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Electric field of an infinite charged rod

  • Thread starter fluidistic
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fluidistic

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1. Homework Statement

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

2. Homework 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:

fluidistic

Gold Member
3,615
94
Nevermind, I got it. E is constant over the cylinder' surface!
 

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