Timeline for Electric field in a hollow object
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12 events
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| Aug 8, 2016 at 7:35 | comment | added | Jeppe Stig Nielsen | The fact that this has confused others before you, can be seen in www.physicsforums.com: Electric field inside a uniformly charged cubical box. Some of the posts in that thread are wrong. | |
| Aug 8, 2016 at 7:32 | comment | added | Jeppe Stig Nielsen | The hollow sphere has much more symmetry than the hollow cube. For the hollow sphere, all points on the boundary are equivalent. Since the full surface integral is zero (total flux through boundary sphere is zero), the symmetry of all points on a sphere leads to the $E$ field being identically zero in the spherical case. In the case of a cubical box, some points are different than others. For example, why would $E$ at a vertex (corner of the cube, there are eight) be the same as $E$ at the midpoint of a face of the cube (there are six)? Or midpoint of an edge of the cube (there are twelve)? | |
| Aug 8, 2016 at 6:56 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 6, 2016 at 14:16 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 29, 2016 at 4:27 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 43 characters in body; edited tags
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| May 29, 2016 at 3:13 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 31, 2015 at 0:43 | answer | added | TheQuantumMan | timeline score: 1 | |
| Mar 20, 2015 at 0:14 | comment | added | MonkeysUncle | If the electric field varies and assumes positive and negative values along the surface you integrate, it can all cancel out to zero. Just imagine the field created by a point charge, and now draw a sphere that doesn't include the point charge. The flux through the sphere is zero in total, but there is still an electric field at all points in space. | |
| Mar 19, 2015 at 23:08 | comment | added | Daiz | Thank you for the hint! This is definitely not true in the second case, because the electric field expands radially. But I still can't grasp it really. So I have no charges in that cube, but still an electric field is created? | |
| Mar 19, 2015 at 22:40 | comment | added | yohBS | Hint: when you write $$ ...=\oint_\Gamma E_i d\Gamma=E_i\oint_\Gamma d\Gamma=... $$ You assume that the electric field is constant over the surface of integration (and perpendicular to it). Is this true in the second example? | |
| Mar 19, 2015 at 22:38 | review | First posts | |||
| Mar 19, 2015 at 23:11 | |||||
| Mar 19, 2015 at 22:35 | history | asked | Daiz | CC BY-SA 3.0 |