Timeline for Why does the electric field only depend on the rod?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 21 at 18:38 | comment | added | PineappleThursday | @MSU What you just said is correct! | |
| May 21 at 18:37 | comment | added | MSU | @PineappleThursday all right, now I get it, thank you very much | |
| May 21 at 18:37 | comment | added | MSU | @LourencoEntrudo so if the rod induces charges in the inner surface of the conductor, in the outer surface there would be also charges that would keep the net charge in the conductor zero? | |
| May 21 at 18:36 | comment | added | PineappleThursday | We are given that the cylindrical crust is electrically neutral to begin with. There is an induced surface charge on the inner surface, but because the crust is electrically neutral there is an equal and opposite charge on the outer surface. Thus, if you draw your Gaussian surface to enclose the entire crust, there is net zero charge. However, if you draw your Gaussian surface to enclose only the inner surface, then there is a net charge contributed by the crust. You will find that this net charge exactly cancels out the charge of the rod. The electric field inside a conductor is zero. | |
| May 21 at 18:34 | comment | added | Lourenco Entrudo | Presumably the rod would induce charges in the inner surface of the conductor, altough the field would be zero inside anyway. I think a more accurate answer would be that the electric field does depend on the conductor, it's just that by its properties (being infinitely conductive), whatever happens outside doesn't care about it, like you described in your answer. | |
| May 21 at 18:32 | comment | added | MSU | But there is an induced charge on the Cylinical crust, why is it electrically neutral? | |
| May 21 at 18:32 | history | edited | PineappleThursday | CC BY-SA 4.0 |
added 127 characters in body
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| May 21 at 18:26 | history | answered | PineappleThursday | CC BY-SA 4.0 |