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[Physics 2 w/calc Uni] Cylinder inside of a cylindrical shell

  1. Sep 10, 2015 #1
    Figure (a) shows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are nonconducting and thin and have uniform surface charge densities on their outer surfaces. Figure (b) gives the radial component E of the electric field versus radial distance r from the common axis. The vertical axis scale is set by Es = 4.8 × 103 N/C. What is the linear charge density of the shell?

    The problem also comes with this graphic (#13 on this page http://www.phys.ufl.edu/courses/phy2049/old_exams/2010f/exam1sol.pdf) --- yes I know the problems are different but the graph is identical

    I have tried to solve this problem multiple times and every time it's wrong. I think my issue lies in the following questions I have in this problem's setup.

    1. It says "narrow charged solid cylinder" is this thing 1 dimensional? 2 dimensional? I'm not sure because in the hint I got it said to use the equation for an infinite LINE of charge.

    2. If it is 2 dimensional, then how is it also solid? A solid cylinder would have electric field lines coming out of the caps of the cylinder as well as the curved part. In the solutions I have seen they completely ignore this.

    3. In the graph, what is being represented? When the value of E goes from positive to negative, what just happened? Did r exceed the radius of the inner cylinder? Or the outer cylinder?

    4. The question asks for a linear charge density, how do I find a linear charge density of an object that is more than 1 dimensional? If I have a cylinder, it has to have a surface charge density, right? I found the following equation to convert from surface charge density to linear charge density, but I'm not sure if it's correct for this application - lambda = [ (surfacechargedensity)*area ]/L

    5. Also, it says that the two objects have charge on their OUTER shells. Does this imply that there are only electric field lines spawning on the outside of the objects? Pointing out towards infinity (or in if negative)?

    If someone could clarify these questions for me I'm confident I will be able to take it from there.

    Thanks!!
     
    Last edited: Sep 10, 2015
  2. jcsd
  3. Sep 10, 2015 #2

    Andrew Mason

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    It is 3 dimensional but you can ignore the diameter of the cylinder. This allows you to ignore any edge effects, so it approximates a line charge. Otherwise, you would not be able to determine the gaussian surface for applying Gauss' law.

    You can ignore the edge effects.

    There appears to be a discontinuity when r = 3.5 cm. Since the charge on the outer shell was on the outer surface, this means that the outer surface has a diameter of 3.5 cm and must contain opposite charge of a linear charge density that is greater in magnitude than that of the inner cylinder.

    Linear charge density is simply the amount of charge per unit length of the cylinder. You don't have to worry about the distribution around the cylinder surface - just the amount per unit length.

    No. But using Gauss' law you can ignore the inward field lines. IF you choose the right gaussian surface (a surface that, by reasons of symmetry, has the same field strength over the entire surface) it becomes apparent that the contributions from all but the enclosed charge will cancel out.

    AM
     
    Last edited: Sep 10, 2015
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