Timeline for Why is electric field zero where equipotential surfaces intersect?
Current License: CC BY-SA 3.0
6 events
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| Dec 21, 2017 at 13:50 | comment | added | Valter Moretti | Yes the equipotential surfaces must enter the spheres: take any point inside the left sphere, there the electric field due to the sphere itself vanishes. The electric field inside the left sphere field is therefore completely due to the right sphere and it is the same as that of a point charge centered outside the left sphere. It is evident that the equipotential surfaces enter the left spheres this way. I was thinking here of superficially charged spheres! If the charge is in the volume? I do not know | |
| Dec 21, 2017 at 13:40 | comment | added | Alfred Centauri | @ValterMoretti, I was just wondering if the equipotentials could enter the spheres and I started to look through Jackson as your comment came in. | |
| Dec 21, 2017 at 13:36 | comment | added | Valter Moretti | Well, now I think that equipotential surfaces different from the separating plane enter the (non-conducting) spheres and my example does not work: when spheres touch together there is only one equipotential surafce through the contact point. So my example does not work. | |
| Dec 21, 2017 at 13:29 | comment | added | Alfred Centauri | @ValterMoretti, OK, so two non-conducting spheres, each with fixed, uniform charge density of opposite sign and identical radii and symmetrically placed above and below the x-y plane along the z-axis but not touching the plane. This smells like a method of images type problem and if so, the x-y plane is the zero potential surface? Then the positive (negative) equipotential surfaces encircle the positively (negatively) charged sphere and, as the spheres are brought closer, those surfaces are 'squeezed' together along the line through the center of the spheres finally touching together? | |
| Dec 21, 2017 at 13:04 | comment | added | Valter Moretti | I understand your point, since the spheres are not equipotential, it is not obvious that there are infinitely many equipotential surfces passting through the contact point...I do not know.... | |
| Dec 21, 2017 at 2:21 | history | answered | Alfred Centauri | CC BY-SA 3.0 |