I was wondering the following. The formula for a cylindrical capacitor is known. However, the formula shows a solid cylinder inside a hollow cylinder. How would having a hollow cylinder inside another hollow cylinder change the capacitance of this cylindrical capacitor?

My understanding leads me to think the formula will remain the same, as the conductors will build the same charge regardless.

Please let me know if this is not the way to think about this problem.


1 Answer 1


You are correct. It makes no difference whether the inner conductor is solid or hollow. All the charge on a conductor resides on its surface. When you place a free charge $Q$ on the inner conductor, that charge will all run to the outer surface. The individual charges repel one another, and they spread out to minimize their potential energy; this configuration places them all on the outer surface. Consequently, it does not matter what the internal structure of the conductor is, whether it is solid or hollow (or anything else).

A similar argument actually applies to the outer conductor. The charge $-Q$ on the outer cylinder will all be located along its inner surface, regardless of how thick the conductor. Thus the capacitance per unit length of two coaxial cylinders is $$C_{0}=\frac{2\pi\epsilon}{\ln(b/a)},$$ where $a$ is the outer radius of the inner cylinder, and $b$ is the inner radius of the outer cylinder. ($\epsilon$ is the permittivity of the insulating material between the cylinders; if there is no dielectric material there, then it takes the vacuum value $\epsilon_{0}$.)


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