Timeline for How to find the electric field of an infinite charged sheet using Gauss’s Law?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 26, 2017 at 14:47 | vote | accept | Sillysack Buttowski | ||
| Sep 25, 2017 at 15:09 | comment | added | Jyotishraj Thoudam | The intuition lies in the fact of infinite plate assumption. Whatever distance you move out from the plate you still get the same area of influence since the plate is infinite. | |
| Sep 25, 2017 at 14:12 | comment | added | ZeroTheHero | @SillysackButtowski Maybe another way to think of this is to realize that, as you move away from the sheet, the amount of charge in a cone of angular opening $d\Omega$ increases as $r^2$ while the field generated by those charges decreases by $r^2$: the two effects exactly balance out. | |
| Sep 25, 2017 at 14:05 | comment | added | ZeroTheHero | @SillysackButtowski I'm not sure what you mean by "verifying this". How can it matter if you are at 1 or 100 meter above an infinite plate? Based only on the plate itself you'd have no way of knowing your "altitude" since everything is exactly the same at all heights in all directions. | |
| Sep 25, 2017 at 13:59 | comment | added | Sillysack Buttowski | Maybe we will never be ever to verify this and I will just have to believe the math. | |
| Sep 25, 2017 at 13:45 | comment | added | ZeroTheHero | $\vec E\cdot d\vec S$ varies in magnitude across the geometrical surface so that the sum (or more properly the integral) is $0$, not because $\vec E=0$. If you think about it, how does the physics change if you are $2m$ or $3m$ above the sheet? How would you know anyways since the sheet is infinite and the surface charge density constant?. | |
| Sep 25, 2017 at 13:40 | comment | added | Sillysack Buttowski | Also, why does the field strength come out to be independent of distance? That is very nonintuitive as the field should vary according to the inverse square law. | |
| Sep 25, 2017 at 13:37 | comment | added | Sillysack Buttowski | "The $0$ results from the geometry of $\vec E \cdot d\vec S$ everywhere on the sphere rather than $\vert \vec E \vert = 0$." What does that mean? | |
| Sep 25, 2017 at 13:15 | history | answered | ZeroTheHero | CC BY-SA 3.0 |