Jackson 4.13 -- Oil-filled Capacitor

Textbook

Classical Electrodynamics (Third Edition) -- J.D. Jackson

Problem Number

4.13

Solution

Jackson4-13.pdf

Other Files

Jackson4-13.tex

jackson 4.13

don't see why you have an h^2 in the gravitational PE. s/b just mgh = pi(b^2-a^2)h rho g, no? I know your answer agrees with jackson but I don't see why the extra h is there.

Clarification

Hi Peter,

Thanks for using the site, and asking this question. Allow me to clarify my thought process. You seem to agree that the gravitational potential is given by

PE = mgh

We are not given the mass of the oil, so let's replace it by density times volume

PE = ρVgh

The volume in question is the larger cylinder's volume minus the smaller cylinder's volume. Remember a cylinder's volume is base area times height.

PE = ρ(πb²h – πa²h)gh

Finally, factoring out some like terms we get

PE = ρπgh²(b² – a²)

Which is what I had in equation 8. To answer your question in words, one h comes directly from mgh, and the other comes from the volume.

Hope that helps, and please follow up with any other questions.
-Josh