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Gauss and E-Field

  1. Feb 19, 2008 #1
    [SOLVED] Gauss and E-Field

    1. The problem statement, all variables and given/known data
    [​IMG]

    My main problem with this problem is finding the electric field at Point 2 (P2)

    2. Relevant equations



    3. The attempt at a solution

    I derived the electric field using Gauss's Law for a cylindrical charge distribution.
    [tex]E (2\pi rL ) = \rho(\pi r^2 L) / \varepsilon_0[/tex]
    [tex]E = \frac{\rho * r}{2\varepsilon_0}[/tex]
    So I found E at P1 to be 0 V/m

    Now onto finding E at P2, according to my equation, it would seem that E is 0, since the radial distance is 0, however that seems strange to me, if the point is at the surface of the charged volume cylinder, wouldn't there be some charge, I know it's not a conductor but still I would think that there would be some charge?


    And since I've got it all out here, is the E field at P3 just the integral [3/2L, L/2] of a charged disk?
     
    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 19, 2008 #2
    The gausses law wont help you much here.
    Try working with Ex = k*q*x/r^3.
     
  4. Feb 19, 2008 #3
    Thanks for your reply,

    I can't use Gauss's Law for finding any of the points? I know usually the cylinder is an infinite length if you want to find E, but even for Point 1 I can't use it?
     
  5. Feb 20, 2008 #4
    Basically not in or outside finite length cylinder the tangential forces will cancel only at the midpoints. Try finding E on axis of a charged disc first.
     
    Last edited: Feb 20, 2008
  6. Feb 23, 2008 #5
    Ok so I've got the equation for a charged disk
    [tex]2\pi k \rho (1 - \frac{x}{\sqrt{x^2+a^2} ) }[/tex] how do I integrate that over the length (L) to get the electric field of the cylinder?

    Would I say that dE = [tex]2\pi k \rho (1 - \frac{x}{\sqrt{x^2+a^2} ) }[/tex] and then integrate dE from 0 to L or do I need to differentiate Edisk before I do that?
     
    Last edited: Feb 23, 2008
  7. Feb 25, 2008 #6
    Just integrate it form 0 to L, since ddQ = 2*pi*rho*a*da*dx and u integrated it along a then your equation should still contain dx.
     
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