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A long cylindrical shell carries positive surface charge $sigma$ in the upper half and negative surface charge $sigma$ in the lower half. The electric field lines a...

Question

A long cylindrical shell carries positive surface charge $sigma$ in the upper half and negative surface charge $sigma$ in the lower half. The electric field lines around the cylinder will look like as follows(Figures are schematic and not drawn to scale.)

A long cylindrical shell carries positive surface charge $sigma$ in the upper half and negative surface charge $sigma$ in the lower half. The electric field lines around the cylinder will look like as follows (Figures are schematic and not drawn to scale.)



Answers

An infinitely long cylindrical conductor has radius $R$ and uniform surface charge density $\sigma$. (a) In terms of $\sigma$ and $R$, what is the charge per unit length $\lambda$ for the cylinder? (b) In terms of s, what is the magnitude of the electric field produced by the charged cylinder at a distance $r > R$ from its axis? (c) Express the result of part (b) in terms of $\lambda$ and show that the electric field outside the cylinder is the same as if all the charge were on the axis. Compare your result to the result for a line of charge in Example 22.6 (Section 22.4).

For this problem. We are given to very large not going to the plastic sheet. Um Each health 10 centimetres the and we know we are given the values for each of the charge densities in their surface in their force in there for um service. These are called sigma one, sigma two, sigma three and CDMA four with this is given values. So we need to find um the magnitude and direction of the electorate fuel at some given points. And the procedure that we are going to follow is that we are going to compress the left slap with the rightist lab in each of the points so that we get a uniform chart density and then optimum and find the electric field magnitude and direct. So to do this, we need first to remember what is the electric field for an infinite plane. So to recall these um, we can obtain theme electric fuel through through guns, lock the guns. Long tell us that the clothes integral of the product between the better of the electric field times. Um a differential of uh R E M of a better area is equal to the and close charge Over Excel. Um zero. So to get an idea of what is this, we're just going to draw uh suppose this is an infinite claim that is or sheet. Want to call it. Um, that is positive church. So to obtain the electric fuel, we use the goals Long. The God's law tell us that we need to use a Gulshan sort of negotiation Yes, ago, ocean surface in this case we will use cylindrical girls in geometry because it is the more convenient. So in this here we will have that. Um because the ship in this case have a positive charge. The electric fuel is upward. So we will have an electric field for this. And um in the front and in the back is also at word. So this is the electric field. Then the electric field is passing through there's surface of the couch in mm geometry. So that this in here corresponds to a picture or perpendicular to the coalition surface. So we obtained it. The problem of that is just these two bettors are parallel to the shorter. So from this we obtain that this is simply the electric field doesn't change. And uh and this is the interval of the story. Um So we know that the area is the integral of the area is just the areas that we obtain them. The electric field times the area. If this is equal to the clothes, the enclosed charged over Excellent zero. Which is the vacuum permissive. Itty. No, we just simply solve for the the electric field. And this will give us the following we note that in this case we are accounting this area and the one at the back. So we have two times that area. And we know dad, the sweet mom surface ah charge density is able to the charge over the area. So in this case we will obtain the disease. The the density charged over two tiny excellent zero. So with this in mind we can use this to obtain in our shit the electric field at any position at any point we want to measure the electric field. So for part A of this problem we are asked um appoint A which is um five centimeters from the left phase of the lane hands sheet. So the position is in here we will have our shit. And this is the point where we want to measure the electric field. So as I said in the beginning, we are going to compress the leftist lamp with the rightist lamp to get a uniform charge density. And then we act him and to find electric field. So in which is a slump islam. D all these mm all the surfaces of the sheets into just one plane. We will have Dad this will contain as we are going to call it. Just like the total surface charge density will be because in here we just need to think that this in here is negative charge. This in here is positive, this is also positive and this also positive. So the electric field and when the surface is negative charge, it's inward. Meanwhile the were positive um Surface sheriff surface the the electric field is outward. This has happened for this and this too. So in this case we will have that we need to multiply the first one. Um The 1st 1 and service uh density chart nine is the over minus the other ones. And this is our next surface density charge with this in mind. We can simply put all the values that we know in here. So we know God Sigma one. It's minus sits with this minus in here. The negative same council. So we have said mom for CDMA two we have minus fine me crow columns over meter square my news seen much three which is to and minus four. So we would obtained a ball you all minus fine me crow columns over meter square. No uh sorry. As we found in the beginning of this problem, the electric field is is this expression is the total in this case the total and surface sure and service Derearge charge over two times and okay. The vacuum permissive. Itty so full obtained out this is equal to Seema over two times that's long too. And the value of the CDMA is find me curl Collins meters squared over two times. Excellent. zero. And absalom cereal has a value of 8.85. 8.85 times 10 to the minus 12 jeremy perimeter. So in this case we will have that. Mhm. Um The magnitude of the electric fuel is 2.82 times 10 to define mittens over columns. And we can see that from here we obtained that it is with direction to the left. So it is pointing to the left Washington. Okay, do the left. No for party of this problem we do something similar. But the point of make sure is 1.25 cm from the inner surface of the right hand sheet. So in this case we will have the point just closer to the inner surface of the right hands sheet. So in this case we need againt just to adjust transform this into something like mhm. In this case we compress these two and these two and we just then up them so we will have something like this. Two to infinite planes want one in one side and one on the order. So in here again we know that this is negative, this is positive, positive. So and here we not thought or these is pointing to the left from this is pointing to the right from this is pointing to the left. Every this punching also to the left the electrical cow that is produced by each of these services. No having that in mind. We're just going to Yeah going to some in here we know that For the right one we have see my one plus CDMA two and for the left hand Should we have 73 plus sigma four and then we will have that in here we will set disease. We'll call these that want so net one is sigma one last signal to knowing that values we know God. Yes my news one micro columns over meters square. And for the net too it's the money to we will have that in this case it goes it is negative because it is pointing to the to the left so we will have this and this is equal to minus seats. Micro columns per meter. It's square. No having all of this. We just simply need to some all of this into the equation for the electric fields that we will have the next Minour a net of this 2nd 1 in this case Plus in this case to over excellent zero. So we will have that disease seven micro colon eatery. You over two times 8 85 times. Well florida ideas perimeter and these. Yes three 3.95 times tend to define mutants um eric column and it is pointing to the left again because we could see that the resulting um there is sultan and surface charge circumstance city charge is well most of them are put into the left. So we will have a negative electric field pointing due to lack. Now for the final question we are asked um the electric field in the point state that is in the middle of the right house shit. So we will have the sheets and we will have a point C in the middle of the right hands. Shit. So again we will try to slap all of this and just The 41 The four Surface is going to be alone. So we will have that in here of this form. So this in here. Yeah left one it's going to be it's gone DM cinema one sigma to platinum three. Yeah in my net want is equal to see my one plus in my two plus in the three and this in here will be just before So we obtain that net cdMA surface um density area which is in Matawan plus sigma two plus sigma three. So having those values we know that that is minus seeds last fight that will give us minus woman last to that will give us a positive number of want seat mom column parameter support. And in the expression for the electric field we will have this network laws the fourth that it is that is in here. But in this case we know that since this is a positive this is a positive surf and is a positive surface we will have and what your field in this direction and the other ones this is positive, positive positive, this is positive, positive positive and we saw that that has a positive direction so that the direction of this is to the right and the direction produced. Bye damn right hand the right hand and she'd is positive. So it is in the negative direction. So we just need to rest these yeah Segments. So in this over two times Excellent zero. So we will have in here juan where minus the semaphore which is war yeah for negro columns per meter squared and these again by Sigma's zero Which is this volume. And from here we obtained about you of one point sense 89 times 10. Two define newtons per cologne. And again it is pointing two the left because the resultant um this resulting quantity of the surface ah density charge is negative so that the direction of the electorate field is pointing to the left to the negative. Yeah.

