- #1
gnome
- 1,041
- 1
Electrical potential is constant everyplace on the surface of a charged conductor.
Also, on the surface of an irregularly-shaped conductor, the charge density is high in convex regions with small radius of curvature (especially, for example, at sharp points), and low in regions of large radius of curvature, and therefore, by Gauss's Law, the field intensity must be high just above a sharp point, and relatively low just above a gently curved region.
So, does it follow that an equipotential surface located just an INFINITESIMAL distance outside the surface of the conductor will be MUCH closer to the surface above a gently curved region than it is in the vicinity of a sharp point, even though the potential is constant everyplace on the surface?
Or, even more puzzling, if the field intensity is very high near a sharp point, would the potential near the point, just above the surface, be higher than the potential at the surface? But how can that be? What would the equipotential surface in that region look like?
Also, on the surface of an irregularly-shaped conductor, the charge density is high in convex regions with small radius of curvature (especially, for example, at sharp points), and low in regions of large radius of curvature, and therefore, by Gauss's Law, the field intensity must be high just above a sharp point, and relatively low just above a gently curved region.
So, does it follow that an equipotential surface located just an INFINITESIMAL distance outside the surface of the conductor will be MUCH closer to the surface above a gently curved region than it is in the vicinity of a sharp point, even though the potential is constant everyplace on the surface?
Or, even more puzzling, if the field intensity is very high near a sharp point, would the potential near the point, just above the surface, be higher than the potential at the surface? But how can that be? What would the equipotential surface in that region look like?
Last edited: