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Calculate electric field for cylindrical charge distributions

  1. May 9, 2012 #1
    (2.16) A long coaxial cable carries a uniform volume charge density ρ on the inner cylinder (radius
    a) and a uniform surface charge density on the outer cylindrical shell (radius b). This surface charge is negative
    and just the right magnitude so that the cable as a whole is electrically neutral. Find the electric field in each
    of the three regions: (i) inside the inner cylinder (s < a), (ii) between the cylinders (a < s < b), (iii) outside
    the cable (s > b). Plot |E| as a function of s.

    2. Relevant equations
    Gauss' law


    3. The attempt at a solution
    I am actually pretty confident at what to do except at one point. I can calculate the field inside the volume cylinder with Gauss' law. But by doing so I am not accounting for the field due to the surface of the outer cylinder. In my solutions manual it indeed seems that the field due to the outer cylinder is omitted - why is that? How can it be zero? I can certainly see it must be zero right in the center of the volume charge cylinder but why is it zero everywhere inside it?
     
  2. jcsd
  3. May 9, 2012 #2

    rude man

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    It is (a) because Gauss says so, and (b) because the electric field due to the charges on the outer surface cancel each other inside that surface. Not hard to show directly.
     
  4. May 10, 2012 #3
    Gauss' law says that the field contributes nothing to the flux integral - how is that the same as saying the field is zero? But how would you show it. It is pretty obvious in the geometric center but what when you move away from that? Then you are closer to the upper part of the outer cylinder. I know for a fact that if you are inside a spherical shell, then the field is zero everywhere inside, but this is not spherical shell - it's cylindrical :(
     
  5. May 10, 2012 #4

    rude man

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    You're smart to question whether the surface integral of flux being zero is the same thing as saying the E field is zero everywhere. But in this case it has to be because there is symmetry all over the circular part of your right circular cylindrical Gaussian surface, and, since the wire is very long (this is key!) there is no flux out the sides of your G. surface, so the field is the same all over the surface. And if it's the same all over the surface then it must be zero everywjere.
     
  6. May 10, 2012 #5
    Thanks for the answer. I had more or less come to the same conclusion after having thought for a good while, but it's nice to get a confirmation of the correctness.
    However, another question did pop into my head in the mean time. You say the flux is 0 out of the sides of the cylinder. I do realize that if you take any point in an infinite cylinder then you have a field going radially out, i.e. perpendicular to the sides. But how does it even make sense to then take of sides, if you are considering an infinite cylinder? It seems a bit ambigious to me.
     
  7. May 10, 2012 #6

    rude man

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    Your Gaussian surface is a small section of an infinitely long r.c. cylinder. It's the sides of the G. surface that have no flux entering or leaving. If the whole cylinder were not infinitely long then your section of it would have some flux leaving the sides.
    Capiche?
     
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