Timeline for Help understanding the divergence theorem as it relates to Gauss' Law
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2017 at 18:10 | comment | added | Bob Knighton | All of the other comments are absolutely right. There is not way to create an electric field of the form you gave without including an appropriate charge distribution. These cannot exist in any region without charged matter in that region. In fact, this is an immediate consequence of Gauss' law, and doesn't disprove it. | |
| Jun 28, 2017 at 18:09 | comment | added | Bob Knighton | @jphollowed The field lines in a small region away from a charge do not behave like the vector field $\textbf{k}$ you described. In a small region away a charge, electric fields appear constant. However, the electrostatic potential does appear linear. This is just like the gravitational field near the Earth. It is approximated as a constant $g$ since we are only considering a small observable region away from the center of the gravitational field, so it looks like a constant; however, the potential energy appears linear. | |
| Jun 28, 2017 at 17:36 | answer | added | Gonenc | timeline score: 0 | |
| Jun 21, 2017 at 21:12 | answer | added | tparker | timeline score: 1 | |
| Jun 21, 2017 at 20:32 | answer | added | user87745 | timeline score: 1 | |
| Jun 21, 2017 at 20:13 | answer | added | J. Murray | timeline score: 0 | |
| Jun 21, 2017 at 18:43 | comment | added | Pawr | In fact, Gauss' Law proves that there must be a source/sink inside, considering the divergence is non-zero | |
| Jun 21, 2017 at 18:25 | comment | added | J. Murray | @jphollowed The field E = (x,0,0) cannot exist if there is no charge inside the volume. Remember, div E = rho, so in this case, rho = 1, not zero. | |
| Jun 21, 2017 at 18:22 | history | edited | Qmechanic♦ |
edited tags
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| Jun 21, 2017 at 18:19 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
Minor fixes.
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| Jun 21, 2017 at 18:18 | comment | added | user97626 | @hyportnex Exactly, but then I see a disagreement in the statements of Gauss Law and the more general divergence theorem. Gauss' law clearly says "The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity"... But in the case that I have described there is no enclosed charge. Again, imagine a very small region near a point charge acting as a vector field sink. The field in this small region would be, to good approximation, parallel lines. But the region would contain no charge. | |
| Jun 21, 2017 at 18:15 | comment | added | hyportnex | since the $\text{div}$ is uniformly $1$ across the region the flux integral is not zero but equal to the volume of integration | |
| Jun 21, 2017 at 18:10 | comment | added | user97626 | @hyportnex But then what about Gauss' law in the context of such a field? | |
| Jun 21, 2017 at 18:09 | comment | added | user97626 | @HritikNarayan No you don't. If I have one single point charge (or mass) that is acting as a vector field sink, then if I focus on a very small region near the sink, the field lines will appear parallel, and will have an approximate form similar to what I state for k. This is an approximation that we make all the time when dealing with gravity on the Earth, rays from the sun, etc. | |
| Jun 21, 2017 at 18:09 | comment | added | hyportnex | as you defined it $\text{div} \textbf{k} = 1$ | |
| Jun 21, 2017 at 18:08 | comment | added | Hritik Narayan | You need a charge distribution for a field like that, and that distribution is given by Gauss's Law. | |
| Jun 21, 2017 at 18:04 | history | asked | user97626 | CC BY-SA 3.0 |