Timeline for What is the charge density in the proof of Earnshaw's theorem?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2015 at 22:01 | vote | accept | ari | ||
| Feb 8, 2015 at 13:57 | comment | added | glS | 1 comment up: it is because you are looking at the divergence of the field where the charge is (in my notation this is $\textbf x$). 2 comments up: It depends on what you are calculating. If you want the electric field generated by a charged sphere, then you have to consider the charge density $\rho$ of that sphere as a source. If you instead want to know how a charge behaves in the electric field generated by the other charges, only these other charges must be considered as a source. Otherwise you would be analyzing the behaviour of a charge due to its own field. | |
| Feb 8, 2015 at 10:34 | comment | added | ari | Is it because we are only looking at the divergence at exactly that point? | |
| Feb 8, 2015 at 10:03 | comment | added | ari | I think I am starting to see your point. But why can that equation ever not be zero? I am thinking about a charged sphere again. There are no sources at a point $r>R$ ($R$ - Radius of sphere) and still one uses the mentioned Maxwell equation with $\rho \neq 0$. Or is that because by using Gauss we are now looking at a Volume? If so, why don't we in the Earnshaw problem as well? | |
| Feb 8, 2015 at 0:39 | comment | added | glS | @ari there are no source at the point $\textbf x$. Simply because we are considering point charges, and the charge $q$ at the point $\textbf x$ is not considered as source for $\textbf E$. | |
| Feb 7, 2015 at 19:03 | comment | added | ari | Why are there no sources of $E$? If $E$ is induced by the other charges, why are those not considered sources? | |
| Feb 7, 2015 at 18:10 | history | answered | glS | CC BY-SA 3.0 |