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Electric Field in Cylindrical Insulator

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    We have a solid cylinder of radius R0, and it has a uniform charge density p. The length is much greater than its radius, so it appears almost infinite.


    2. Relevant equations
    Find the electric field inside the cylinder where r<R0 and outside the cylinder where r>R0


    3. The attempt at a solution

    I know that you need to use Gauss's law on this and I believe that for inside the cylinder it will be E= p/(pi*r^2*ε0)

    Will it be the same for outside the insulator?
     
  2. jcsd
  3. Sep 20, 2010 #2

    collinsmark

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    Hello felguk,

    Welcome to Physics Forums!
    Yes, that's right you do need to use Guass' law. :smile:
    Not quite.

    Let's take a look at Guass' law.

    [tex] \oint_S \vec E \cdot d \vec A = \frac{Q_{enc}}{\varepsilon_0} [/tex]

    Choose your Gaussian surface such that the electric field's magnitude is constant over the Gaussian surface, and perpendicular to the surface (parallel to the surface's normal vector). (The part of the problem statement, "The length is much greater than its radius," allows to make an approximation: you can ignore the end-caps of the cylinder.)

    With that, the left hand side of the equation is simply the electric field's magnitude E, multiplied by the surface area of the Gaussian surface. Here the surface area of the Gaussian surface is the surface area of a cylinder of radius r (ignoring the endcaps). Of course you don't know what E is yet, but that's what you are solving for!

    The right hand side of the equation is proportional to the total charge enclosed in the Gaussian surface.

    So you need to find what the total charge is within the cylinder, when rR0. The total charge Qenc within the Gaussian surface (where rR0) is the volume charge ρ multiplied by the volume of a cylinder with radius r.
    No, it is not quite the same. Use the same approach as the above, but when calculating Qenc, you need to use the radius of the physical cylinder, not the radius of the Gaussian surface.
     
  4. Sep 20, 2010 #3
    Thanks Collinsmark

    I believe I have it down correctly now,

    for inside the cylinder I ended up with E= pr/2ε0

    and for outside the cylinder I ended up with E= pR02/r2ε0
     
  5. Sep 20, 2010 #4

    collinsmark

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    Great job! :approve:
     
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