Timeline for Determining Electric Field Inside Long Cylinder (Using Gauss' Law)?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2018 at 17:36 | comment | added | BioPhysicist | This needs to be changed to say the dot product is $0$ if $E$ is parallel to the surface. Not perpendicular. This is because the area vector is perpendicular to the surface. | |
| Oct 4, 2015 at 5:54 | comment | added | A4Treok | Yeah, that's a good point as well. Better explanation of why the ends contribute nothing. Thanks for the help! | |
| Oct 2, 2015 at 9:33 | comment | added | Bobson Dugnutt | As I see it, it has more to do with the fact that the electric field has no component along the axis of the cylinder; the endcircles of the cylinder in the illustration is only the Gaussian box (which we can choose to be any form) - the cylinder itself extends to infinity and therefore doesn't have such endcircles. | |
| Oct 2, 2015 at 7:39 | comment | added | A4Treok | Wow, I blanked hard. Physical answer is that the charges on either side of the cylinder's lengthwise section will cancel out their respective (x,y) coordinate forces. All force is in z direction because of symmetry. | |
| Oct 2, 2015 at 4:03 | comment | added | A4Treok | Right, but what's the physical reason for this? Why can't the surface charge of the cylinder have E on the outside edges? | |
| Oct 2, 2015 at 3:08 | review | First posts | |||
| Oct 2, 2015 at 3:25 | |||||
| Oct 2, 2015 at 3:03 | history | answered | Bobson Dugnutt | CC BY-SA 3.0 |