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Peinta SerPSE923P03. Notes Ask Your Teacher uniformly charged disk of radius 35.0 cm carries a charge density of 7.00 10-3 C/m2 . Calculate the electric field on th...

Question

Peinta SerPSE923P03. Notes Ask Your Teacher uniformly charged disk of radius 35.0 cm carries a charge density of 7.00 10-3 C/m2 . Calculate the electric field on the axis of the disk at the following distances from the center of the disk: (a) 5.00 cm MN/C(b) 10.0 cm MN/C(c) 50.0 cm MN/C(d) 200 cm MN/CNeed Help?eiSubmit Answer3440 Prcgress

Peinta SerPSE923P03. Notes Ask Your Teacher uniformly charged disk of radius 35.0 cm carries a charge density of 7.00 10-3 C/m2 . Calculate the electric field on the axis of the disk at the following distances from the center of the disk: (a) 5.00 cm MN/C (b) 10.0 cm MN/C (c) 50.0 cm MN/C (d) 200 cm MN/C Need Help? ei Submit Answer 3440 Prcgress



Answers

A uniformly charged disk of radius 35.0 $\mathrm{cm}$ carries charge with a density of $7.90 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2} .$ Calculate the electric field on the axis of the disk at (a) $5.00 \mathrm{cm},(\mathrm{b}) 10.0 \mathrm{cm},$ (c) $50.0 \mathrm{cm},$ and $(\mathrm{d}) 200 \mathrm{cm}$ from the center of the disk.

