us how much is charges closed over four pi. Absolutely no r squared. So cute. Enclosed for our less than or not is given by deed charged density tigers How much, boy? And we've gotten through or cool. So if we plug that into our equation right here we see that e now equals ro e r Over three slums often. Pardon me, this is for ours. Greater than are not, but less than arm one. I'm coming. So the enclosed charge in this case is just all of the charge included in the small sphere. So we drove us out here. We're not including any of the charge from conductor yet so cute and closed his road E times 4/3 pi are not cute. Plug that into our equation for the electric field and we're left with wrote e r not cubed over three Absalon not r squared. Cool. Now let's move on to part. See, So this is for our greater than our one, but less than are too. And this puts us inside of the conductor, so we don't even have to do anything. We automatically know that the electric field is zero Part D finally asks us no for our greater than our two. So now of total enclosed charge is cue from the conductor. Plus, however, much was enclosed in our small sphere, so that's the total in charge of quote charging closed. So our electric field now becomes que plus wrote e times for third. I are not huge all over. Four pi slow our school. Or, if we want to simplify this a little bit, we can right Q Over or pie. Absolute murder plus Road E r. Not cubed over three slung mud all times one over r squared. Cool. And that's it for this problem.

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.


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