Timeline for Are electric field lines always conserved?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2016 at 4:20 | vote | accept | Shodai | ||
| Mar 31, 2016 at 4:20 | |||||
| Mar 23, 2016 at 22:13 | comment | added | Timaeus | @JohnRennie Almost all the field lines come from infinity or the positive charge, and almost all of them terminate on the negative charge. But in your answer you make is sound like they start on the negative charge and terminate on the positive charge. Which could be confusing to beginning students. You could stay there must be field lines that end on the -6q that don't start on the +1q charge (instead of saying there must be field lines that start from the -6q that don't end on the +1q). | |
| Mar 23, 2016 at 18:04 | comment | added | John Rennie | @Benjamin: I'm not sure i understand what you're asking. Maybe you should work this up into a new question or maybe ask in the chat. | |
| Mar 23, 2016 at 18:02 | comment | added | Benjamin | This is true because we believe in Gauss's Law. However, if we think more about QFT approach there would a contradict where some lines (yet better fields) are just missing or in excess. So, some fields are just free streaming in Universe without ever "noticing" the presence of the other charge. Am I making the situation more complex that it is? | |
| Mar 23, 2016 at 17:57 | comment | added | John Rennie | Obviously in the real world the universe is charge neutral (probably) so flux lines won't really go to infinity, but in the idealised model of a single isolated charge they would. | |
| Mar 23, 2016 at 17:56 | comment | added | John Rennie | @Benjamin: suppose I choose a Gaussian surface big enough to enclose both charges, then the total flux through that surface is $-5q/\varepsilon$ i.e. there must be a net flow of flux lines through this surface, and there are no other charges in the system outside the surface. So $-5q/\varepsilon$ worth of flux lines must go off to infinity. I don't understand why this is controversial. Suppose you had just one charge, then that would have flux lines going off to infinity because there are no other charges for the lines to end on. | |
| Mar 23, 2016 at 17:41 | comment | added | Benjamin | Hi John, I am still not pesuaded that field lines will NOT be equal. Are you saying that for different charges, the lines can be non-conservative? Where those extra lines come from or go to? Are there any source at infinity from which or towards which these line propagate? | |
| Mar 23, 2016 at 10:21 | comment | added | John Rennie | @user36790: you did upvote Anubhav as well, yes? :-) | |
| Mar 23, 2016 at 10:20 | comment | added | user36790 | You are so modest. +1. | |
| Mar 23, 2016 at 10:16 | history | answered | John Rennie | CC BY-SA 3.0 |