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Cylinder rod charge density (Gauss Law)

  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Theres a cylinder rod inner radius a outer raidus b.we want to find electric field between a and b,like point r (or radius r) a<r<b.
    2. Relevant equations
    Gauss Law
    3. The attempt at a solution
    I am trying to find Q enclosed but something make me confused.I am taking an integral and I found ##E=\frac {Q} {2π(b^2-a^2)ε_0}## is this correct ?
     
  2. jcsd
  3. Mar 11, 2017 #2

    TSny

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    Your result for E doesn't look correct.

    The problem doesn't give any information about the length of the cylindrical rod or how the charge is distributed inside the rod.

    When using Gauss' law, clearly state the geometry of the Gaussian surface and show your steps in arriving at the result.
     
  4. Mar 11, 2017 #3
    Its lenght is infinite and charge distribution is uniform.I found now ##E=/frac {pr} {2e_0}##
     
  5. Mar 11, 2017 #4
    I was thinking it was wrong too cause it was looking so weird.
     
  6. Mar 11, 2017 #5
    If last result is true I ll be happy
     
  7. Mar 11, 2017 #6

    haruspex

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    That makes no sense. Did you mean ##E=\frac {\rho r} {2e_0}##? ##E=\frac {\pi r} {2e_0}##? Or....?
     
  8. Mar 11, 2017 #7
    oh sorry I typed from my phone . ok İts ##E=\frac {\rho r} {2e_0}##
     
  9. Mar 11, 2017 #8

    haruspex

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    I think that would right for a=0. Clearly it must depend on a though (e.g. consider r=a).
     
  10. Mar 11, 2017 #9
    I ll share the whole question to make things clear.Its confusing like this
     
  11. Mar 11, 2017 #10
    Cylindrical Shell: Consider an infinite cylindrical shell with inner radius a, outer radius b and uniform volume charge density ρ. Consider a point at a distance r from the central axis. Compute the electric field at this point
    (a) when the point is inside the cavity of the cylinder (r < a),
    (b) when the point is within the shell (a < r < b),
    (c) and when the point is outside (b < r).
    (d) Sketch the E vs r graph. (Here E is the radial component of the electric field.)
     
  12. Mar 11, 2017 #11
  13. Mar 11, 2017 #12
    I ll share my steps
     
  14. Mar 11, 2017 #13
    I changed my approach.

    Lets take a lenght ##L## horizontal so Total Flux will be ##E(2πrL)##

    ##Q_{enc}=\int ρ \, dv##
    ##dv=π2rLdr##
    ##ρ## is constant so
    ##Q_enc=ρ\int_f^g π2rL \, dr##
    What should be ##f## and ##g## ?
    ##f=a , g=r## ?
     
  15. Mar 12, 2017 #14

    haruspex

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    Yes.
     
  16. Mar 12, 2017 #15
    ok thanks.
     
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