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Gauss Law? Average charge density

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi. A cylinder of radius r & length L whose charge density distribution is given by
    ρ = C/2 * r3
    where r = radial distance in cylindrical coordinates
    C = constant
    show that the average charge density ρbar = a3 C / 5


    2. Relevant equations
    Gauss differential law div E = ρ / ε0

    div E = 1/r * ∂/∂r * (rEr) + 1/r ∂E∅/∂∅ + ∂Ez/∂z


    3. The attempt at a solution

    Hmmm! Not really sure where to begin with this.
    Not sure if it's to do with stuff as "difficult" as equations above.
    * * Scratches head. Help please! * *
     
    Last edited: Apr 6, 2013
  2. jcsd
  3. Apr 6, 2013 #2

    Andrew Mason

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    What is a?

    Also it is not clear whether: ρ = C/(2*r3) or ρ = (C/2)*r3

    What does average charge density mean? Can you determine the total charge in the cylinder?

    AM
     
  4. Apr 6, 2013 #3
    Ah. Detail.
    Get it right, me.

    a is radius of cylinder
    r is radial distance in cylindrical coordinates
    first equation is
    ρ = (C/2)*r3

    As far as "average" goes I'm assuming to be the mean.

    Hope this clears things up.
     
  5. Apr 6, 2013 #4

    Andrew Mason

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    Ok. But give us an expression for it.

    AM
     
  6. Apr 6, 2013 #5

    Andrew Mason

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    Hint: average charge density is: total __________/ total _________

    AM
     
  7. Apr 7, 2013 #6
    I'll try total charge / total area.
     
  8. Apr 7, 2013 #7

    Andrew Mason

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    Why area? This is a solid cylinder. Hint: what are the units of mass density? By analogy, what are the units of charge density?

    AM
     
    Last edited: Apr 7, 2013
  9. Apr 8, 2013 #8
    Aha! It's volume.
    Density has units of mass per unit volume such as g/ml
    so charge density must be units of charge per unit volume such as coulombs/ m3 for volume such as the Gaussian surface (cylinder)
    Thanks I shall post again soon.
     
  10. Apr 8, 2013 #9

    Andrew Mason

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    I am not sure why you are referring to Gauss' law here. You are not calculating the electric field so there is no need to think of the cylinder surface as a Gaussian surface for this problem. You are just trying to find the total charge and total volume.

    AM
     
  11. Apr 8, 2013 #10
    Ah! It all clears.
    Clouds have gone. It's not just about algebra after all.
    ρ = charge density distribution which is distribution at a single point, so have double integral
     

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  12. Apr 8, 2013 #11

    Andrew Mason

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    You don't really need the double integral. You know that the cylindrical surface with radius r and dr thick has volume 2πrLdr. So you just have to integrate from r=0 to r=a to find the charge.

    [itex]q = \int_0^a ρdV =\int_0^a ρ2\pi rLdr [/itex]

    Then it is just a matter of dividing by the total volume.

    AM
     
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