1. Limited time only! Sign up for a free 30min personal 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!

Limits and infinity problem

  1. Aug 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi, I need to figure out what happens to this equation in the limits

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

    in the two different cases

    that z>>L

    and when L -> infinity

    (note this equation was derived from finding the electric field da distance z, above the midpoint of a straight line segment of length 2L, which carries a uniform line charge of [tex] \lambda [/tex]

    3. The attempt at a solution

    for the case when, z>>L I can see how the L term becomes insignificant in the square root on the bottom,
    and so the equation would just become
    [tex]\frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z^2} [/tex]

    but for the case when L approaches infinity, what do I do???
    the squareroot of a L^2 +z^2 == L?
    does that mean the L can just be canceled out?
    [tex] \frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z *L} [/tex]
    and the equation becomes?
    [tex] \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{z} [/tex]

    i'm not sure if i'm allowed to since the Z was the distance from the midpoint of the line,

    the first one makes sense since it just becomes a point charge of 2lambda L
    but the second case, i'm not too sure what it becomes
  2. jcsd
  3. Aug 21, 2010 #2


    User Avatar
    Homework Helper

    Re: Limits

    consider the form of the field produced by an infinite conductor, which should agree with your 2nd result
  4. Aug 21, 2010 #3


    User Avatar
    Homework Helper

    Re: Limits

    it helps to re-write the expression in the following form so it is clear when you take the limit

    first as follows
    [tex] E
    = \frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z \sqrt{z^2+L^2}}
    = \frac{\lambda}{ 2 \pi\epsilon_0} \frac{ L}{z^2} \frac{1}{\sqrt{1+\frac{L^2}{z^2}}}

    \approx \frac{\lambda}{ 2 \pi\epsilon_0} \frac{ L}{z^2} (1-\left(\frac{L}{z}\right)^2 + O\left(\frac{L}{z}\right)^4)

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook