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Ch 22: Gauss' Law

Chapter 22, Problem 40

A very long conducting tube (hollow cylinder) has inner radius A and outer radius b. It carries charge per unit length +α, where α is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length +α. (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

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everyone in this problem. We have an access of an infinite cylindrical conducting shell. Okay, that runs a long line of charges with a net uniform linear charge density of four columns per meter. Ok, so net charge density, the show has an inner radius R one in outer radius R. Two. And a linear density of four columns per meter. Okay, we're asked to find the charge per unit length on the inner and outer surfaces of the show. Alright, so we're trying to find a charge per unit length. Okay, so that's lambda. And recall that lambda is going to be equal to Q over L. Okay, so the charge divided by the length. L. Alright, so in order to find lambda, we need Q. And we need L. Okay, we know L. Because we're given these radius is R. One and R. Two. Okay, so what about Q? Well, let's recall Kaos is law. Okay, this is that says that by E. Is equal to Q. And closed divided by epsilon? Not Okay. Which is equal to e. The electric field dot A. The area. Okay, now we're thinking about this. We are inside a conductor. Okay, On the in the shell of a conductor, what that means is that we have no electric field? Okay, so E is zero. If E is zero, that means that Q enclosed must be zero as well. Okay, so Q. And close this zero. So we have no charge enclosed in that conducting show. So Q. and closed must be zero. Alright, well if queuing closes zero. That tells us that the charge Q. Both in er and outer on the inner and outer surface. Zero. Okay. Alright. So let's start with part one then. Okay we have lambda is equal to Q. Over L. Okay now lambda on the inner radius is gonna consist of two components. Okay, it's gonna consist of lambda on the inner shell. Like we want to find. Okay, as well as lambda, the charge per unit length. A long. The it's called line charge. Okay. Alright now this is equal to Q. Which we just found to be zero because our electric field is zero inside of our conductor divided by the length R. One. Okay well zero divided by R. One is just zero. So lambda of the inner shell that we're trying to find? Mhm. Plus well what is the line church? Well we're told that the line charge has a uniform linear charge density for columns per meter. So we have lambda plus four columns per meter is equal to zero. Okay well this tells us that lambda on the inner surface is going to be negative for columns per meter. Okay, so we found part one lambda. The linear charge density on the inner surface. Now, what about the outer surface? Okay so let's move over to part two here and again, we're gonna have lambda equals Q. Over L. Now what do we know about the inner and the outer surfaces. Okay well we know that the shell has a uniform linear density of four columns per meter. Okay, so we don't even need this equation. What we know is that the net charge is going to be four coolers per meter. Okay, so lame the net. Well that's just gonna be the sum of the two lambda of the inner shell plus lambda of the outer shell. We know that this is four columns per meter. Given in the question. We just found lambda inner to be negative for columns per meter. And then we have a lambda outer which we're trying to find and what we get. We add the negative for the left hand side. We get that lambda of the outer shell is equal to eight columns per meter. Okay, so we have our two linear charge densities. Or sorry, linear charge. We have lambda inner. Yes, negative four centimeters per meter. And lambda outer is eight centimeters per meter. Okay, so the charge per unit length is going to be c negative four columns per meter and eight columns per meter. Thanks everyone for watching. I hope this video helped see you in the next one.