Timeline for On proving that charge is linearly proportional to potential for a conductor
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 30, 2019 at 21:42 | history | edited | Bob Jacobsen | CC BY-SA 4.0 |
edited body
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| Aug 30, 2019 at 19:40 | comment | added | Bob Jacobsen | If something is true for an arbitrary function of position then it’s true for a constant function of position I.e not depending on position. | |
| Aug 30, 2019 at 19:20 | comment | added | Hilbert | In the question, $\phi _{0}$ (or $Q_{0}$) is the constant potential (charge) on the surface of the conductor, it is not a function of position. | |
| Aug 30, 2019 at 19:17 | comment | added | Bob Jacobsen | @hilbert I think you're saying the same thing I'm saying, so I don't see where the disagreement lies. Perhaps I wasn't clear. I've added a more formal paragraph to try to address that. | |
| Aug 30, 2019 at 19:16 | history | edited | Bob Jacobsen | CC BY-SA 4.0 |
Added a more formal paragraph
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| Aug 30, 2019 at 19:07 | comment | added | Hilbert | I respectfully disagree, I don't think it has to do with the equations themselves. We specify $Q_{0}$ or, equivalently, $\phi _{0}$ before solving Maxwell's equation, they are boundary conditions. The question can be reformulated as, given a boundary condition $\phi_{0}$ (equivalently $Q_{0}$) on the surface of a conductor, and another boundary condition $\phi_{1}$ (equivalently $Q_{1}$), why does one have $Q_{0}/ \phi_{0} = Q_{1} / \phi_{1} =C$? | |
| Aug 30, 2019 at 2:34 | history | answered | Bob Jacobsen | CC BY-SA 4.0 |