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Nonuniformly Charged Semicircle

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A thin glass rod is a semicircle of radius R, see the figure. A charge is nonuniformly distributed along the rod with a linear charge density given by [tex]\lambda=\lambda_{0} + sin(\theta)[/tex], where [tex]\lambda_{0}[/tex] is a positive constant. Point P is at the center of the semicircle.

    GIANCOLI.ch21.p50.jpg

    Determine the acceleration (magnitude and direction) of an electron placed at point P, assuming R = 1.9 and [tex]\lambda_{0}[/tex] = 2.0 [tex]\mu C/m[/tex].

    2. Relevant equations
    Coulomb's Law [tex]F_{e}=kq_{1}q_{2}/r^{2}[/tex]
    Electric Field Equation [tex]E=F_{e}/q[/tex]
    Newton's 2nd Law

    3. The attempt at a solution
    I attempted to use Newton's 2nd Law to solve this problem. I know the mass of an electron, and since gravity can be ignored, I knew that all I needed to do was to find the electric force on the electron to find the acceleration. I know the the force points straight up do to the positive charge above and the negative charge below, but I'm having trouble quanitfying the forces. I attempted to use Coulomb's Law by plugging in the charges for just one electron and proton, but I know that's not right. Please help!
    (Also, let me know if the image disappears, it should work out fine though...)
     
  2. jcsd
  3. Sep 3, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi saxyliz ! Welcome to PF! :smile:

    (have a theta: θ and a lambda: λ and a mu: µ :wink:)
    Are you sure it isn't λ = λ0 times sinθ (that seems more consistent with the image)? :confused:
     
  4. Sep 3, 2009 #3
    Re: Welcome to PF!

    Oh yes sorry! It's times! Silly me!
     
  5. Sep 3, 2009 #4

    tiny-tim

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    :rolleyes: :smile:
    ok, then … yes, the force will point up, so you only need the up-components.

    But you must slice the semicircle into bits of length dθ, and plug in the charge for that. :smile:
     
  6. Sep 3, 2009 #5
    So, dq = [tex]\lambda d \theta[/tex]? Then I plug that into Coulomb's Law to solve for dE, but how do I solve for the components?

    To be more specific, what I get is:

    [tex]dE=k\lambda d \theta /R^{2}[/tex]

    Which is the same as

    [tex]dE=k\lambda_{0} sin(\theta) d \theta /R^{2}[/tex]

    Do I use a sector for the components? Like polar coordinates?
     
    Last edited: Sep 3, 2009
  7. Sep 3, 2009 #6

    tiny-tim

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    Hi saxyliz! :smile:

    (what happened to that θ and λ i gave you? :confused:)
    No, dq = λ sin θ dθ.

    For components, you use cos of the angle as usual … what is worrying you about that? :confused:
     
  8. Sep 3, 2009 #7
    Sorry! I'm totally new to this and didn't get what you meant in the first place (about the θ, etc.)

    Why isn't it λ_0 ? Wouldn't it be sin^2 then?

    I think I get what you mean about the components, give me a second.

    EDIT:
    I got it!! Thanks so much for your help!
     
    Last edited: Sep 3, 2009
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