Electric Potential within coaxial cylinder

AI Thread Summary
The discussion centers on calculating the electric potential in a system of coaxial cylinders, specifically for regions inside and outside the cylinders. The potential for r < a is derived as [(lambda)/(2(pi)(epsilon))]*[ln(b/a)], with the reference point at radius b. There is confusion regarding the integration limits, as the potential inside the smaller cylinder is questioned, particularly if it should be integrated from b to 0 instead. It is clarified that the electric field inside a conductor is zero, indicating that the potential remains constant within the hollow conductor at equilibrium. The distinction between the inner cylinder being a metal and its conductive properties is also addressed, emphasizing that being metal typically implies it behaves as a conductor.
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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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