Timeline for Why can't we have an $E$-field like this one inside a conductor?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2019 at 18:39 | comment | added | Michael Seifert | @Hilbert: That's correct; in general, you need to make an additional assumption of symmetry in order to find $\vec{E}$ using Gauss's Law. For example, it seems reasonable to assume that the electric field produced by a spherical shell should itself be spherically symmetric. If you make that assumption, then the electric field must be purely radial and have a magnitude only depending on $r$. For a spherical surface, this implies that $\int \vec{E}\cdot d\vec{a} = 4 \pi r^2 |\vec{E}|$. Then you apply Gauss's Law and find that $|\vec{E}| = 0$. | |
| Aug 14, 2019 at 18:36 | comment | added | Hilbert | Moreover, the $E$-field inside is null because of the shell theorem, the only remaining problem is to prove it is zero for any arbitrarily shaped conductor. | |
| Aug 14, 2019 at 18:30 | comment | added | Hilbert | @MichaelSeifert With hindsight, I think the argument using the small Gaussian sphere is not an argument. I said, "If ones takes a Gaussian sphere with a slightly smaller radius, they could argue that the field cannot exist since there is no charge enclosed inside the surface", but then a zero net flux doesn't mean in any way that there is no field inside the surface, it means only the net flux is null. | |
| Aug 12, 2019 at 19:17 | vote | accept | Hilbert | ||
| Aug 12, 2019 at 19:12 | answer | added | dmckee --- ex-moderator kitten | timeline score: 1 | |
| Aug 12, 2019 at 18:36 | comment | added | Hilbert | @dmckee "Why do you limit this analysis to surface charges?" The conductor is a hollow sphere with negative charges on the surface. | |
| Aug 12, 2019 at 18:23 | comment | added | Michael Seifert | Which is to say that if you believe the argument about the "slightly smaller Gaussian sphere", then your second Gaussian surface is irrelevant; you've already proven that there is no field inside the shell. (The argument about the second surface fails because all it tells you is the difference between the interior & exterior field, not the value of either one.) | |
| Aug 12, 2019 at 18:22 | comment | added | Michael Seifert | Concerning your arguments about Gaussian surfaces, I'm reminded of the old joke: "At a conference, a mathematician proves a theorem. Someone in the audience interrupts him. 'But, sir, that proof must be wrong. I’ve found a counterexample.' The speaker replies, 'I don’t care — I have another proof for it.'" | |
| Aug 12, 2019 at 18:20 | comment | added | dmckee --- ex-moderator kitten | "because had there been any field within it, the charges on the surface would start to move" Why do you limit this analysis to surface charges? Even though any excess charge will migrate to the surface, there are free charges in the bulk, as well. | |
| Aug 12, 2019 at 18:14 | history | asked | Hilbert | CC BY-SA 4.0 |