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Gauss' Law: Cylindrical Symmetry

  1. Sep 22, 2008 #1
    Say we are looking at a positively charged rod with uniform charge density and a radius of R.

    When using Gauss' law and taking a cylindrical surface we use the formula
    E = lambda/2*pi*epsilon*r

    When we derive this equation we are assuming R is significantly smaller than L and so we consider the charged body to be similiar to an rod that is co-axial to the Gaussian cylinder.

    What if we want to consider what E is inside the cylinder at say r=R/2

    I have read on another forum that we would consider the electric field inside the rod as 0 but that doesn't make sense because a charged rod with uniform charge density will have some electric field inside the rod as long as we are not right in the center where the field would cancel each other out.

    If we do consider the EF inside the rod to be 0 then we must be assuming that the diifference between the magnitude of the EF caused from the opposite sides of the rod is insignificant but I do not see this assumption clarified anywhere.

    Could someone please clarify this for me?

  2. jcsd
  3. Sep 22, 2008 #2


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    If you assume an ideal scenario in which the uniformly charged body is infinitely long, the cylindrical symmetry is perfect, and you can apply Gauss's law and the Gaussian cylindrical surface for any radius, even one inside the charged body.

    The assumption that R must be much less than L is then the condition for using the above ideal scenario to model a real world scenario where L is never infinite.
  4. Sep 22, 2008 #3


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    Staff: Mentor

    Remember Gauss's Law says: the total flux of E through a closed surface equals the total charge enclosed by the surface, divided by [itex]\epsilon_0[/itex].

    1. What is the total charge enclosed by a cylinder of length L and radius r, and with a uniform charge density [itex]\rho[/itex]?

    2. Assuming that E is directed radially outward, and uniform over the outer surface of the cylinder, because of the cylindrical symmetry, what is the total flux through the surface, in terms of the unknown value of E?
  5. Sep 23, 2008 #4


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    "I have read on another forum that we would consider the electric field inside the rod as 0 but that doesn't make sense".
    It doesn't make sense.
  6. Sep 24, 2008 #5
    Well it might make sense. If the rod is not perfectly insulating, then it is a conductor. All charge appears on the surface of conductors, since in electrostatics, there is no E-field inside a conductor.

    For the purpose of this question, I think you would be expected to assume a uniform charge density within the (insulating) rod.
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