Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook