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New intregral same old problem

  1. Mar 3, 2005 #1
    A stright line segment of length 2L is placed on the z-axis with its midpoint at the origin. The segment has a linear charge density given by...
    [tex] \lambda = \frac{aQ}{z^2 + a^2} [/tex]
    where Q and a are constants, a>0 find the electrostatic potential of this charge distribution at a point on the x-axis in Cartesian coords.

    [tex] \Phi = \int_{-L}^{L} \frac{k_e dt \frac{aQ}{t^2+a^2}}{\sqrt{x^2 + t^2}}[/tex]
    [tex] = k_e Q a \int_{-L}^{L} \frac{dt}{(t^2 +a^2) \sqrt{t^2 + x^2}} [/tex]

    what do now?....no idea what subsitution will make this work,any suggestions?
     
    Last edited: Mar 3, 2005
  2. jcsd
  3. Mar 3, 2005 #2

    dextercioby

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    Yes,u may put your integral under the form
    [tex] \Phi=2kQa\int_{0}^{L}\frac{dt}{(t^{2}+a^{2})x\sqrt{\left(\frac{t}{x}\right) ^{2}+1}} [/tex]

    and do the substitution:
    [tex] \frac{t}{x}=\sinh u [/tex]

    Daniel.
     
    Last edited: Mar 3, 2005
  4. Mar 3, 2005 #3

    dextercioby

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    Okay,if you do that substitution,u'll end up with these integrals (actually the second).

    Daniel.
     

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  5. Mar 3, 2005 #4
    ya....instead of that good idea (which it is) how about...
    [tex] = k_e Q a \int_{-L}^{L} \frac{dt}{(t^2 +a^2) \sqrt{t^2 + x^2}} [/tex]

    [tex] = \frac{k_e Q a}{x} \int_{-L}^{L} \frac{dt}{(t^2 +a^2) \sqrt{(\frac{t}{x})^2 + 1}}, \ \frac{t}{x} = tan(u) [/tex]

    [tex]t^2 = x^2 tan^2 (u), \ dt = x sec^2 (u) du [/tex]

    [tex]= \frac{k_e Q a}{x} \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{dt}{(x^2 tan^2 (u) +a^2) \sqrt{tan^2 u + 1}}[/tex]

    [tex]= \frac{k_e Q a}{x} \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{x sec^2 (u) du}{(x^2 tan^2 (u) +a^2) sec(u)}[/tex]

    [tex]= k_e Q a \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{sec (u) du}{(x^2 tan^2 (u) +a^2)}, \ c= \frac{a}{x}[/tex]

    [tex]= k_e Q a \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{sec (u) du}{x^2( tan^2 (u) + c^2)}[/tex]

    [tex]= \frac{k_e Q a}{x^2} \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{sec (u)}{(tan^2 (u) + (\frac{a}{x})^2}[/tex]


    [tex]=\frac{2 tan^{-1}(\frac{L\sqrt{x^2-a^2}}{a\sqrt{x^2+L^2}}{1}[/tex]

    does this all look legal? if theres an x^2 in the last one, idk why its there it's not supposed to be...
     
    Last edited: Mar 3, 2005
  6. Mar 3, 2005 #5

    dextercioby

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    Correct that latex errors.It looks good.That substitution leads to the same answer.I still prefer hyperbolic trig.functions.

    Anyway,your problem is solvable.It depends on u which eq.u prefer most.

    Daniel.
     
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