Okay, so we're doing Chapter 22 problems. 27 here. So we have to thin concentric Spiric ALS shells of Rady. I are one are two as seen in this figure. Where are two is obviously greater than our one and each contained surface charge densities signal one and Sigmund to respectively. So now we want to determine the electric field for these different scenarios. First, we're going to say that our is less than or equal to her one or between zero and are one. So let's do this case first. So now we should see that a galaxy and surface would include no charge here because the only charge is outside the surface on this on the spherical shells. So the charge and closed zero we know that there's gonna be no electric field here. Cool. So now we can easily move on. So now, for part B, we're gonna do the case where are is greater than our one, but less than are too. So now if we draw a galaxy in service out here, the charge enclosed is gonna be all the charge on the first surface. So the charge and closed on the first surface is given by charged, enclosed as the surface area times the surface density and the surface area of a sphere is four pi r squared. So that's the surface area 10 times it by surface density charge so interesting to me. So if this is the total charge enclosed, we draw a calcium surface. So let's go ahead and write. That's how we know that the electric field is gonna be perpendicular, the normal vector at all points. So that means we can easily write this cell as the magnitude of e times four pi r squared in this corner. Miguel says law is equal cute, enclosed, over, absolutely rearranged for E. And we see this is Q been closed over four pi absolute, not our square. Awesome. So now we just plug in What are cute? Enclosed is that is four pi r one squared. So we see the four pies cancel out on our left. This signal one are one squared over, absolutely not r squared. So that's what the electric field is in between the two shows. So now you can we want to part see, and we're going to do this one for our greater than our two. So for this case aren't cute. Enclosed is now on the first surface, plus the 2nd 1 So this is Sigma one are one squared plus sigma to our two squared all times for pie So we can put that back in our equation. And we know that the Gaussian surface out here is gonna close all of those charges. So this just becomes signal. One are one squared plus sigma to our two squared all over Absalom, not R squared. Awesome. Let's move on to part Dino. Right, Actually, let's come back to me. Box this and you guys write it down. Cool. This equals E. Okay, so now moving on to part D party Says what? Under what conditions were equal zero for our greater than our two. Well, we already know the equation was given as signal one Arland squared, plus sigma to our two squared over R. Oops square. Absolutely not. So the only way for this to equal to zero is if the numerator equals to Syria or sigma one are one squared equals negative signature R squared. So yeah, this is the This is a condition that must hold for eating the zero in that range. Party says on what conditions will e V zero for the range we did in part beef So are between our one and are too. So for this one, our equation waas e equaling signal one are one squared over up Sloane are too. And the only way this can be zero All of these things are non zero, but signal one could be non zero if there was no charge. So the only way this could be zero is if Sigma One is also zero, which means no charge on the first show. Another way we could make this zero is if we put a charge of negative for pie one squared signal one at the center of the shells. That could also make the zero because this opposite would just cancel help. Exactly the case here, but not adding anything to the problem that wasn't already included. You would have to have signal one equal zero

