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Improper integral help: coloumb's law

  1. Jul 5, 2008 #1
    1. The problem statement, all variables and given/known data
    i was deriving an infinite line of charge formula by coloumb's law:
    so i got stuck with this integral (since it is in the maths forum)
    [tex] \vec{E}_{\rho} = \int_{-\infty}^{\infty} \frac{\rho_L \rho dz}{4\pi\epsilon_o ({\rho}^2 + z^2)^{\frac{3}{2}}} [/tex]


    2. Relevant equations
    where
    [tex]{\rho}_L = [/tex] linear charge density
    [tex]{\rho} = [/tex] direction perpendicular to z axis in cylindrical coordinates


    3. The attempt at a solution
    so when integrating (using trigo substitution):
    [tex] \vec{E}_{\rho} = \frac{\rho_L\rho}{4\pi\epsilon_o} (\frac{1}{{\rho}^2} \frac{z}{p^2 + z^2})_{-\infty}^{\infty} [/tex]
    this is where i got stuck
    no matter how i use lhopitals rule in this equation:
    \lim_{z \infty} \frac{z}{({\rho}^2 + z^2)^{\frac{1}{2}}
    it keeps going back

    because the infinite is supposed to add to equate to 2
    the answer which is
    [tex] \vec{E}_{\rho} = \frac{\rho_L}{2\pi\epsilon_o\rho}[/tex]
     
    Last edited: Jul 5, 2008
  2. jcsd
  3. Jul 5, 2008 #2

    HallsofIvy

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    I think you need to show your work here. This is not at all what I got.

    You don't need to use L'Hopital's rule- just divide numberator and denominator by z2. Of course, then the limits are both 0 which is not what you want.

    I would prefer to change the limits of integration with each substitution.
     
  4. Jul 5, 2008 #3
    yeah typo:
    [tex](\frac{z}{\sqrt{p^2 + z^2}})_{-\infty}^{\infty}[/tex]

    the problem is differentiating [tex]\sqrt{p^2 + z^2}[/tex] when i keep using LHR it
    keeps on going back

    edit:
    wait did you mean this?
    [tex] \frac{\frac{z}{z}}{\frac{\frac{\sqrt{p^2+z^2}}{z^2}}}} [/tex]
    i tihink i get it now

    just divide both sides by z
    the denominator you divide by sqrt(z^2)

    thx alot halls
     
    Last edited: Jul 5, 2008
  5. Jul 5, 2008 #4

    HallsofIvy

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    You say you used a trig substitution so why worry about the integral as a function of z at all? Change the limits of integration as you change the variable.

    When I did it, my first substitution was [itex]\rho tan(\theta)= z[/itex]. z will go to [itex]-\infty[/itex] and [itex]\infty[/itex] as [itex]\theta[/itex] goes from [itex]-\pi/2[/itex] to [itex]\pi/2[/itex]. I then integrated the resulting trig function by letting [itex]u= sin(\theta)[/itex]. While theta goes from [itex]-\pi/2[/itex] to [itex]\pi/2[/itex], u goes from -1 to 1 so I only had to evaluate the "u" formula between -1 and 1 without having to go back to the original variable.
     
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