Timeline for Facing a paradox: Earnshaw's theorem in one dimension
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 4, 2019 at 17:25 | comment | added | knzhou | @SRS It's not quite correct because things aren't defined, but if you smooth it out to some charge density, then it's perfectly well-defined. You get a stable equilibrium, which again does not violate Earnshaw's theorem. | |
| Apr 4, 2019 at 16:44 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Apr 4, 2019 at 16:37 | comment | added | SRS | I have to think more about it and I'll get back. | |
| Apr 4, 2019 at 16:33 | comment | added | BioPhysicist | @SRS The potential at a point charge is not defined (or you could say infinite) | |
| Apr 4, 2019 at 15:37 | comment | added | SRS | I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou | |
| Apr 4, 2019 at 14:58 | comment | added | knzhou | @SRS Yes, that's true. | |
| Apr 4, 2019 at 14:57 | comment | added | SRS | Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^{\prime\prime}(x)=0$? | |
| Apr 4, 2019 at 13:58 | history | answered | knzhou | CC BY-SA 4.0 |