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Nonuniform Volume Charge in a Cylinder

  1. Aug 25, 2012 #1
    The problem states:
    "An infinitely long cylinder of radius a in free space is charged with a volume charge density p(r)=p0(a-r)/a where (0<=r<=a), where p0 is a constant and r the radial distance from the cylinder axis. Find the charge per unit length of the cylinder."

    Q = ∫ p dv

    My attempt:

    dv = ∏(r^2)h dr

    ∫0 to a ( p0(a-r)/a) h∏r^2 dr

    After integrating I get

    Q = p0∏h * (a^3)/12

    Could someone please confirm if this looks correct?

    Thanks!
     
  2. jcsd
  3. Aug 25, 2012 #2

    vela

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    It's not. The units don't work out.
     
  4. Aug 25, 2012 #3
    Thank you.

    Should the units be C/m^3?

    Is the integral set up correctly? If not any advice as to where I have gone wrong?
     
  5. Aug 25, 2012 #4

    vela

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    Q is charge, right? So the units of Q should be simply coulombs. Try tracking down where the units went awry in your calculation. That's a good way to identify where the problem is.

    Also, note that the question is asking for Q/L, where L is the length of some section of the cylinder. Your final answer should have units of coulombs/meter.
     
  6. Aug 26, 2012 #5
    Yes I know that Q is the charge but I can't seem to figure this out.

    In the problem a volume charge density was given. The only definition I can find is:
    p=dQ/dv. Knowing this I can integrate the volume charge density given to get the charge, however, I'm unsure how to come to charge per unit length.
     
  7. Aug 26, 2012 #6
    Hello ,
    Please tell me what element are you integrating ?
     
  8. Aug 26, 2012 #7
    Hi I was integrating the volume charge density:

    dv = v dr

    ∫ p dv = ∫ p πr2l dr = p0∫((a-r)/a) * ∏r2l dr.

    This should give me the charge I believe but I don't know how to get charge per unit length.

    Thanks!

    Edit: Integrating from 0 to a.
     
  9. Aug 26, 2012 #8

    vela

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    This is obviously wrong. The lefthand side has units of length3, since it's a volume, whereas the righthand side has units of length4.
     
  10. Aug 26, 2012 #9
    You need to take an infinitesimal element is such a way it is symmetric to the body , with a variable good enough to integrate .
     
  11. Aug 26, 2012 #10
    Alright. So the statement below should be the correct statement?
    v= ∏r2h
    dv = 2∏rh dr

    I assumed I would just substitute this into the integral for dv and integrate with respect to r from 0 to a.

    Doing this I get ∏ρ0ha2/3.
     
    Last edited: Aug 26, 2012
  12. Aug 26, 2012 #11

    vela

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    Good. So you have in a length h of the cylinder, a total charge ##Q=\frac{\rho_0 \pi h a^2}{3}##, so what's the charge per unit length?
     
  13. Aug 26, 2012 #12
    I may be over thinking this but the unit length is given by λ = E 2∏ ε0r?

    Is there an easier way to get the charge per unit length? Thanks for your patience I am new to this.
     
  14. Aug 26, 2012 #13

    vela

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    I kinda get the feeling you're just pushing symbols around without understanding what they mean. Say you're trying to explain your calculation to someone else, and he or she asks, "What does Q stand for?" And you say "charge." They respond by asking "charge of what specifically?" What's your answer?
     
  15. Aug 26, 2012 #14
    I would say your feeling is correct. I'm really am trying to understand but am having a hard time doing so.

    The total charge over a length l?
     
  16. Aug 28, 2012 #15

    vela

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    What is l the length of?
     
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