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Electric field of a wire segment

  1. May 16, 2005 #1
    This is a solved example in Griffith's book.
    Question: Find the electirc field a distance 'z' above the midpoint of a straight line segment of length 2L, which carries uniform charge [itex]\lambda[/itex]

    Solution: Let me work out the steps and show you where my problem is(I have avoided showing some of the steps).
    The coordinate axis has been set up taking the midpoint of the wire as the origin. Magnitude of the electric field is given by:
    [tex] E = \frac{1}{4\pi \epsilon_0} \int_{0}^{L} \frac{2\lambda z}{(z^2 + x^2)^{3/2}} dx[/tex]

    HERE, Griffith has not elaborated the evauation of the integral under the bracket and directly given the solution. I have tried my level best to solve it but couldn't suceed. So can someone show me the intermediate steps involved in evaluating the integral:
    [tex]\int_{0}^{L}\frac{1}{(z^2 + x^2)^{3/2}} dx[/tex] ([itex]\lambda[/itex] & z are constants)

    For your convenience, the final solution is:
    [tex]E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z\sqrt{z^2 + L^2}}][/tex]
  2. jcsd
  3. May 16, 2005 #2
    to deal with the integral [tex]\int \frac{1}{(z^2+x^2)^{3/2}} dx[/tex]
    try putting [tex]x = z \tan\theta[/tex]

    and using the identity [tex]1+\tan^2\theta = \sec^2\theta[/tex]
  4. May 19, 2005 #3
    Thank you for the guidance. Here is my evalution of the integral:
    [tex]\int_{0}^{L} \frac{z^2\sec^2 \theta d\theta}{({z^2\tan^2 \theta +z^2})^{3/2}}[/tex]
    [tex]= \frac{1}{z^2} \int_{0}^{L} \cos \theta d\theta[/tex]
    [tex]= \frac{1}{z^2} [\sin \theta] \mid_{0}^{L}[/tex]
    [tex]= \frac{1}{z^2} \frac{x}{\sqrt{z^2 + x^2}} \mid_{0}^{L}[/tex]
    [tex]=\frac{1}{z^2} \frac{L}{\sqrt{z^2 + L^2}}[/tex]
  5. May 19, 2005 #4
    Sticking to the original question, electric field is:

    [tex]E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z\sqrt{z^2 + L^2}}][/tex]
    How will the formula be modified if [itex]L\rightarrow \infty[/itex]
  6. May 19, 2005 #5

    Doc Al

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    Staff: Mentor

    Consider that [tex]\sqrt{z^2 + L^2} = L \sqrt{1 + (z/L)^2}[/tex]

    Now can you see what happens when [itex]L\rightarrow \infty[/itex] ?
  7. May 20, 2005 #6
    Wow!! Thanks Doc Al :smile:.

    [tex]E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z L\sqrt{1 + (z/L)^2}}][/tex]

    So,[itex](z/L)^2\rightarrow 0[/itex]

    [tex]E = \frac{2\lambda}{4\pi \epsilon_0 z}[/tex]
  8. Jul 14, 2011 #7


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    Please note, that the solution in a limit of L→∞ is nonphysical. The total charge of such system is infinite. That leads to another singularity: the electrostatic potential at any point of the space is also infinite.
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