Integral problem on electric potential

[venom_h]

Homework Statement

A long metal cylinder with radius a is held on the axis of a long, hollow, metal tube with radius b. The inner cylinder has positive charge per unit length $$\lambda$$, and the outer cylinder has an equal negative charge per unit length. Calculate the potential V(r) for r<a

Homework Equations

Va-Vb = $$\int$$E.dr, where E can be found by Gauss's law

The Attempt at a Solution

My only problem is why the limit for the integral is from a to b even though r < a ??
How does any point r < a experience a potential outside??
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[learningphysics]

If you take the potential at infinity to be 0... then the potential at any radius r is,

$$V(r) = -\int_{\infty}^r \vec{E}\cdot\vec{dr}$$

assuming b is the outer radius, and a is the inner radius

$$V(a) = -\int_{\infty}^a \vec{E}\cdot\vec{dr} = -\int_{\infty}^b \vec{E}\cdot\vec{dr} - \int_{b}^a \vec{E}\cdot\vec{dr}$$

 

[m2003]

Response]

Thank you for your question. The integral limit from a to b is actually the correct range for this problem. This is because the electric potential at a point r is determined not only by the charges located at that point, but also by the charges located outside of that point. In this case, even though the inner cylinder has a radius of a, the charges on the outer cylinder with radius b still contribute to the electric potential at r < a. This is because the electric field lines from the outer cylinder can extend into the region inside the inner cylinder, affecting the potential at that point. Therefore, the integral must take into account the contributions from both cylinders, resulting in the limits from a to b. I hope this explanation helps clarify the concept.