Timeline for Why is electric flux through a cube the same as electric flux through a spherical shell?
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| May 29, 2020 at 13:34 | comment | added | Joe Iddon | @satan29 No, my answer wouldn't apply if it were not an inverse square law. Conical surfaces closer to the corner of the cube are further from the center of the cube than surface elements at the middle of the faces, so even though the angle changes, this distance must be accounted for too. Since the conical top surface has spread out in proportion to $r^2$, but the electric field has decayed in proportion to $\frac{1}{r^2}$, this distance does does not affect the flux. Hence $\frac{1}{r^2}$ behaviour is necessary. See feynmanlectures.caltech.edu/II_05.html#Ch5-S8 for more detail. | |
| May 29, 2020 at 7:25 | comment | added | satan 29 | I remember reading that the gauss law depends on the inverse square relation in the coulumb force. Had the coulumb force be proportional to say, 1/r^3, Gauss law wouldnt work. However, Your arguement doesnt account for this fact...Had the coulumb force be 1/r^3 I can still use your arguement and claim that the flux depends only on the enclosed charge. However, This is not the case. | |
| May 12, 2020 at 19:36 | vote | accept | kamer_kane | ||
| May 11, 2020 at 20:44 | history | edited | Joe Iddon | CC BY-SA 4.0 |
added 2 characters in body
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| May 10, 2020 at 14:04 | history | answered | Joe Iddon | CC BY-SA 4.0 |