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Homework Help: Concentric cylinders E field

  1. Sep 24, 2008 #1

    Can someone help?

    I have 2 concentric cylinders which are both conductors. The inner conductor has linear charge density of 6 n C/m. The outer conductor has no net charge. The inner conductor has R of 0.015m, the distance between the inner conductor and the inner wall of the outer conductor is 0.03m, and the outer conductor has a radius of 0.065 m. I am to find the E field at all Rs.

    So within the inner conductor, the R = 0.
    outside the inner conductor but before you reach the inner wall of the outer conductor, the R is E*2*pi*R*L = 6e-9 * L/epsilon. So you simply that and you get 108/epsilon.
    IN between the inner and outer wall of the outer conductor the E field is again 0.

    But what about the E field outside the outer conductor? I know that it has to do with the ratio of the density, but I am not sure what I am doing wrong but I get the wrong answer when I say that the outer density is proportional to the radius ( = 6 * 0.065/0.015)? Any help would be great. THank you.
  2. jcsd
  3. Sep 24, 2008 #2


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    Your question does not appear to be properly stated. The inner cylindrical conductor has a non-zero radius, so one would expect a surface charge density, but instead you gave a linear charge density with units C/m. Is this problem part of a larger question or is it standalone?
  4. Sep 25, 2008 #3

    This is the question as is given in the book. But couldn't I derive the surface density from the linear density by the following: sigma*2*pi*R = linear charge density? Where sigma is surface charge density and R is the radius of the cylinder. This is the problem. My problem is I don't see why the outside cynlinder should have an E field that is 1.46 times that of the inner E field?
  5. Sep 25, 2008 #4


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    The problem with this is that it is stated at the outset that the inner cylinder is a conductor, which means we would expect the charge to be uniformly distributed over the cylindrical surface. Hence I expected a surface charge density. But instead the linear charge density is given. I don't know how to interpret this.

    Secondly, assuming your interpretation of this is correct, I don't see what you mean. What does "1.46 times of the inner E-field" mean? It's clear, since the outer cylinder is uncharged and hence does not affect the E-field of the configuration that the E field outside is 6n/(2pi*epsilon*r). It's the same in the empty space between the two conductors, only that the r value is different. So what does "1.46 times" refer to?
  6. Sep 25, 2008 #5
    That is what I would expect too. But the answer in the book is that the E field between the inner cynlinder and the inner wall of the outer cynlinder is exactly as you provided, which is 108 N/m^2. But then the answer they give for the E field outside the outer cylinder is for some strange reason 158 N/m^2?

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