Timeline for Potential on conducting shell without image charge
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 16 at 18:05 | history | edited | Jos Bergervoet | CC BY-SA 4.0 |
spelling error corrected (to see if this reopens voting...)
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| Jun 15 at 14:38 | comment | added | Jos Bergervoet | PS: and your assumption that the charge on the outer shell is $2Q$ is wrong. It is only $+Q$ if the total charge on the sphere is $0$ and it's only $\frac53 Q$ if the potential of the sphere is $0$. From your question it appears that the condition to be used is the former, but strictly speaking this information is missing. And the claim by @khaxan is also wrong. There a factor $\frac{11}{6}$ is missing! (But khaxan would be correct if the potential of the sphere was $0$, only that is inconsistent with "find the potential on the sphere". So I think total charge on the sphere should be $0$.) | |
| Jun 15 at 13:22 | comment | added | Jos Bergervoet | With the title now changed (to exclude using images) I think the most elegant way is to use 3D conformal mapping to a spherical void where the point charge is then exactly in the center. By symmetry the charge distribution on the void surface is then uniform. (For the outer region we can invert the ball to a void. 3D conformal mappings are less versatile than their 2D counterparts, and need scaling of the potential with the Jacobian, but for spherical shapes and point charges they are well-suited!) Alternatively you could just solve for the charge distribution by "mathematical brute force". | |
| Jun 13 at 4:20 | history | edited | zxayn | CC BY-SA 4.0 |
changed focus of question
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| Jun 9 at 1:56 | history | edited | zxayn | CC BY-SA 4.0 |
Afterthought
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| Jun 5 at 17:42 | vote | accept | zxayn | ||
| Jun 13 at 4:13 | |||||
| Jun 5 at 14:12 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 4 characters in body; edited tags
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| Jun 5 at 8:42 | answer | added | Jos Bergervoet | timeline score: 1 | |
| Jun 5 at 8:14 | comment | added | Vivek Kalita | The outer surface charge density would be uniform. Consider this: You have a charge q inside, an amount -q would get accumulated in the inner surface of the conductor, these two together would cancel each others field, and so the outer surface experiences no other forces. Thereby they arrange themselves in the most uniform way possible. | |
| Jun 5 at 4:50 | comment | added | khaxan | Electrostatic potential at the centre of a spherical shell is kQ/r even if charge is unevenly distributed on the surface | |
| Jun 5 at 4:10 | comment | added | zxayn | Longest question I've ever written :p | |
| Jun 5 at 4:09 | history | asked | zxayn | CC BY-SA 4.0 |