In this question, we're told that a uniformly charged ring of Radius Sensei meters has a charge of 75 micro columns and were asked to find the electric field on the axis of the ring at various distances away from the center of the ring. So let's start off with a diagram here. So what we're gonna do is we're gonna take a look at the ring edge on meaning that the ring here you can kind of think of is coming into and out of the page. And we're just looking at, you know, a side view. So this is our ring. And let's take this as the center of the ring. And that means that the distance from the center to the outside of the ring here is 10 centimeters as given in the problem. And so basically what we're being asked to find here is the electric field at some points above the center of the ring. And what's being changed from eight a to B to C to D is the distance between the center of the ring and the point p. So basically what we can think of as the why value here what is the separation between the center of the ring and the point P. That's what's being changed. So what I'm gonna do is I'm going to find the electric field as a function of y. So I'm just gonna do, ah, general focus on creating a general equation for the electric field, and then we'll just be able to plug in the Y value that we're giving each separate part and will be, and we'll be done. So what we want to do initially is think about two little bundles of charge that are directly across from each other on the ring and think about how their electric fields will interact. So the electric field at Point P from D Q on the left will point up into the right and the electric field from Dick you on the right will point up into the left. And because of the geometry and the same symmetry of the situation, what's going to happen here is that the X component of D E, from the charge on the right will cancel out with the X component of D from the charge on the left. So the D E. E X is here are completely equal and opposite to one another. And so this is going to mean that the electric field from the total ring is not going to have any component in the X direction here. Because if you take any piece of charge on the ring, there's going to be a charge little piece of charge directly across from it that has an equal and opposite contribution to the electric field in the X direction. So if you're looking at the entire ring, the all of the little contributions of the electric field in the extraction are going to cancel each other out. So we don't need to worry about the electric field. In the extraction, there is none, and so we're going to concentrate on the electric field in the Y direction. So what is the electric field in the Y Direction from de que So the tiny contribution to the electric field in the Y direction from D. Q is gonna be equal to K D. Q. R. Squared Times Coast data and the fate of that I'm referring to is going to be the data in between the electric field vector and the vertical from that individual point. So this is the angle to which I'm referring Teoh. And so if we use co stato, we're taking the vertical component here of the electric field. So this gives us an expression for the electric field from one tiny bundle of charge. What we want is the total electric field from the ring. So what we're going to do is we're going to integrates all of the tiny little contributions to the electric field over the entire ring. Now, the good thing about this inter girl is that everything inside the integral is a constant, except the de que right K is a constant are is a constant. Every single piece of the ring is a distance r away from point P, so that doesn't change. And the angle is always the same as well, so everything can be pulled outside of the integral except the de que And so we're just integrating de que over the entire ring. So we just get the total Q Okay, so our expression for the electric field from the ring is gonna be K Q R squared Coast data Now, to make our calculations a little bit easier, what I'm gonna do is instead of using coast data, I'm going to rewrite this in terms of wise and ours. So let's take a look at our picture here. So if this angle is is data, then this one here is also fate up. And that means that this angle here at the top of this triangle is also data as well. And so what we can see is that CO state is actually equal toe. Why, over our and so we can simplify this formula even further to give us que que are over r squared times y over our. So we get cake. You Why over r cubed. So this gives us a really nice compact formula to work with when we apply the different Y values from part A too deep. So let's go ahead and do that. So this is the formula that we're using. This is the entire electric field from the ring. So for part A, we've got a why value of one centimeter. Okay, that's the distance from the center of the ring to the point that were interested in calculating the electric field up. And instead of one centimeter, I'm gonna use 0.1 meters. Everything should be in standard units. The radius we can calculate that's going to be given by Pythagoras theorem. So if you look at this triangle here, then we can calculate the are the distance. Ah, from the cue to the point p, we can calculate that using progresses the're, um so that's going to be at the square root of 0.1 squared. That's the radius of the circle plus 0.1 squared Thought is the Y value. And so that's going to give us a value of R of zero points. 100 four 05 actually will change that, too. Then it's important to keep those decimal places here because we are going to be cubing are so that does make a difference. So let's plug these values into our expression. So we've got that the electric field is equal. Teoh que que take you is gonna be the same for every single part here. So I'm going to write that down, especially separate. And then why over r cubed so cake you is going to be equal. Teoh. 2 674,050 That's just multiplying k by the 75 my curriculums. And then we've got a Y value of 0.1 meters and the R value Cute. And so that gives us a final answer for the electric field off 6.64 times 10 to the power seven Newtons per column. Okay, so we're just gonna repeat that exact same process for all of the different Y values that are given. So we've got next week. God's a value of five centimeters or 0.5 centimeters. All right, meters, rather and we're going to calculate the radius again. I'm not gonna write dot All out. It's a 0.1118 keeping a lot of decimal places again. And so the electric field is going to be 674,002. 50. Remember, that's que que and we're gonna multiply that by 0.5 divided by 0.1118 cubed. And so the electric field. The value for the electric field here is going to be 2.41 times tend to the seven Newtons per Coolum and for part C, we've got a Y value of 30 centimeters or 0.3 meters. The our value you can calculate. Then the distance between the the ring and the point is going to be 0.316 meters. And so the electric field is going to be 66 74 to 50 times 0.3, divided by 0.31 six. Cute. And that gives us e final value for the electric field of 6.39 times tend to this six Newtons per Coolum. So you notice that we're getting a smaller and smaller electric field as we go away from the center of the ring which makes complete sons. And then we've got for parts de. We've got a why value of 100 centimeters, a k a one meter and the radius is or sorry. The distance between the ring and the point P is about 1.5 and so we get a value for electric field and the final answer here will be 6.64 times 10 to the six Newtons per column. So this is our final answer for part D here

