Timeline for Does a super-conductor have a perpendicular electric field like a 'Perfect conductor'?
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Jul 31 at 22:23 | vote | accept | InexperiencedCoder | ||
Jul 31 at 22:23 | comment | added | InexperiencedCoder | @JánLalinský Okay, that makes a lot more sense. I never really accounted for them being separate ideas. Thanks! | |
Jul 31 at 18:04 | comment | added | Ján Lalinský | @InexperiencedCoder On the surface, the Ohm law $\vec{j} = \sigma \vec{E}$ does not hold. So only inside the conductor, zero current implies zero electric field and zero charge density. On the surface, we can have zero current and non-zero electric field and non-zero charge density. | |
Jul 31 at 17:29 | comment | added | InexperiencedCoder | @my2cts That's internally though? There's still charges on the surface, which have electric fields with a vector perpendicular to the wire. - I may be getting myself confused now with ignorable details. | |
Jul 31 at 15:54 | comment | added | my2cts | In a conductor ${\vec j} = \sigma {\vec E}$. So ${\vec E}=0$ implies ${\vec j} =0$. | |
Jul 31 at 15:40 | comment | added | InexperiencedCoder | @my2cts I thought that it is the resistance in the static copper wire which causes internal fields, for example, that stop all of the charges making their way to the surface like in a 'Perfect conductor' | |
Jul 31 at 15:28 | history | edited | InexperiencedCoder | CC BY-SA 4.0 |
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Jul 30 at 22:22 | comment | added | my2cts | ‘ copper, gold, aliminium has resistance … therefore there remains a field within the conductor.’ Only if there is a current. | |
Jul 30 at 21:13 | history | edited | InexperiencedCoder | CC BY-SA 4.0 |
added 300 characters in body
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Jul 29 at 22:43 | answer | added | Ján Lalinský | timeline score: 2 | |
S Jul 29 at 20:50 | review | First questions | |||
Jul 30 at 0:32 | |||||
S Jul 29 at 20:50 | history | asked | InexperiencedCoder | CC BY-SA 4.0 |