Timeline for Dipole moment of a single point charge
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2020 at 15:02 | comment | added | Emilio Pisanty | @Paddy I disagree - this answers the question. You're obviously welcome to your opinion, as well as any on-site actions you consider relevant, including downvoting and flagging as not-an-answer. Or, you know, providing an answer of your own, instead of limiting yourself to criticizing others. I for one am completely uninteresting in litigating four-year-old answers to mediocre questions. | |
| Nov 6, 2020 at 10:09 | comment | added | Paddy | Visiting this question 4 years after the question was posted - The answer, although not incorrect, is not really an answer. A comment at best. | |
| May 18, 2016 at 7:59 | comment | added | user103515 | I think this link physics.stackexchange.com/questions/17063/… gives a more elegant explanation to my question | |
| May 17, 2016 at 13:36 | comment | added | Emilio Pisanty | @user103515 The explanation is that you set too much stock on under-explained sections of Wikipedia. Doing your due diligence would have sent you to this reference which makes it clear that this is a separate quantity that only makes sense for composite systems. If you have further queries arising from a lack of due diligence, feel free to ask them elsewhere - I will only reply again if there's anything actually interesting. | |
| May 17, 2016 at 12:37 | comment | added | user103515 | Kindly refer to en.wikipedia.org/wiki/Dipole and see "Quantum mechanical dipole operator" See the expression for CHARGED DIPOLE. If we apply this formula to a single electron or any single point charge the the dipole moment p becomes 0. What is your explanation on that? | |
| Apr 29, 2016 at 13:36 | comment | added | Emilio Pisanty | @auxsvr Hence the careful qualifying of the entire post to apply only to charged systems where relevant. | |
| Apr 29, 2016 at 13:17 | comment | added | auxsvr | If the moments up to order $n-1$ vanish, the moment of order $n$ is independent of coordinate system. Perhaps you should add "in general" to the first sentence? | |
| Apr 29, 2016 at 11:45 | comment | added | Emilio Pisanty | Because the system is neutral, in which case (as opposed to a system with nonzero total charge) the dipole moment is indeed independent of the coordinate origin. That's a simple calculation which you should be able to perform. | |
| Apr 29, 2016 at 10:34 | comment | added | user103515 | The dipole moment of a physical dipole (+q and -q separated by a distance d) is always charge times distance of separation between them (q*d) and is always independent of the coordinate system. Why is it so? | |
| Apr 29, 2016 at 9:18 | history | answered | Emilio Pisanty | CC BY-SA 3.0 |