Quick question, i think, link to problem included

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The discussion centers on understanding charge density in relation to a problem involving a charged cylinder and an outer sphere. The question arises about why the radius used for calculating charge density is that of the outer sphere rather than just the charged cylinder. It is clarified that while the outer shell has a net charge of zero, it still possesses surface charges—negative on the inner surface and positive on the outer surface. The relationship between surface charge density and electric field is emphasized, with the formula surface charge density = ε0E being highlighted. Overall, the inquiry seeks clarity on the application of charge density in this specific electrostatic scenario.
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Homework Statement


http://www.physics.wisc.edu/undergrads/courses/spring09/248/HWSolutions/HW6Solutions.pdf

My question regards to part b of the problem on the 1st page.

Why is the radius to get the charge density the radius of the outer sphere and not just the radius of the cylinder that's actually charged? Like when the radius is more than the radius of the inner cylinder, that means there's no charge right so shouldn't the maximum radius be 1.5cm in this case?
 
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Although the net charge of the outer shell is zero, there are charges on its surfaces, negative on the inner surface and positive on the outer one. The surface charge density is equal to ε0E (E is the magnitude of electric field at the surface).

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