How Does Gauss' Law Apply to an Insulated Cylindrical Shell and Rod System?

[Faiq]

Homework Statement

Question
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An infinitely long insulating cylindrical rod with a positive charge $\lambda$ per unit length and of radius $R_1$ is surrounded by a thin conducting cylindrical shell (which is also infinitely long) with a charge per unit length of $-2\lambda$ and radius $R_2$.

Find the radial component of electric field at all points in space.

https://prnt.sc/iaeer9
Confusion
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I was able to calculate the electric field outside the cylindrical shell.
However a problem I am having is calculating the electric field inside the cylindrical shell.

I have some trouble reconciling the fact that the electric field inside the cylindrical shell is independent of the charge present on the cylindrical shell, as evidenced by Gauss law since constructing a Gaussian surface inside the cylindrical shell will result in enclosing of the rod.
This implies that whether the shell is present or not, the electric field inside the shell is only due to the rod.
Why is that the case?

My argument against the above reasoning is Gauss law is inapplicable in this situation since the shell is insulating and hence the charges present on it will contribute to providing an electric field. Consequently, the electric field at any $r$ such that $R_1<r<R_2$ will not only be affected by the rod but also by the electric field produced by the shell.

 

[kuruman]

Consequently, the electric field at any rr such that R₁<r<R₂ will not only be affected by the rod but also by the electric field produced by the shell.

Suppose the inner rod is removed and you have just the outer charged cylindrical shell. What do you think the electric field is inside the shell? Can you draw electric field lines? Remember they have to start at positive charges and stop at negative charges.

 

[jtbell]

My argument against the above reasoning is Gauss law is inapplicable in this situation since the shell is insulating and hence the charges present on it will contribute to providing an electric field.

Yes, the charge on each small piece of the shell produces an electric field at a point inside the shell. Nevertheless, the sum of the contributions at that point from all pieces of the shell is zero. For a spherical shell, this is part of the famous shell theorem which was first proved by Isaac Newton:

https://en.wikipedia.org/wiki/Shell_theorem

I've never seen it done for an infinitely-long cylindrical shell, but I expect that a similar integral would give the same result for that case.

 

[NFuller]

This implies that whether the shell is present or not, the electric field inside the shell is only due to the rod.
Why is that the case?

This a special case of Newton's shell theorem. https://en.wikipedia.org/wiki/Shell_theorem

The theorem states that for fields with $1/r^{2}$ dependence, the field is zero for any point within a uniform spherical shell. The theorem also works for an infinite cylinder since this geometry simplifies to a 2-d sphere (a circle). The link above describes both a geometric and calculus based proof of the theorem.

 

[Chandra Prayaga]

What did you get for the electric field inside the conducting material? The problem asks for the electric field for all values of the radius.

 

[rude man]

Appeal to Friend Gauss!