An infinitely long cylindrical conductor has radius r and uniform...

Question

An infinitely long cylindrical conductor has radius r and uniform surface charge density σ. (b) In terms of σ, 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 λ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

Accepted Answer

Hey everyone. So this problem is working with electric fields, let's see what they are asking us. We have a long rod of a given radius and a certain charge. We are asked to determine the expression for the magnitude of the electric field induced by the rod at a distance lowercase R. Which is greater than the radius of the rod. Given my uppercase R. As a function of the uniform charge per unit area stigma. And then that's part one and part two is of the uniform of the linear charge density given by lampa. So the first thing we need to recall for this problem is that the electric flux is given by by equals Q over epsilon, not where Q. Is our electric charge. So from here we can recall. So now we need to get this in terms of electric field. So from here we can recall that phi can be rewritten as two pi R L. E. Where this lower case R. Is the distance from the axis of the rod. And then Q. The electric charge is given by the uniform charge per unit area. So sigma times the area. The area there for the rod you can recall is two pi our uppercase R, which is the radius of the rod times L the length of the rod. And that's over epsilon. Not. So here we can see a couple of things canceled two pi l. Length of the rod. And so we are left with our electric field magnitude E. Is given by sigma are over epsilon not lowercase R. And so that is the answer to part one of our problem. So when we look at our potential solutions, the only one that is correct is C. But we're gonna go through Part two just to make sure that we are on the right track. Make sure Part two matches our answer as well. So part two is asking for the magnitude the electric field. But in terms of λ so landing, which is the linear charge density. So too solve for ian tarp in terms of linear charge density, we need to recall that linear charge density is given as lambda equals Q over L. So we're gonna solve we're gonna rearrange that for Q. So that's lambda L. And we had already established in the first part of the problem that Q can also be given as the charge per unit area. In this case sigma times the area. So that is too high are out. Alright, so those two charges are equal to each other because it's the same rod. So we can rewrite this as sigma times two pi R. L equals L. Times lambda. The length of the rod canceled. And we are going to solve this for sigma so that we can plug this back into our electric field in attitude equation. So sigma is gonna equal lambda over two pi R. All right. So we're gonna plug this back in here for sigma. So the electric field attitude in terms of lambda is going to be Land over two Pi R. Then the rest of that was our and then epsilon, not lower case are uppercase R. S. Cancel. And we are left with the electric field magnitude equals land over two pi epsilon, not lower case R. And that is the answer to the second part of this problem. And let's check our answers and yes, that does match. So the correct answer for this problem is C. So that's all we have for this one. We'll see you in the next video.

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