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!

Electric field of an infinite charged rod

  1. Sep 8, 2009 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    I tried to derive the electric field of an infinite (non conductor and conductor, I believe it is the same) charged rod.

    2. Relevant equations

    [tex]\oint \vec E d \vec A = \frac{Q_{\text {enclosed}}}{\varepsilon _0}[/tex].


    3. The attempt at a solution

    I could do all, except at the end... when he wrote that [tex]\oint \vec E d \vec A=E 2 \pi rL[/tex]. I understand that [tex]\oint d\vec A = 2\pi rL[/tex], but I don't understand how he could pass the [tex]E[/tex] outside the line integral, as if [tex]E[/tex] was constant. Because it isn't constant, [tex]E[/tex] depends on [tex]r[/tex].

    I would have understood this step if we were to derive the electric field due to an infinite charged plane, where [tex]E[/tex] does not depend on the distance between a charged particle and the plane.

    Can you explain me why does [tex]\oint \vec E d \vec A=E 2 \pi rL[/tex]?
    Thanks in advance!


    Here's the link : http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html [Broken].
    (look at the very bottom of the page)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 9, 2009 #2

    fluidistic

    User Avatar
    Gold Member

    Nevermind, I got it. E is constant over the cylinder' surface!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electric field of an infinite charged rod
Loading...