Okay, is there were doing Chapter 22 problems 33 here. So it says a long cylindrical show of radius are big are not and think a little low with our not much smaller than Elsa's very long rod. And it possesses a uniform's surface charged Density Sigma. So the charges the uniform across the surface of the cylindrical show. So for part A, we want to determine the electric field at points are greater than our not so outside of the cylinder. So if we're follow, exactly are similarly how example. 20 to 6 goes. We can draw a galaxy and cylinder around our why air here around this electric cylindrical show of radius. Big are, and we should see that the electric field is going to be distributed perpendicular to the surface at all points. And that's what we need for Galaxies law to apply. So it's right out gases little here, so we need the electric field. If it's perpendicular to the surface at all points, we then can pull it out and evaluate just the surface area, and this then becomes e the surface area of our cylindrical show. Then at this point is given by two pi are times the length l and this equals the charge Enclosed over Absolutely not. From that we can rearrange and solve for what he is. And this is given as the surface density times the area enclosed surface area enclosed here over to pie. Absolutely not big R l awesome. So now for are greater than are not. We can see that the area enclosed is just the surface complete surface area. Just two pi r. Not so. If we plug that in for our area enclosed E then becomes Sigma or not over Absalon, Not our as our two pies and medals cancel out. And since this is positive, we already know it's gonna be perpendicular to the surface at all points. So that's radio and positiveness tells us it's radio outward. Awesome. So there's your answer for part A. So now, for part B, it asks What is the electric field for inside of the cylinder? And at this point, we can see the enclosed area zero, because if we draw our calc and cylinder inside, there's no there's no area there. There's nothing no area with charge accumulated. So if area enclosed zero, you know that the electric field is also zero moving on the part. See, it says, compared to the result for a long line of charge, which is example six of gases law. So the field for our greater than are not due to a shell is the same as the field due to a long line of charge. If you substitute land for two, pi are not signal. So if we replace that, then we should see that e well, First of all, right out was Sigma are not over. Absalon are so if we replace signal there, look, if we replace Lambda with that, then Sigma are not as Landover to pie and we plug it in and get him over to pie. Absolutely R, which is exactly what it is for a line charge. So it's pretty similar if we substitute that. And when we're talking about distances far away, it really does act like that, and that's usually what we can do. We could approximate the surface charge of this cylinder as a linear charged up city, but when we're expecting a closer recon, see that the answer's really do compare Similarly so

So we have a cylindrical rod off land Elf are not. We want to find a field, so let's go in the field for an hour. That is greater than our Not that is any point outside this cylinder. So for that we'll use a costume service that looks something like this. It's another cylinder in closing our cylinder. Why do you use this? The reason is because our system is cylindrical is symmetry. We expect the electric field Toby a radial e outwards on, depend only on the distance from the cylinder it is. We don't expect it to depend on who were these out on the cylinder. So sit on, such as Gaussian surface on these two services, the two circular surface onto ends. The electric field would be worth a call to the coast interfaces, and there will be no contribution to the flux, whereas the center circle the center surface, the electric field will be perpendicular to the area on. Hence, our we could find the electric field body easily. No, let's say that this really is off the cylinder is our as we've seen already. The flex on the two side service's will be zero so the total flux will be only do tow center surface. Let's say this Gaussian surface has excellent on you know that the radius is our the really is off this circular surface off the cynical surface would be x times the rate, the circumference of a circle. This would be to buy our This will be the area we need to multiply with electric field and we expect the electric field with seem all over the cylinder because of our political symmetry. We know that this is going to charge in closed be weighted by That's it. I'm not so if he extend this so if we extend this cylindrical surface all over the our cause in service all over the cylinder on we ignore the changing effect and we put X is equal to hell. We will see there e in tow to bite our and is equal to queuing. Yes, it or not. Now the charge enclosed is simply the charged density times the volume enclosed and you know that the volume in close will be the volume wealthy broad that is enclosed. That would be the area off the road that is fire north square times the distance so that would be road times. I are not square. Not so canceling you Common terms. We see that the electric field Israel are not scared by tow, vessel or not are And it's a directive Really upwards for Barbie be considered Art is less than are out Then we use the same gods and surface and we have The flux is equal. Do in tow Ellen to two pi r But this is a call to charge. Enclosed, derided by accident I'm in this case the charge in close will be. Let's say this is our cylinder with radius are not Let's say we're considering the distance. Distance are so we considered another surrender in this radius. So they were charged in clothes will be rolled times I ask Weird s it on? I'm sorry. And so that is rope. I ask where l by SL or not? No canceling. The comment comes. You can see that our is out until get scattered giving us He's a weirdo. Hello. I'm sorry. This is an r squared So through our by a salon Not again. This field is also director really over words


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