1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E field above a plane

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    This is problem 47(chapter 21) in the text book - Physics for engineers and scientists (Giancoli)
    Uniformly charged wire has length L, where point 0 is the mid point. Show that the field at P, perpendicular distance x from 0 is

    E= (lambda/2*pi*epsilon_0) *(L/x*sqrt(L^2+4x^2)

    2. Relevant equations

    3. The attempt at a solution

    I tried solving it, I got E = - (lambda/2pi*epsilon_0)*[1/sqrt(L^2+4x^2)-1/2x)
    Is something wrong with my integration?

    My attempt is correct until

    E= lamda/r^2*4pi*epsilon_0 * integration (cos theta)dl

    After this, I use cos theta = x/r (r is the hypotenuse = sqrt (L^2+4x^2))

    I am trying to do this instead of taking r=x cos theta.
  2. jcsd
  3. Feb 3, 2009 #2


    User Avatar
    Homework Helper

    Looks like something wrong with the "(cos theta)dl."
    I'll use A instead of theta. And z in place of your l, running from -L/2 to L/2.
    I get some constants times integral of cos(A)/(x^2 + z^2)*dz
    Since tan(A) = z/x, I can use z = x*tan(A) to simplify the integral.
    And dz = x*sec^2(A) dA
    After the dust settles on this change of variable, I get integral of cos(A)dA.

    Not the difference from your integral: I have dA where you have dl
    I end up with the given answer.
  4. Feb 3, 2009 #3
    If I use r=cos theta/x, then i get the correct result.
    I am trying to get there without that substitution and just cos theta = x/sqrt(x^2+(l/2)^2)
    huh. Doesn't work if I use the integral but does work if I take l/2 as y initially. The whole term becomes (x^2+y^2) and then apply limit to get answer.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?