Electric Potential within coaxial cylinder

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SUMMARY

The discussion focuses on calculating the electric potential within a coaxial cylinder system, specifically addressing the potential in three regions: inside the inner cylinder (r < a), between the cylinders (a < r < b), and outside the outer cylinder (r > b). The potential for the region inside the inner cylinder is derived as [(λ)/(2πε)]*[ln(b/a)], with the reference point set at the outer cylinder. The confusion arises regarding the integration limits and the nature of the inner cylinder, which is confirmed to be a metal but not a hollow conductor, leading to a zero electric field inside a conductor at equilibrium.

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Homework Statement


A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow metal tube with radius b. Thew positive charge per unit length on the inner cylinder is \lambda, and there is an equal negative charge per unit length on the outer cylinder. Calculate the potential for r < a; a < r < b; r > b.


Homework Equations


EA=q/epsilon
Va - Vb = \intE.dl


The Attempt at a Solution


I really just need help figuring out why the answer for when r < a is:
[(lambda)/(2(pi)(epsilon))]*[ln(b/a)]
The reference point here is b. If we are looking for the potential INSIDE the smaller cylinder with radius a, then why are we only integrating from b to a? Shouldn't it be from b to 0? Or is the potential inside the smaller cylinder constant? Why would that be?
 
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Or is the potential inside the smaller cylinder constant? Why would that be?

Well what IS the potential inside a hollow conductor at equilibrium? The electric field is zero, right? so...
 
The smaller cylinder is not hollow and I don't think its a conductor (and its supported on an insulating stand anyway)... unless being "metal" is synonymous with conductor? E inside a conductor is 0, but I don't think the smaller cylinder is a conductor.
 

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