(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 [tex]\lambda[/tex], 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.

2. Relevant equations

EA=q/epsilon

Va - Vb = [tex]\int[/tex]E.dl

3. 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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# Homework Help: Electric Potential within coaxial cylinder

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