Timeline for Charge density of an uncharged sphere
Current License: CC BY-SA 4.0
10 events
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| Mar 19, 2021 at 10:15 | comment | added | Roger | @jensenpaull I'm not sure I understood everything. But what you are saying is simply that the fact that $\rho= 0$ and $\vec J\neq 0$ are used in the same situation is bothering you because they are not compatible. I would propose then that you add a link to those papers to your answer, so that people can see exactly what you mean. If I missed anything please be a bit more specific with everything and try to use $\rho$ instead of p and $\vec E$ instead of e. | |
| Mar 18, 2021 at 22:35 | comment | added | jensen paull | and p is zero, then their cannot be a current density. so im assuming the assumption that all of the papers that i have read are simply saying this: These equations solve maxwells equations acknowledging that there IS a current and a charge density p. but the charges themselfs dont have fields of their own so therefore saying that (div e =/ p/epsilon. )as p is non zero ( as this would come from a direct consequence of substituting ohms law into the equation which does by DEFINITION define a non zero p function in the medium and thus artificially removing the fields of the charges themselfs | |
| Mar 18, 2021 at 22:34 | comment | added | jensen paull | so the e assumption that p is zero for em waves inductors is wrong..... so solving the field.equations for those wouldnt actually tell you what would really happen. much like the free space wave equation. the solution to the wave equation for that is also not an accurate picture of what would happen in reality. it is only when u add the source terms and solve that ,would get realistic solutions. so by substituting ohms law into maxwells equations which defines a quantity that is p dependant - | |
| Mar 18, 2021 at 22:23 | comment | added | jensen paull | to your answer, i get thats what we mean. but what we mean is inherently different to what the math tells you , if it simply isnt true. then solving the field equations will lead to different results to reality and thus pointless. I still dont get how if p is zero and j depends on p being there then how in the case of em waves in conductors u set p to zero. as if p is zero the math simply isnt telling what really happens. when there is J in e.g wires then there must ALWAYS he an associated charge density else the quantity doesnt make sense.so - | |
| Mar 18, 2021 at 21:20 | history | edited | Roger | CC BY-SA 4.0 |
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| Mar 18, 2021 at 21:10 | comment | added | Roger | @jensenpaull I will extend my answer then. | |
| Mar 18, 2021 at 20:17 | comment | added | jensen paull | i get that people say its zero as 0/volume = 0 so on AVG its zero. but when doing calculations about.current density. this APPROXIMATION is just clearly false... | |
| Mar 18, 2021 at 20:14 | comment | added | jensen paull | in the real world its best decribed as a density function that is e.g positive then neg then pos then neg etc etc... the only way that you can TRULY say that Rho = 0 is if there is no positive or negative charge in that region at all.... | |
| Mar 18, 2021 at 20:14 | comment | added | jensen paull | We are in agreement on your second statement. I understand why if the Integral of rho dv = 0 then there can still be a current density. But... if rho is zero EVERYWHERE how can there be a current density? as every paper on EM waves set charge density to be zero yet still have a current density. A conductor that is half negative and half positive on either sides in my mind , is still an uncharged conductor. as its integral is zero - | |
| Mar 18, 2021 at 20:00 | history | answered | Roger | CC BY-SA 4.0 |