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The figure below represents a negative point charge with Qpoint 7.5 nC located at the center of a conducting spherical shell whose inner radius is Rz =4.5 cm and ou...

Question

The figure below represents a negative point charge with Qpoint 7.5 nC located at the center of a conducting spherical shell whose inner radius is Rz =4.5 cm and outer radius Rz = 11 cm . The conducting spherical shell has total charge Q shell 21.8 nC . Based on this information, answer the following questions_ Note that if your answer is negative, you need to include the minus sign in the boxes provided! If your result is positive write the numerical value only:a) For r = 3.5 cm , E, 7 Nlc b) F

The figure below represents a negative point charge with Qpoint 7.5 nC located at the center of a conducting spherical shell whose inner radius is Rz =4.5 cm and outer radius Rz = 11 cm . The conducting spherical shell has total charge Q shell 21.8 nC . Based on this information, answer the following questions_ Note that if your answer is negative, you need to include the minus sign in the boxes provided! If your result is positive write the numerical value only: a) For r = 3.5 cm , E, 7 Nlc b) Forr = 10 cm , [ez] Nlc c) For r = 16 cm, E3 - Nlc d) How much charge is induced on the inner surface of the spherical shell? Write the numerical answer in nano Coulombs nc inner surface e) How much charge is present on the outer surface of the spherical shell? Express your numerical answer in nano Coulombs nC outer surface (n = 10- k=9X 109 Nm?C



Answers

A conducting spherical shell with inner radius $a$ and outer radius $b$ has a positive point charge $Q$ located at its center. The total charge on the shell is $-3 Q,$ and it is insulated from its surroundings (Fig. P2..46). (a) Derive expressions for the electric-field magnitude in terms of the distance $r$ from the center for the regions $r < a$ $a < r < b,$ and $r > b .$ (b) What is the surface charge density on the inner surface of the conducting shell? (c) What is the surface charge density on the outer surface of the conducting shell? (d) Sketch the electric field lines and the location of all charges. (e) Graph the electric-field magnitude as a function of $r .$

Gases all can be used to determine the electric field for very symmetric uh objects such as fears. So this is an example of using gases. Law with a conducting sphere. That is a shell with a hollow in. It will quickly review gases law. It says that the electric flux through a closed surface is proportional to the enclosed charge. You almost never have to do the integral on the left hand side. Instead you have a symmetry which allows you to assume that a component of the electric field is uniform over a symmetric surface. And here, what we're talking about is a radio component of the electric field. In a spherical geometry with a surface area equal to that of an arbitrary sphere of radius R. That you get to draw wherever you'd like. And so that radio electric field times four pi R squared is equal to the n close charge over epson. Not. So now for a look at a spherical conducting shell. So it has a finite thickness. And what we're going to do is remind ourselves that conducting objects have all their charge go to the surface in electrostatic six. And that is because the charges are free to move and they will move as far apart from each other as they possibly can get. Um so that's the charges resigned on the surface. And also the electric field inside the conductor, the material Of the conductor equals zero. And again, that's because charges can quickly move out of that volume and get as far away as they possibly can. So let's imagine a situation in which we have a neutral shell to begin with. And we Sabot with -3 times. Big, cute charge. Yeah, nothing else going on. That charge will uniformly distribute it itself on the outer surface. Remember it's trying to get away as far as possible from themselves. Now let's assume for the more there's a whole uh that you can drop in some charge some point charges. So you carefully lower a positive Q into the whole um let's talk a little bit about what happens that positive Q. Is going to draw in some negative charge from that conductor. Remember Charges are free to move. So you will have -6. Will come zipping in to the inner surface to match that positive Q. And what does that leave on the outer surface. Um what it leaves is minus two Q. So you haven't put any more charge on that outer sphere or that uh conducting shell but it has rearranged itself in response to the charge inside. And now we can take a look at little gal siem surfaces. Now this is starting to get like a Russian doll type situation. But I'm drawing here in green, a calcium surface in the hollow gap inside the conductor and we can draw one outside just for fun. But what Galaxies law tells us is that on each of these galaxy um surface is the only thing that matters for the electric field is the charge inside. So for our less than a, what we have is the radio electric field times for pie that arbitrary are of that surface is equal to the positive Q. Divided by the constant of proportionality. Also known as the Perma Tive Itty of Free Space ε zero. So what the electric field looks like on the inside is a radio field that points outwards. Um And has. Uh huh. One over r square dependence to it. So it points out we'll call that a positive radio field and you'll see why in a minute Um if ours in the shell, that's the easy one, we're inside the conductor. But now we can see kind of what's going on to neutralize the field or make it disappear. Um That inner shell surface that I drew in green in there has a net Q. Enclosed of minus Q. And plus Q. Which is zero. Now, just because your surface encloses no charge, that does not mean your electric field necessarily has to be zero. But because of the symmetry of the situation, there's no way you can have the electric field Be anything other than zero. That plus the property of conductors. Makes you feel pretty confident that the electric field is zero in there. Yeah. And then finally are greater than B. We have a uh enclosed charge A minus C -2 Q. So that's minus three Q plus Q. And notice that that is the surface charge that's sitting on the outer surface. Um And if you didn't know any better, you wouldn't know what's going on with the shell and its interior, you would just see the electric field from that outer surface. Um And so that is uh huh. Two Q over four pi absolute not R squared is equal to the radio electric field. And that is an inward field which when we graph it will look negative here, I'm just doing the magnitude of the radio field. But of course it will point differently. And so finally we're going to make a graph and there's one more thing we want to talk about with conductors in their electric fields. So you want to make a plot of a sketch of e radio versus radial distance. Um It blows up at the origin because of the point charge, it dips down to zero. Maybe I should use a different color. The electric field disappears, turns into zero at R equals A. And it becomes negative at R equals B. It doesn't become infinitely negative. It takes on the value that it would have from a point charge at the origin of two Q. We'll just kind of sketch it And here I want to show something that is the nature of conductors versus insulators. Is that the normal component normal to the surface of the electric field is discontinuous on a conductor surface and looking at the derivative of that component, you can tell whether the surface charges negative or positive. So he dipped down means that there was negative surface charge, and then at the second radius it dips down again. So both surfaces display on the shell that is display negative surface charge the surface charge behavior.

For this problem on the topic, of course is law. We are shown in the figure a conducting spherical shell that has an inner radius A. And outer radius B. It has a positive point charge Q. At its center and the total charge in the shell is minus three Q. We want to find expressions for the electric field magnitude in terms of the distance R. From the center. For the region's less than a. Between A and B and greater than B. Want to find the surface charge density on the inner surface of the conducting shell and on the outer surface of the conducting shell only want to sketch the electric field lines for and the location of all charges as well as graph E. As a function of our. Now we'll use the gulf himself as there is a sphere of radius R. And as a point charge at its center for our less than a. We have the electric field E. To be one Mhm. Mhm. 1/4 pi Absolutely not times the charging close which is Q Over R squared. And this electric field is readily outward. Now the charge enclosed there is Q. Which is the charge of the point charge. And for the region between A. And B. A less than our less than B. We have easy equal 20. Since these points are within the conducting material. And lastly for the region are greater than B. We have the electric field E. To contain all charge. So it's 1/4 pi. Absolutely not times two Q of uh R squared since the total enclosed charges minus two Q. And so this if the total charges negative, this electric field points readily inward. And so we have the electric field for each of the regions for part B. Since a Gaussian surface with the areas are for a less than an hour less than be Mustn't close the net charge because easy equal missing. Close zero net charge because he is equal to zero inside the conductor. The total charge on the inner surface is minus Q. And so the surface charge density on the inner surface sigma, his minus the total charge Q. Over the area for pie A squared. And then for part C. The net charge on the shell is minus three key minus three Q. And so there is minus Q. On the inner surface. There must be minus two key on the outer surface. And so the surface charge density in the outer surface sigma is equal to minus two Q. Of uh four pi B squared. And lastly, we want to draw the electric field lines and graph E as a function hours are. And for party, the electric field lines point inward on the outer shell and outward onto the inner shell from the center. And we have a graph of the electric field E against our

The electric grid that you're looking for. He's 1/4 by absolutely not times Cuba, where our sweat it was. Now we're going do substitute what is cute. So one of our for by absolutely not parents who Underwood are Time's up charge He's contains the service charge density of sigma over the surface whose area is for by are split So that big, dumb sigma over Absolutely not Now this means that the outward filled is negative Sigma over to it's the not all it is Sigma over to absolutely not in the in ward get