Okay, so this is the equation. But I did interview over Uniformed Ring with the charge, which is? You could okay X over a squared plus expert to apart are all 3/2 times charge Q. He's a radius would just give it a 10.1 centimeter was 0.1 meter. The axe is the distance from the center of ring in in question, a scheme as one point, although centimeter, or just 0.1 meter and K is the cooling constant, which is April 9 and has tend to all nine. Utah has meter square, but coolants work and trying to is human. 75 point of micro code, which is 75 times 10 to the power next six school. So that's flying all these values back into the equation. Don't have, you know after a few, uh, is equal to Oh, sorry. A point and I times sends a power nine Yutong times meter square, Pokhran square, 10 times zero points here along meter over 80 square, which is 0.1 meter square plus zero going Joe, one meter square. Angela Power Truth, uh, 3/2. Okay. And has charged que, which is 75 times, tend to apartment six cool, and it will give us hope, the natural view for questions about 6.64 times 10 to the power of a six Newton Ochola. Therefore, question B. We use the same conclusion. Well, this time. The axe, which is the distance finest endeavoring scheme, is 5.7 year become very meter is still 0.5 meter, so that's trying to determine that review police case, which is easy. Go to eight Going on, uh, times sensual power when I noontime times meter square for cool elsewhere, 10 times 0.5 meter over a square was just 0.1, you know, square plus 0.5 Peter and then square into a power of ah 3/2 ends has charged was just 75 times attention part next six cooler, and this will give us The latter view is about 2.41 times central power of seven. News on the cool, which is 24 0.1 times tens of part six new temper cool. Okay for questions. See, we is the same question in this high. The This is a promise and a vibrant scheme is 30.1 centimeter, which is 0.3 meters. So you will be able to a Boeing on I times tend to the barn I. Newton Times Meter Square But courthouse square n times joe 0.3 meter older, uh, 0.1 meter square plus 0.3 meter square to a power off 3/2 end times sand. If I central governing six cooler and it will give us the electric views about, UH, 6.4 times 10 to the power of six Newton cool before the last question. The distance from the center of the ring is keeping it 100 centimeters, which is one meter. So we can now we can determine the electric you for this case, which is, uh, e you go to a point and I times says the ball when I Newtown times meter square, cool on square and in times one meter over a square, which is 0.1 meter square, us one meter square and answer the part off 3/2, 10 times 75 Task potential Barbara are over next six. Cool, and this would give us the that refuse. About six point 64 times tends to the power of fire, Newtown. Very cool. And these are the answers for this question.

Okay, so let's consider in this problem a fin. It's vertical show. This is Onley, just the shell. Which means that the charge is distributed on Lee. Are the surface okay? Not Onley, other surface. Not ah, in the entire circle as it has a right here. So it's just the surface that we have here is just the shell. And we have to calculate in this problem the electric future generated by the shell in the point insight, this show and the second point, this is the first point and in the second point outside the business radical show. So we have two problems in here. First of all, we have to solve for from the insight, then have to solve from the outside. And we're going to use to describe these the girls law equation. So let's begin to calculate the first case. So in the first case, we have the points inside this burger shell. The inside case is the easiest one. Because when we consider dig ocean law, let's describe here, Uh oh, gosh, influx. So the flux, the electric flux according to the go slow, is going to be cool. The charge inside the Goshen surface divided by absolute zero, which is also equal the integral off e d a. Okay, the closing to go off e d. A. So if we look to the right equation here, we're going to see that the electric feuds inside this is farcical show. Well, depend on the charge inside the surface and we do not have any charge in Sargis Verger show. Therefore, we can see that the first electric field inside is going to be zero. Just considering that there is no charge around us. A Gulshan surface insides Jesus for Cochell. Now, when we consider ago ocean surface outside is this vertical show. We have the charge of this worker shell inside and therefore we goingto have electric field. So in deeds, condition tool is the second situation where we want to cooperate with the radios outside. We're going to use that switch in here. We're going to use the integral the closing to girl off E. He, uh, is equal the charge in sight, divided by absolute zero since the electric field, despite although with the area, we can also say that this is were simplified to just yeah, a equals the charge inside divided by absolute zero. Ah, we know that the area the guy ocean surface Gosh on the surface, this is going to be the guardian surface has an area It's vertical area or four pi r square and this is equal the charge insides divided by absolute zero. Therefore, the electric field outside this worker shell is going to be the charge in sight, divided by four by absolute zero are square. Okay, so let's calculate this well, the electric field that we want in the point outside is going to be the charge, which is 32 times 10 to the minor. Six. Because it's micro columns divided by four by the multiplies absolute zero, which is 8.85 for two times 10 to the minus 12. They're multiplies the radios, the radios, this 20 centimeters which is going to be 0.2 meters the square. That is correct. Let me see. Ah, we have to calculate to 20 centimeters from the center of the charges distribution. OK, Is that exactly exactly this? Okay, so then let's calculate this. We're going to discover that the electric field as a value off seven point 19 time stone to the Six Newtons for Coolum directly, readily outward. So do you lines of the electric field points outwards? Do you use for nickel show Okay, Because it's positive. And while Dusty final answer. The first situation. We have electric field Nico zero because there is no charge inside the ocean surface. And in the second situation we have 7.19 times 10 to the six that's defined. Arrested to this problem. Thanks for watching.

In this question, we have ah, uh, charge disk with a non uniform surface charge density, which is six and at times are a And then, uh, we are interested to find out, uh, actually passing the question. First, we need to find an electric potential at Point X, um, which is ex a distance X away from the center of the disk. And then after that, we need to find an electric view and find out electric few and, uh, very far distance. Okay, for X is much, much larger than a Yeah. So, uh, to do part A, we are given a continuous charge distribution. So we are going to break the charge this into smaller pieces, and the smaller pieces will be small, uh, 10 rings. Okay, so I'm okay. You'll be like, thin rings. Okay, So And if you refer to the book, see the electric potential due to the ring will be Q four pipes or not hours times square of our square plus x square. But since you are looking at small little rings, it's a small lettering, but, uh, brings with infinite testimony mark charge. Uh, a cube. Okay. So, um and then we have the distances where we ask where plus x squared. So not that in this question, the access a constant We are going to integrate respect to our way. He too is our customer amount. Charge is equal to Sigma D A. Yeah, and then see my c might not. Times are over a and then the d a is two pi r p r. Yeah. So the two pi r is a circumference of the ring, and then the r is with of the ring. Okay, so, uh huh uh, circumference times a small length small, which is a small area that then ring covers. So, uh, you can put everything together, So the ring is they are not are a I'm super ir on the r. You're right by for pie. Absolutely not. Square root square, I think. Square. Yeah, And then if you want to find ah, the electric potential. Okay. You do that desk. Yeah, you'll be some ish Integrate devi ring. Okay. And okay, so from the both expression, you can see that uh, two pi and four pi can be simplified and then you can actually pull out the Constant Sigma over two X on a A and then you have great from zero to a Our square by square root are square X where we are. Yeah. So, um, we have this integral that we need to do. Family cannot do it easily, so we need to do some substitution. So if you look at the diagram, Okay, I'm going to call this anger data. Yeah, um and so if you from the diagram. Okay, find data, yes. Or or work square root by square plus x square. And then we also have tangent data, the sequel to our over X. Okay, so, yeah, this is distance X. So we have are it goes to X engine data. So I'm going to do I'm going to use substitution. All right? It goes to X tangent data. So if I do, the are you will be x times you can square data e data. And then I also need to change the limits. So when are it goes to zero? Yeah, uh, to increase zero. Yeah. And then when? All right. Good to a Hey, that would be our engine. Uh, they were X. Yeah. So I'm going. I'm just going to evaluate this, uh, integral first. Okay. Okay. So, um, so this integral from zero to a r squared, divided by the square of our square pass X where we are after the substitution, you look like Yes. And then, uh, so you have our square divided by the square of our square. Uh, class X square. So some you have one part of it as a sign. Data. Okay, so you have. And then the other parties are so the prices extension data and then signed data. And then you put the r replace that. They are your data. You have X sequence, square data. Uh, data. Okay. Okay. So, simplify. This is what you get for other X square, because they are constant. And then so tangent data is signed over design. So you get, um, you can actually get secret data and June Square. Yeah. Yeah. So second square data is one of the core science square combined in one of the science data. Get tangent, and then you have everyone second data. Okay, so this, uh, thing can be integrated using trying use a friend so far. Yeah. And you get, um, engine data. Second data minus hyperbolic. tangent. Hyperbolic tangent. Uh, science data A from zero to attention. Ah, over X. Okay. So, yeah, this is, uh, scary part, But you can do it using Wolfram Alpha, So substitute the