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2 A rod of length L is charged with +QCharge 0~L/JLIZa. Find the potential at distance x from the middle of the rod on its axis shown in the figure: b. Find the ele...

Question

2 A rod of length L is charged with +QCharge 0~L/JLIZa. Find the potential at distance x from the middle of the rod on its axis shown in the figure: b. Find the electric field at the same point using two different methods.

2 A rod of length L is charged with +Q Charge 0 ~L/J LIZ a. Find the potential at distance x from the middle of the rod on its axis shown in the figure: b. Find the electric field at the same point using two different methods.



Answers

a. Use the methods of Chapter 29 to find the potential at distance $x$ on the axis of the charged rod shown in FIGURE EX30.44. b. Use the result of part a to find the electric field at distance $x$ on the axis of a rod. (FIGURE CANT COPY)

Okay. And this from were asked to find the electric potential, not the electric field. Let your potential of a uniformly charged rod of length to L as a function of why the perpendicular distance away from this rod. So here's the ride. It has length l on either side of a point, and we're looking for Joel as fictional. Why? So we're going to have to do this via integration. The is going to be as a function of why is going to be one over for pie? Absolutely not. Integral over small pieces of charge over the distance are from the charge to the point we're interested in. So at this point, we can express this little piece of charge this infinite test, one piece of charge in terms of the total charge. Q. As little charge D Q equals the total charge. Q. Over to Little L Times. A little if intestinal piece of in the X axis are uniformly charged right is centered on the X axis, so we used the differential in that T X. It's also worth noting that the distance are from the charge to our point of interest is this can be related to Why an ex, respectively? You can get that by Pythagoras. Is there, um, using this by constructing this triangle? So with these two elements, we can begin with the doing the integral he has a functional. Why is one of her four pipes 19 integrating from negative l l on both sides of the run r D little D Q becomes big Q over to little L T X are actual differential differential. We divide all that by Route X squared plus y iceberg. Okay, we can put pull out some constants, one of her four pipes or not Q. Over to l. We can keep the integral. So what's left over is D X over Route X squared plus y sward. So at this point, we can do the interval by any means necessary. The result will end up the yang. All are constants. Of course, Cuba tour to L. And the result of our integration gives us a natural log of root. X squared plus y squared was X We've found with this from negative else l and with a bit of simplification, we got Q over eight pi upsilon, not little L multiplied by the natural Log L Square. This is a little out squared plus y squared. Let's let well, all over roots. Well, scoring plus y squared minus l That's it. We have the electric potential as a function of why.

