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Gauss's Law Clarification

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    problem.jpg

    2. Relevant equations

    [tex]\Phi = \int{d\vec E \cdot d\vec A}[/tex]
    [tex]\Phi = \frac{q_{enc}}{\epsilon_0}[/tex]

    3. The attempt at a solution

    I'm having a problem with this problem. I don't really understand the givens.

    If we're dealing with thin-walled cylinders, we're dealing with Area, not with lengths as if we had a line of charge. Why am I given Q/L as sigma instead of Q\A?
     
  2. jcsd
  3. Sep 5, 2009 #2

    Doc Al

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    Staff: Mentor

    What matters when applying Gauss's law is the total charge contained within the Gaussian surface. For this problem, what would you choose as your Gaussian surfaces?
     
  4. Sep 5, 2009 #3
    Cylinders on the common axis so that i have an electric field penetrating the shell of the cylinders (as opposed to the top and bottom which have zero electric field penetration).

    EA=Q/e0 would become
    E(2piRL)=Q/e0
    E=Q/2piRLe0

    Oh at this point, could I substitue in the Q/L into the equation here making it

    E=sigma/2piRe0
     
  5. Sep 5, 2009 #4

    ideasrule

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    Homework Helper

    Yup, you've got it. Just plug in the numbers and you'll be done.
     
  6. Sep 5, 2009 #5

    Doc Al

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    Staff: Mentor

    Good.

    Q/L would be lambda the charge per unit length, not sigma which is the charge per unit area.
     
  7. Sep 5, 2009 #6
    Oh, yes, lambda. My fault.

    Thanks!
     
  8. Sep 5, 2009 #7
    For the second part of this, I'd create a Guassian cylindrical shell in the same way.

    Can I use:

    [tex]E_{inner}=\frac{\lambda_{inner}}{2\pi R \epsilon_0}[/tex]

    [tex]E_{outer}=\frac{\lambda_{outer}}{2\pi R \epsilon_0}[/tex]

    and add them together? I know that the electric field created by them will be:


    [tex]E_{inner}=\frac{5E-6}{2\pi(.86)(8.85E-12)}=1.05E-4[/tex]

    [tex]E_{outer}=\frac{-8.5E-6}{2\pi(.86)(8.85E-12)}=-1.78E5[/tex]

    E_inner has an electric field pointing outward at +1.05E-4 N/C. And E_outer is producing an electric field pushing the opposite direction of -1.78E5 N/C. Can I say that the total electric field at r=.86 is -1.78E5 N/C?
     
  9. Sep 5, 2009 #8

    Doc Al

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    Staff: Mentor

    Definitely.

    You can even just use Lambda_net = Lambda_inner + Lambda_outer.
    I didn't confirm your arithmetic, but the radius needed is 8.6cm = 0.086m.

    You definitely have the right idea to use superposition to find the net field.
     
  10. Sep 5, 2009 #9
    Thanks for the .086 error. Problem worked out fine.
     
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