okay, so the lower limit is zero. So when data is zero, everything is zero. So we don't need to care about the lower limit. You need to care about the upper limit. Okay. So, tangent, uh, they don't get your ex. So second data is gonna work with science data, and then you go back to the diagram here. The coastline data is equal to, um X over square root of our square plus x square and seconds later, you just to take the reciprocal. So you get a square root of a square plus x square. Because this time the R is a and then the Rubber X Okay. It's also half price. Okay? Yeah. And then, uh, apart at the bank, you just be a over square root a square plus x square. Okay, is then you multiply in the X squared over two, so you have half or a over two square root of X squared plus x square minus X squared over two. Ah, hyperbolic attention. Right. Okay. Hyperbolic tangent of a divided by the square root of X squared plus x square. Okay. Uh huh. Okay, so this is the integral, and then you put back into the electric potential. So you have someone on over two x on a You were too. Square, square, square pass, X square minus X square, too. I have a public pension. Uh, in your words were of a square. Thanks. Great. Mm. So this is the answer for heart aid? Yeah. Now you proceed to find an electric fuel. Okay, So e equals two negative DVD X. Yeah. So the constants just you mean and change, and then the inside able to, uh, you have so differentiate the square root of X squared plus X, where you're going to get to eggs. So get the half from the power. Then you are going to get You are going to reduce the power from half to buy one. So we got minus the power to the negative half, so I'm going to put that Yeah, uh, denominator. Okay, then the next time you are going to do product rule, so try to see that ex uh, have a bolic have abolished tension, uh, square root of X squared plus x square, and then you need to differentiate a hyperbolic tension. So this is, uh, one over one, minus the square over the square, past X square, multiply. And then So this is the Yeah, I need to do change room, okay? And then differentiate the inside Respect to X. You get, uh, a times minus half the myself is the power. And then you're also going to get two x from the university of of a square plus x square, and then you Then you reduce the power from minus half, too. Ministry have. So it's where? Plus X square. Mhm. Negative tree. Not negative, but this tree are mhm. He looks scary, but yes, it's actually quite good, because it can simplify it to nice tones. Leader. Okay, just keep going. He saw the two and the two cancer. So we have a X or two. Where is square pass? X square, minus X hyperbolic pensions. Yeah, We were school of the square plus X. Where he here he does. There's a minus sign here and then becomes a plus. Either two cancels you have X squared over two. Okay, so the term here, uh, I'm going to combine the fraction in the bottom. So you have one minus a square by a square plus x square. Right. So one minus a squared B but by a square Class X where this is, uh, square plus X squared, minus the square right by a square plus x square. So if you do the reciprocal, you get a square plus x square by x square, and then you have a X B bye bye, uh, square past X Where the tree house. So we can see that the X square councils or X squared plus x way. This one together half. And you see that is the same Lazar from the first time so you can actually combine them. Do you have a x rabbi square off a square classic square minus X hyperbolic tension or squared off the square pass x way, Then access a common factor. So I'm going to take it out, and at the same time, I'm going to change that sign. Okay, Like multiply the negative side inside. They are not X over to Exxon or a hyperbolic tension in a square of n squared plus X square minus a over square off E x or square Rocky square, past X square. Mm. Yeah. Okay. So this is, uh, 22 of e. Okay? Yes. Yeah. If you want a vector notation, This is how it looks like. Okay. See, now you move on to Passy. What happens when, uh, access much, Much larger than the radios. Okay. All right. So, um