Once law in order to figure out the electric field but thin rod. And we're finding the electric field off to the right of that rod which is lying along the X. Axis. So in order to use columns law form of the electric field, we really want to think about that rod being broken up into tiny chunks of charge. We'll call them the queue. Okay. And since um we have a linear uh huh. Rod, we are going to write the Q. As a product of linear charge density or lambda is the if it's uniform, it's the total charge on the rod divided by its length, which they are calling little L. Um But what you're doing with columns law is you're starting with the form for the electric field magnitude. Uh And here we're going to assume that the electric field is pointing entirely along the extraction because of symmetry. That is there is symmetry going all the way around the wire such that the Y and Z components should cancel each other out. So what really finding is the X component of the electric field. And if we do that um And use column saw that should be KQ total over um Mhm. Ex separation. But we'll call it R squared. It will be a separation and X squared which will work out. But usually what you do to break up this up is you right? The electric field as a sum of a lot of electric fields due to little tiny chunks of charge. So your Q total become stick you. And here Dick you is lambda times some linear um differential which is going to be along the extraction because we are going to integrate over the source and use X as a way to count which source charge. Which little infinite decimal source charge we're looking at um so we can rewrite are integral this way and are what the R is in that formula is it is the separation between our observation point which I'll just call x minus our source point. Now those are all negative values but we'll just call those X source um and it's the magnitude of that. Now we don't worry too much about the magnitude because we're going to square the quantity so we don't have to worry about negative values getting us messed up and the dx Ok, so it's important to realize the observation point is fixed and we're not letting that move around in the integral, it's the source point that's moving around. And so we have fairly simple relationship for a tiny differential part of the E X field X component of the field. Now this is where we want to integrate over the source points and what you're really saying is that the electric field you can use superposition um the superposition idea that electric field from many many points is just the some of the all these little electric fields to add it up. So we're going to integrate that on both sides. And yes that is a exact differential on the electric field side. And we'll pull out our electric constant in our lambda. And then are integral over X. Is going to go from minus L. Uh The X coordinate where the wire starts all the way up, 20 where it ends. And then the X. Source over whoops talking so much. I forgot the X. Squared idea. Yes the denominator denominator is squared. Yeah. Okay. And um you can do that integral using a substitution such as you is X minus X. Source anti you with minus front front is the X. Source. So I'll just do that integral carefully to make sure I have it right. Uh And then I have D you Times You to the -2. I know I'm being a little sloppy the way I write my integral. Maybe I should be a little bit more conventional. Usually they like the D. U. Holding the functioned in and I've included the negative sign. So that's you to the -2. Do you? What is the lower limit? It is X minus L. X. Plus L. Sorry And the upper limit is just X. Okay. And that's a fairly easy integral. It's just due to the -1 With a negative sign that cancels the one out in front and we can mess around with this a little out Bradley. I don't know how much people need to do this but we can get common denominator and also put in for the lambda. Uh Let's see That would be X plus l minus X. All over X times X plus L. And yes, there are some pretty nice things that will cancel. Okay. And what's nice as we can see, the final answer definitely varies as a one over a distance squared. Um, the other thing to realize is that there are things you don't want to do with that formula. So notice that it blows up At x equals to zero. That is always true electric field that's sitting right on top of a charge source will be infinite more or less if you use columns law. So don't use that point, but it is valid for just about any other point on the axis. And also notice if the X. Is much much greater than L. So you're really, really, really far away from the rod than the E. X. Actually looks like a elect an electric field for a point charge. It goes strictly As a one over distance squared. So both of those are some things to recognize about that relationship

Here. This is the rod Central line that's plus at over two. And that's minus and over to at a distance off. Why? And they re y used to points which with have we will make angle theta so the some of the electric fields will be given by this. That's far this d e. That's got these de inside them and this to be terror. So we have de ingress to G X adom was I did a no, we have so charge here is de que and dick you is given by dander. Do you are you? And on the separate Bagram Lambda varies like this. And over here and and the word do the Lambda values ease a l over to So Dickie liquids them Ready y which is okay. Why do I No does, George, We can just integrate on the half of it and just matter by Edwidge do So that comes out to be to a times half. Why square or elsewhere before? So that is a it spread over four. So we find that a equals four cube over Esquire. Now we go back to this equation we have de it was two times G absolutely is one over for by Yes, the not times Get a cue over. Let's start this part to be X expert. That's why square And then there's this post sine theta part who scientist will be given by why you over that? So we have why on the upstairs and this becomes part of 3/2. I'm sure you understanding by now because you have 20 of problems like that. Now we have to substitute educ you so this will come out as 1/4 by It's not a X. Why de y over exploit. That's why spread 3/2 Now from here we have to integrate to find a total electric field. E equals indication of d e. It was Let's take all the constant terms outside the integration sign which is whatever four by yes, they're not times a eggs. The rest is why dependent. So you will be integrating. Why, dear i over expert plus y squared thanks to that three over do but the introduction them it should be negative. Other work to do positive that there were two. But we simply want to integrate on one side and master it are indeed grand by twice doing the abolition electric filled with come out as also, we have to substitute the A. As for Gyu over l squared and then the final electric filled with come out to be one of our four by FC, not eight. Q over s squared times one minus X over. Spread it off. Exports us, especially over full.

