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GL
University of California, Berkeley

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Conducting Spherical Shell

Gauss's law for magnetism states that the magnetic flux through a closed surface is proportional to the product of the surface area of the closed surface and the magnetic field strength inside the surface. The law is named after the German mathematician Carl Friedrich Gauss, who published the law in 1835.

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Video Transcript

this video will explore how charges spread over a conductor. And in the first case here we have, ah, conducting shell. It's got an inner radius of a and an outer radius of B, and the cavity is centered at the center of the shell. It's also got a total charge of Q. So first we want to know what does the chart. What does the electric field look like everywhere in space? Well, we've already discussed that inside a conductor electric field is always zero. And so right away we can say e for are greater than a and are less than B is simply zero. Now how about inside the cavity? Well, we could draw Gulshan surface and we see that Phi equals integral of e dot d A, which is equal to Q and closed over Epsilon. Not and there's no charge in there. So this has to be zero, which means our electric field is zero or more appropriately when our is less than a equals zero. So now we draw Argh! Ocean surface on the outside and I'll leave the work to you. But we know that again by this formula here, the total charge enclosed is the charge on the conducting shell, which is Q and so we get that e dot d a. Is equal to Q over Epsilon. Not and again, The electric field is perpendicular to the area of our gout to the surface area of Goshen surface, and it's always pointing Radio Lee outward. So this is simply e times Thean a Gral of D A. On the radius of the sphere is four pi r squared, and we get back to our familiar formula for the electric field of a point charge when we're outside of the conducting sphere. Spherical shell of Q over four pi Excellent, not R squared. But now the more important question becomes, How is this charge distributed in this conductor? It's got two surfaces and a total charge. Q. So how was that charge? Q Located. So we'll clean this up a little bit and consider this on each case. We know that inside the cavity there is no charge in the electric field. Is zero further? We know that inside the conductor itself in this guy ocean surface, there is no charge because the field there is zero. So if there's no charge in the cavity. There can't be any charge on this inner wall. So we can say that our surface charge density that the surface were at the inner cavity where R equals a zero, and that extends all the way out to the edge of the outer cavity, since there can't be any charge inside, otherwise we'd have an electric field inside the conductor. So we conclude that all our charge has to be distributed over this outer surface. And we already knew that in a conductor charge will distribute to the surfaces. But in this case, we've proven that it's gotta be the outer surface, and I'm stressed that it's Onley. In this case, we're not only in this case, but its particular to this case in this manner, so we can write that are surface charge density at R equals B is the total charge over the area, which is Q over four pi B squared. But now let's consider a slightly different case. We have the exact same conducting shell. It's got an inner radius of a and outrageous of be spherical in shape. So this time, instead of charging it, we put a point charge of Q at the center of its cavity, so at the center of the entire show. First question. What is the electric field everywhere in space? Well, again, without having to do any work, we know that inside the conductor, the electric field has to be zero. Now. Inside the shell, we have Argh! Ocean surface here of some radius R. And the charging close is simply Q. And we know we'll get the familiar expression for a point. Charge electric field as such. And if we go outside well again, the total in charges Q. At the center of this Goshen surface. And so we know this applies for inside the cavity and outside of the entire conductor, and you can work through that. But it's a situation we've seen many times. More, interestingly, is now how is charged distributed in this conductor. Well, we know that the electric field on some Gulshan surface inside the conductor has to be zero, which means the net charge inside has to be zero from the simple fact that the flux, which is equal t e d. A equal to Q enclosed or epsilon, not there is no charging closed because there is no field inside a conductor. That means we have toe have some kind of charge inside the conductor or inside of this guy ocean surface of negative que. And as we know, charging a conductor always goes to the surfaces and spreads out evenly over symmetric surfaces. And so we get little charge around the inner cavity of this shell. The total charge of Q. So we can say the surface charge density on the inner shell is negative. Que over four pi. Hey, squared charge per unit area. Now what happens to a situation where we go outside of the entire thing? Well, we've just shown that there's gotta be a charge of negative Q on the inner shell, which means right now we have zero total net charge. Also, we know this entire shall This entire conducting shell is uncharged. And right now we have a total charge of negative cue. So that means that somewhere out here we need a charge of plus cute to counter act that negative que and again in a conductor. These charges will spread out all over the surface like so and so we see that the surface charge density at the outer shell is positive. Que over four pi b squared. And just to double check our total charge, well, we have our point charge. Q. We have the charge on the outer part of the shell, which is also que, and we have a charge on the inner part of the shell, which is negative, que? Which gives us a total charge of Q as stated in the problem. So we can see that when you place a charge inside the shell of a cavity of a conductor, the charges within the conductor will move around, which they can do very easily because it is a conductor in order to make sure that the electric field within the connector remain zero. And so there's no there's no charge on the conductor. There is net charge on the surfaces of it. Clean things up a little bit because I want to consider a special case, particularly what happens at our boundaries. We know the charge on the inner shallow, this conductor or the inner surface becomes Q over four pi epsilon doubt a squared, and at the outer surface we get something very similar but would be squared instead of a squared. So let's consider this first. Well, I can actually rewrite this. And sorry, let me put in the vector directions because that will matter in a moment. I can actually write this Sigma of r equals B over Epsilon. Not conversely, over here, I can also write this as Sigma of r equals a over epsilon, not. And let's put in the radio directions and you'll notice Epsilon, not Epsilon. So articles A is negative, which points us in the right direction. It's pulling us toward the surface, which is Radio Lee outward that way. So the negative sign accounts for the direction of the electric field. And so we see one question is always well, what happens right at the border of a charge conductor or the charge conducting surface and free space. What happens to the electric field? Well, at that point, the component is simply the perpendicular component of the electric field is simply the charge density on that surface over epsilon Not, and this will always be able to. This is something will always be able to check by checking our boundary conditions. Uh, whenever we do a problem like this,

GL
University of California, Berkeley
Top Physics 102 Electricity and Magnetism Educators
Andy C.

University of Michigan - Ann Arbor

Zachary M.

Hope College

Jared E.

University of Winnipeg

Meghan M.

McMaster University

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