- #1
Pdawgg
- 5
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Here is the question that we have been debating for the past day or so:
Take a conducting cubic box and center a conducting sphere with charge +q inside of it. Will the induced charge density on the sphere be uniform or not?My gut instinct is no. The cube and sphere have different symmetries, so if the density were to be uniform, that would have to be true in general for spheres inside of conducting cavities.
On the other hand, if you imagine replacing the sphere with a point charge, the induced charge on the cube will be identical in that situation. It seems like you could make the argument that this is just an "image charge"-like configuration that matches our boundary conditions so we are done. However, I think that this actually does not quite work since we are either fixing the potential on the sphere (erroneously) or changing our charge density in the region of interest.
So I am leaning towards non-uniform, with that reasoning. Any thoughts on this? It's a surprisingly difficult conceptual question, and I would love a fully satisfactory answer...
Take a conducting cubic box and center a conducting sphere with charge +q inside of it. Will the induced charge density on the sphere be uniform or not?My gut instinct is no. The cube and sphere have different symmetries, so if the density were to be uniform, that would have to be true in general for spheres inside of conducting cavities.
On the other hand, if you imagine replacing the sphere with a point charge, the induced charge on the cube will be identical in that situation. It seems like you could make the argument that this is just an "image charge"-like configuration that matches our boundary conditions so we are done. However, I think that this actually does not quite work since we are either fixing the potential on the sphere (erroneously) or changing our charge density in the region of interest.
So I am leaning towards non-uniform, with that reasoning. Any thoughts on this? It's a surprisingly difficult conceptual question, and I would love a fully satisfactory answer...
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