Timeline for What does it mean for nature to be discontinous?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2020 at 16:53 | comment | added | FoundABetterName | Let us continue this discussion in chat. | |
| Jun 22, 2020 at 16:22 | comment | added | Trebor | @DinoManPhyLab Nope, Gauss Law does not require any of those. | |
| Jun 22, 2020 at 15:39 | comment | added | FoundABetterName | No i mean that Gauss Law assumes symmetry and idealistic conditions which you said aren't possible like infinite point charges making a sphere and proposed a different method which included blurring the shell a bit so i meant to ask do the results from these 2 methods differ? | |
| Jun 22, 2020 at 15:21 | comment | added | Trebor | @DinoManPhyLab Well, Gauss Law is a law, so if you use it correctly the result will be correct. | |
| Jun 22, 2020 at 9:30 | comment | added | FoundABetterName | Yes i've learnt to calculate the field with Gauss Law but what i was trying to ask is that are the results obtained by Gauss Law and the other method you mentioned quite similar and if they differ is there a significant value of electric field inside the shell? | |
| Jun 22, 2020 at 7:33 | comment | added | Trebor | @DinoManPhyLab If the whole thing is still spherical, you can still calculate the electric field with the Gaussian theorem, which I suppose you have already learnt. Go ahead and try it! | |
| Jun 22, 2020 at 7:31 | comment | added | FoundABetterName | Just one more clarification is needed so does this mean if i get a numerical solution there will be electric field all inside the shell as well and i presume not at the centre due to symmetry or will it be there at the centre too? also what will be the expected deviations (in terms of magnitude of field) from the Gaussian results? | |
| Jun 22, 2020 at 7:26 | vote | accept | FoundABetterName | ||
| Jun 22, 2020 at 7:23 | history | answered | Trebor | CC BY-SA 4.0 |