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Gauss Law Problem with Cylinders

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    An infinitely long, non-conducting cylinder of radius R carries a uniform volume charge density ρ. Find the magnitude of the electric field for 0 < r < R


    2. Relevant equations
    EA=Qin/εo


    3. The attempt at a solution
    I am debating whether the answer should be ρr^2/2εoR or ρr/2εo.
    I said E(2πRL)=(ρπr^2L)/εo and solved to get ρr^2/2εoR.

    I was confident in my answer but then found the same problem in the book, but it asked to find E at distance r from axis, where r<R. The answer to this was ρr/2εo. Can someone explain the difference between these two problems, if there is one. They may be exactly the same and my calculation is wrong, but I am confused over the two
     
  2. jcsd
  3. Sep 21, 2010 #2

    ehild

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    If you want to find E at r<R your Gaussian surface is a cylinder of radius r.

    ehild
     
  4. Sep 21, 2010 #3
    Since r<R, R will never appear in the equation since my dA vector relates to r and Qin also relates to r. I would simply replace R with r in my equation and solve, yielding ρr/2εo. Is this logic correct?

    I think I just failed to remember that in the basic form of the equation, you are integrating over the Gaussian surface. But does Qin relate charge enclosed by the Gaussian volume drawn or by the entire volume in question?
     
  5. Sep 21, 2010 #4

    ehild

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    Qin is the charge of volume enclosed by the Gaussian surface.

    ehild
     
  6. Sep 22, 2010 #5
  7. Sep 22, 2010 #6

    ehild

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    In that formula at hyperphysics, E is given in terms of lambda, the charge per unit length. In your problem, the charge per unit volume (rho)was given. Compare your new result with that you cited from your book in your first post.


    ehild
     
  8. Sep 22, 2010 #7
    When I calculate it, I get ρr/2εo.
    But the website above says something different, so I am clueless as to what to put if this problem were on my test tomorrow(today lol).
     
  9. Sep 22, 2010 #8
    We posted at the same time. I did not notice that. I'll change my calculation and see what I get
     
  10. Sep 22, 2010 #9

    ehild

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    See my previous answer.

    ehild
     
  11. Sep 22, 2010 #10
    I've concluded that the calculation is correct for my problem and it is ρr/2εo.
    But, I don't understand where they get the R^2 in the bottom of their equation on hyperphysics. It has to come from the A part of EA=Qin/eps, but how? Shouldn't A always be in terms of the gaussian surface, which has nothing to do with R.
     
  12. Sep 22, 2010 #11
    nevermind, found the link that explains it. I would never come up with that on my own but at least I now understand the difference and why it is there
     
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