or X Much, much larger than a. So we have a or X Be much, much smaller than one. Okay, so, um, all right, there's no X here. Sorry. Because the access be taken up. Yeah. So, um, so far, so a or square it off. Uh, a square class X square is going to be much, much smaller than one. Yeah, because the denominator is going to be X. Okay. So, um okay, so we are going to have, uh We also need the relationship for, uh, hyperbolic tangent X in your case on small X. Yeah. Thanks. Much much. Uh, less than one gets X right, X plus no minus. Uh, no, essentially a plus. Yeah. Plus, uh, one third. Thanks, Q. All right. So, um above expression e We have e equals two sigma, not X over two x on a, um, hyperbolic tension. He over square it off. Yes, square plastic square minus a square off a square past x square. Yeah. So, uh, so the hyperbolic tension we get over square a square plastic square last one third. Oh, yeah, Cute is I buy you square pass X square with three half and then minus You will square off a square plastic square. Yeah, so this time is gone, And then we have seen a lot of X over to exile. Not a one third. I'm going to pull out X Square Insight. Okay. And then you have one plus a square or x square. The negative three half. Okay, So if you look at this, uh, expression we're here. So we have this term that is very small. Okay. And in this expansion, we only we will just take one out, because the next time will be time with a square times x square and you get into the fire divide by X to the fire, which is so small that you don't need come anymore. So, uh, we're here. You can just approximate this into single no X divide by two X l a one did. Thank you. Excuse me. Okay. Yeah. So you can't cancer. One of the things we're here. Okay, so you reach this step, uh, seem a lot, uh, a square. All right. Six. Absolutely not, uh, x squared. OK, so until now, we need to find a total charge. So because of what we expect to get, uh, electric fuel due to a point charge when When is very far away. Okay, so, yes. Have a separate step. We're here to the charge on the desk. It is is equal to, uh, sigma and the a A. This is, uh, Simona. Times are a two pi r. You are from zero to a You got all the constants in mind all the time to pay, uh, divide by a and then your leverage integral off square. We are from zero to a So you get some are not, um, two pi. Oh, a times I wanted to thank you. Okay, so you get to debt to pay over key. I think my knife you square. Yeah. And this is our cue. Yeah. So, um so check you over two pi. Is he going to Single? Earner is square. So you just replace uh, g q or what? Uh, six x or not X square. And then then you have your two pi. Yeah, and then you simplify too. Get you over for pay. Absolutely not X square. Yeah. This is what we want to get. Okay. Yeah. So this is the electric fuel. Yeah. You, too. On charge. Yeah. Which is why we expect together for X much, much larger than be So that's all for this question.


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Constants Visible light passes through dilfraction grating Ihat L Inas 900 slits per centimeter; and the interference pattem observed On screen that is 2.46 m from the grating Part A In the Iirst-order spectrum maxima Ior two dilferent wavelengths are separated on the screen by = between these wavel...
5 answers
The infrared spectnm of the compound with the mass spectrum shown below has medium-intensity peak = at about [650 cm" There also C-H out-of-plane bending pcak near 960 cmMost prominent peaks: M(S6} 41 'Repriaia courier oithe Natiqnal Jnninue 0t SJJdans Jla Technolop US Depurtrnt 0} Commene Nof copwrightable UHcd SWhich of the following structures consistent with the data?PreviousNext
The infrared spectnm of the compound with the mass spectrum shown below has medium-intensity peak = at about [650 cm" There also C-H out-of-plane bending pcak near 960 cm Most prominent peaks: M(S6} 41 ' Repriaia courier oithe Natiqnal Jnninue 0t SJJdans Jla Technolop US Depurtrnt 0} Comm...
5 answers
L-1 300 = ke2tt3 where k = (s-2)4
L-1 300 = ke2tt3 where k = (s-2)4...

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