Conductor with shell problem getting zero as answer?

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SUMMARY

The discussion centers on a physics problem involving a cylindrical cable with a copper outer tube and an inner wire. The inner wire has a charge density of -2λ, while the outer tube carries a total charge of +λ. The surface charge density (σ) at the outer surface of the tube, located at 4R, was calculated incorrectly, resulting in a value of zero. The correct approach involves applying the formula σ = q/A, taking into account the distribution of charge across the surfaces of the tube.

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AGGENGR
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1. Problem Statement:

The figure shows a cross-section view of a very long cylindrical cable. There is an outer tube
made of copper, inner radius 2R, outer radius 4R. The inner copper wire has radius R and is
concentric with the tube. The inner wire has charge density –2λ (per unit length), while the tube
carries total charge +λ. [You need to show solution method to get full credit on this problem!]
(a) How does the charge arrange itself on the outer tube? On outside of course!
b) What is the surface charge density (σ) at the outer surface of the tube, at 4R? Getting zero for this one?

2. Homework Equations
sigma=q/A

3. Work.
So I used the above formula and got zero for some odd reason. I calculated the sigma inner and that's what giving me zero. Work is attached in jpeg.
 

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Hi. Check... \frac{\lambda - 4\lambda \pi R}{8 \pi R} \neq \frac{\lambda-\lambda}{2}.
 
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Ahhh makes sense now. Thanks!
 

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