We are asked to prove that the capacitance of two infinitely long uniformly charged conductors with coaxial cylindrical surfaces is given by \begin{equation} C= \dfrac{2\pi\epsilon_0L}{\mathrm{ln}\left(\dfrac{b}{a}\right)} \end{equation}

Now, I was successful in proving the relation but I have a conceptual doubt.

Doubt: If the the inner cylinder was hollow (It was solid in the given problem), would the capacitance be different?

My guess: I guess it won't be different. Because the capacitance depends on the potential between the surfaces and does not care what happens beyond the surfaces.

Is this reasoning correct? If not, please explain where is this logic flawed?


For conductors all excess charge resides on the surface. Therefore, if your conductor was hollow the charge distribution would not be any different than if you had a solid cylindrical conductor. Your results will not be any different.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much $\endgroup$ – Nirmal Padwal Jun 4 '19 at 8:57
  • $\begingroup$ @NirmalPadwal Glad I could help $\endgroup$ – BioPhysicist Jun 4 '19 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.