Okay, so this is the grab. True. So as we can tell you that review due to the rock toe the point peace, you have two components, which is the accident point. And why camo components here? Okay. In anglers, they and we know e which is you introduce. You okay? A task. You of our square. So there will have these equal to Katie time. Stick you over our square. I am a nolanda, which is you Go to Q R. L a lot. I still in your charge. Density. Ok, terrible accusing Yolanda. 10. Sell antique use evil to London and Tokyo. Therefore, these UK heat has plundered density are over our square and the two components The X component is equal to que Linda Dio over our square testicle Santa and the y component the you wise, you look a done that er over password and science data front graph. We can tell the court Santa Justin people out, which is the This is here Look and d is the distance family or is important to the point p Okay, so this in between the raw an appointment should be squared with the square plus elsewhere. Okay, So therefore the satellite is equal to our over square with the score plus elsewhere in San and I should be over a school of the square plus elsewhere. Now that's trying to determine the e X first, which is Katie under Dror Square test L, uh over square the square Pascal sports therefore will have yeses Boudreau to the L and then inside. And there goes Casey on the Al Dia Dia are square times square with the square plus elsewhere and we know are square with 60 distance between the rock and the point Peace thescore passed elsewhere So therefore we have in a growth zero to l inside of his ked under ality out over a D Square plus elsewhere to a par 3/2. Okay, so then we'll have k e blunder times in the girls go to hell and then I'll deal over de Square plus elsewhere to the power of ah 3/2. Okay. And this is equal to can you under times negative one over de square plus al square to the power of ah 1/2 and that ranges from zero to l. Therefore, we'll have que Linda times next just one over the square plus elsewhere. So the power of 1/2 along, plus one over D squared to the power of 1/2. Okay. And this would give us okay. You under tires? No. You want over the square. Thus I'll square. So the power of 1/2 I am plus over the So if we take the negus I out, we'll have a nexus. Katie Lunda, One over de Square plus elsewhere to a part of 1/2 and then minus what d and we know London, which is that in your charge density. You see, you go to charge over the let's. Okay, so cure Bell. So therefore, half negative. Can you que over times one over the square, plus elsewhere to a parable. Why half minus while over D Okay, Hoop. Uh, they make it clear, and I will determine the why components, which is d y and this is equal to can you under de over de square. Plus, I'll square to the power of 3/2 and entice deal. Okay, So therefore will have you are just simply equal to integral. Josie off and then exile in ago is K E lender D over these were last elsewhere to the power of 3/2 and then dia it can take que Linda and d out now have k ee times Lunda and then times D and intense The integral I was inside now is ah, deal over these square plus elsewhere. So the power off 3/2. Okay. And this will give us que onda de times over. These were times the square plus software to the power of who I have. And the range is Joe to l. So they're full left. Okay, landlady times, uh, over he square times, thescore plus elsewhere 10 times. What have I mean to hardwire my okay and then minus zero is just simply give us okay, Lana de times. Oh, over these were times the square lost all square to the part of 1/2. As you can tell, d come castle here and I went deer. And therefore we have que Linda times out over day times D square plus elsewhere till a part of 1/2. And then we know London is the in charge density which is even charge over, then sell no half. Um que e times Q R and sometimes over de honesty square us L Square to a part of 1/2. And as you can tell Oh, and I'll come because so. Well, okay. They will have you are is just simply you go to okay e que Over, um, de times uh, de square plus l Square to the power 1/2. And for question, be was asking us for having through the review, but its components is if, ah, he's waving a gun now. Okay, so remember the first component, I say the X component was equal to take a look was equal to next of Katie Que over l and times one over the's square because ah, Coop plus elsewhere to a part of my half and then minus lower the Okay, So now even these waving other l. That means that would be behind sergeant question here, which is negative. Katie Que, Over under. Okay, well, times over D squared to the power of one. Okay, Because since he's is so much larger than l so d squared plus elsewhere, which is a last minute you go to these were ok and minus Wildwood easier. No. Have not a K times q of well times one over the okay, Because he squares with power. Why have a STD? And then my knees were already, As you can tell, this turn here is equal zero. It's a letter for the answer is zero for the X components. Let's take a look at the white components you are. Remember, uh, you are a sequel to K e que Over the times the's square Plus elsewhere. So the power of why have Okay, it seems now these much bigger than l know have. Okay, thank you. Over de ties de square. So the power of 1/2 Okay, because remember, the square plus elsewhere is approximately equal to these for these case, Okay, since the is much greater than l never wear just in Crete que e que over de square. Okay. And these are the answers for this question. Uh,


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