Electric field of an infinite charged rod

[fluidistic]

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

$$\oint \vec E d \vec A = \frac{Q_{\text {enclosed}}}{\varepsilon _0}$$.

The Attempt at a Solution

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

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

Can you explain me why does $$\oint \vec E d \vec A=E 2 \pi rL$$?
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)

 

[fluidistic]

Nevermind, I got it. E is constant over the cylinder' surface!

 

[tomcue]

I understand your confusion about the step where E is taken outside the line integral. This step can be better understood by considering the physical meaning of the integral and the properties of an infinite charged rod.

The integral \oint \vec E d \vec A represents the flux of the electric field through a closed surface. In this case, the closed surface is a circular loop around the charged rod. Now, since we are dealing with an infinite charged rod, the electric field will have the same magnitude and direction at all points along the rod. This means that the electric field is constant along the surface of the loop.

Now, the surface area element d\vec A is perpendicular to the electric field at every point along the loop. This means that the dot product \vec E \cdot d\vec A will always give the same value (E times the area of the loop) regardless of the position on the loop. Therefore, we can take the electric field outside the integral as a constant factor.

In summary, the reason why \oint \vec E d \vec A = E 2 \pi rL is because in the case of an infinite charged rod, the electric field is constant along the surface of the loop and the surface area element is always perpendicular to the electric field. This allows us to take the electric field outside the integral and simply multiply it by the area of the loop.

I hope this explanation helps to clarify the concept for you. Keep up the good work in your studies!