Case over this problem. You're given a sphere, and we want to know with a conducting shell surrounding it. So here's this fear. And then here is the show. No, it's not really important of the whole thing in there. Let's just do that. And, uh, you the radius of the inner spheres, eh? This is B. This is C and but write out all the variables. There's Q one on the inside, and that's five um, fundo columns. So 5.0 times. 10 to the minus 15 Crooms Um, que tu So that's Q one Q twos that total charge on the outer shell and that's minus Q one a is two centimeters. So 20.2 meters never too early to convert something two meters, in my opinion, be used to a so I'll just read it out. It's quaint. 04 meters and see is two point for a Oh, I'm just gonna leave it in terms of a MME. And then for completeness soldiers, right? This is to pay. Great. So, yeah, we want to find the electric field at different points. So finding it in the center in each case you're gonna, um, the idea is to draw a galaxy and surface. And in the case of the sphere of the galaxy and surface is always a spherical shell. Um, around where? And you wanted the galaxy and surface to intersect where you're concerned with the field. So for, um, for are our equals zero your gals, Ian's here would be a point, Really? And so it wouldn't encapsulate any charge. So the field there is, ah, groups for a night A equals but e sorry, it's an ugly e cool zero. But now, as you go further out, you're gonna start encapsulating more charge. So I'm just gonna go ahead and draw pictures. It's just sort of one of the earlier one. So here is that close up of that sphere, and then we want we're concerned with the radius, which is, um, half of the radius of the sphere. So it over to so are is a over two point. 00 and so here's your Gaussian surface. Um, so this is like a radius are. And, um, now we wanna apply calluses law, which says he and a girl of ee dot d a equals Q and closed over epsilon. That's an l That's just randomly high over Absalon night. Um, and the this integral in the case of the choice of this fear, um, spherical shell of radius are it's just you times the surface area of that fear. So it's for pie. Well, the prize struggling. It's another leg. Four pi r squared and then that's equal to the human closed over Absalon. Not so how much charges enclosed here? So it's how much charge would be sort of within this red circle. And so it would be the percent of charge this enclosed or it's a percent of the total charge, um, which is proportional to the percent volume, um, within this fear. So that's just gonna be q one, not keep the Epsilon. Not, um, come on. 10 times the ratio of the volumes. The volume of a sphere is 4/3 pi r cubed and so and we want to set that we wantto divided by the volume of the total sphere to get the proportion charging closed so we can cancel the poor thirds and the pies and we get caught. So too next I do some algebra to determine what ah formula for ye at that particular are. And I would say it's Q one over four Pi upsilon nine. A queue of times are so the field just gets bigger, bigger as you, um, go further out as long as you stay within the the sphere. And then if you plug in numbers to that. So you're looking like you're ours too, eh? Well, I'm your end. The two centimeters, um, maybe I get when I go. It which was 0.563 Newtons per Coolum. Okay. And then for see, at this point, we're going to the edge of the circle, right? And saying our equals a Yeah, so are equals a And so we're still kind of technically within the sphere so we can apply this same formula. Um, and so you can say that he is just If our we could just say that our is equal to a So hee is Kyu won over for pie El Salon. I, um a squared right e did our equals saying canceled out the days. Um so if you put that into computer, you'll get what I got, which is 0.112 noons per cool. It makes sense. It just a bit bigger than, uh, the the a bit further end where there's less charged contributing to the field. Great. And then for D, it's, um, you want to know at 1.5 s? Oh, that's still not up to be. So it's between A and B. Ah, and then again we would draw our galaxy and surface. This time we would draw lit I R equals 1.5 a, and the only charge that would be enclosed would be Q one. And so you can say that E equals Q. If you go too fast, you will be punished, says my tablet Kyu won over for pie. Absolute night r squared again on DSO this time where this R is the 1.5 a so plugging that into a computer. I got 0.5 moons for Coolum, so it kind of starts to like sink back down again like it's a mazar gets bigger, it goes up, then six pack down as you get further away. Hey, almost done. So Hee is asking at 2.38 So that's beyond because B is to a and so at that point you could draw your calc and surface and since, um, and you ask how much charges enclosed within my surface? Andi, you would count the charge from the inner, too, in the charge from the outer, and that total is zero charge. And so there's no charging clothes. Therefore, the electric field is zero. And for F, it asks, Um, at 3.5 a. Oh, my bad. I, um that that shot should be the reasoning for F. So 2.3 a is not, um, it's It's in between B and C because, see, is 2.4 a. And then we were concerned it's 2.3 a. So it's between B and C. But the electric field is still zero because the the shell is a conductor. And as a rule, uh, the electric field is always zero in a perfect conductor. So the answer is no change, uh, Fergie and asks what the charges on the inner surface. So, um, I've done this really someone before I think a few questions back. So I'm just gonna quote a result that, um, you have the charge in the middle, and it's going to draw the exact opposite charge on that inner shell. So that's minus Q. One. And this is always the case because you can draw the galaxy and surface within the conductor, and then the A total charge. The electric field field a zero and therefore the total charging clothes must be zero, So the minus Q. Must balance the positive cue in the center and for each. How much are just left on the outside. So in principle, you could have trudged left on the outside. But it's the problem, states. It was only minus Q one total on the outside. We already have minus Q one on the inner surface, so there's none left for the outer, so that zero.


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