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In the following exercise:

enter image description here

We are asked to calculate the electric field at a certain distance from an exis where a rod is located. Previously, we are asked to calculate the charge density a cylindrical thin conductor of radius R which axix happens to meet the already named.

I calculated the electric field using Gauss' theorem and taking only into account the rod, and I obtained the electric field of the solution.

So that is my question, why does the cylindrical crust do not influence the electric field? Is it because the crust is conductive?

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Gauss's law says that the electric field depends only on the charge enclosed of your Guassian surface. In this problem, the cylindrical crust is electrically neutral. Thus, if you draw your Guassian surface to enclose the entire electrically-neutral crust, there is net zero charge on the crust and thus there is no change to the $Q_{enc}$ term in Guass's law.

\begin{equation} \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \end{equation}

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  • $\begingroup$ But there is an induced charge on the Cylinical crust, why is it electrically neutral? $\endgroup$
    – MSU
    Commented May 21 at 18:32
  • $\begingroup$ Presumably the rod would induce charges in the inner surface of the conductor, altough the field would be zero inside anyway. I think a more accurate answer would be that the electric field does depend on the conductor, it's just that by its properties (being infinitely conductive), whatever happens outside doesn't care about it, like you described in your answer. $\endgroup$
    – Lourenco Entrudo
    Commented May 21 at 18:34
  • $\begingroup$ We are given that the cylindrical crust is electrically neutral to begin with. There is an induced surface charge on the inner surface, but because the crust is electrically neutral there is an equal and opposite charge on the outer surface. Thus, if you draw your Gaussian surface to enclose the entire crust, there is net zero charge. However, if you draw your Gaussian surface to enclose only the inner surface, then there is a net charge contributed by the crust. You will find that this net charge exactly cancels out the charge of the rod. The electric field inside a conductor is zero. $\endgroup$
    – PineappleThursday
    Commented May 21 at 18:36
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    $\begingroup$ @LourencoEntrudo so if the rod induces charges in the inner surface of the conductor, in the outer surface there would be also charges that would keep the net charge in the conductor zero? $\endgroup$
    – MSU
    Commented May 21 at 18:37
  • $\begingroup$ @PineappleThursday all right, now I get it, thank you very much $\endgroup$
    – MSU
    Commented May 21 at 18:37

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