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Solid cylinder with nonuniform volume charge density?

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A long, non conducting, solid cylinder of radius 4.2 cm has a nonuniform volume charge density ρ = Ar^2, a function of the radial distance r from the cylinder axis. A = 2.5 µC/m5.

    (a) What is the magnitude of the electric field at a radial distance of 3.2 cm from the axis of the cylinder?

    (b) What is the magnitude of the electric field at a radial distance of 5.2 cm from the axis of the cylinder?


    2. Relevant equations

    e0 Ø = charge enclosed ; e0 is the permittivity constant 8.85e-12 and Ø can be the flux

    e0 E(2π r L) = λ L ; λ is the linear charge density


    3. The attempt at a solution


    I thought that I could just integrate Ar^2 twice, which would give me λ. Then I could multiply by the height of the cylinder, L. After I did all that I arrived at this:

    E=(Ar^3) / (24e0) with e0 being the permittivity constant.



    The correct form would have me do this though:

    charge enclosed = ∫ 2π r L (Ar^2) dr ; from 0 to r

    = 2π L A ∫ r^3 dr ; from 0 to r

    If you integrate that and solve for E you get this:

    E=(Ar^3) / (4e0) <----- This gives the correct answer.


    Can someone explain to me why the method I'm using wrong? Is there some constant I am missing in my method? Or is it just wrong?


    And this is also my first post here on Physics Forums? Please tell me if I am doing something wrong so I can correct it in any future post.
     
  2. jcsd
  3. Feb 21, 2012 #2

    BruceW

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    welcome to physics forums! Or at least, congrats on your first post on physics forums :)

    This is what you should be doing. How exactly did you integrate?
     
  4. Feb 21, 2012 #3
    I would do this:

    ∫ A r^2 dr = A ∫ r^2 dr

    That would give me this:

    (1/3)A r^3

    Integrating again treating A as a constant would then give me this:

    (1/3)A ∫ r^3 dr = (1/12)A r^4


    I could then set λ = (1/12)A r^4 , and I could substitute λ into the equation: e0 E(2π r L) = λ L

    E = (A r^3) / (2π*e0*12) I forgot to divide by 2π. But doing that just makes the denominator bigger. :frown:
     
    Last edited: Feb 21, 2012
  5. Feb 21, 2012 #4

    BruceW

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    The integration is not correct. Remember this rule? [itex]dxdy=r \ dr \ d \theta [/itex]
     
  6. Feb 21, 2012 #5
    No. :uhh: I don't think I've ever seen that.
     
    Last edited: Feb 21, 2012
  7. Feb 22, 2012 #6

    BruceW

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    hmm. I don't think you can do this question without using that rule, so I'm surprised you haven't seen it before. It is essentially an area integral using the cylindrical coordinate system (where the area is perpendicular to the z axis). You could also directly integrate using dxdy, but it would be much more complicated.

    Anyway, if you use that rule (remember that r is inside the integral, along with the function), then you should get the right answer for lambda.
     
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