Electric field of an infinite charged rod

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SUMMARY

The electric field of an infinite charged rod can be derived using Gauss's law, specifically the equation \(\oint \vec E d \vec A = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\). In the discussion, the user initially struggled with the step where \(\oint \vec E d \vec A\) is simplified to \(E 2 \pi r L\). The key insight is that the electric field \(E\) is considered constant over the cylindrical surface, allowing it to be factored out of the integral. This approach is valid due to the symmetry of the infinite charged rod.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and their properties
  • Knowledge of cylindrical symmetry in electrostatics
  • Basic calculus for evaluating integrals
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Homework Statement



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

Homework Equations



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

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 .
(look at the very bottom of the page)
 
Last edited by a moderator:
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Nevermind, I got it. E is constant over the cylinder' surface!
 

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