Integral problem on electric potential

AI Thread Summary
The discussion centers on calculating the electric potential V(r) for a long metal cylinder within a hollow metal tube, where the inner cylinder has a positive charge and the outer cylinder has an equal negative charge. The key equation used is Va - Vb = ∫E.dr, derived from Gauss's law. A question arises regarding the limits of integration, specifically why they extend from a to b when considering points where r < a. The confusion stems from understanding how points inside the inner cylinder can experience potential from the external electric field. The potential at any radius r is ultimately expressed in terms of integrals involving the electric field E, with the potential at infinity set to zero.
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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 = \intE.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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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}
 
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