Meir Achuz said:
The electric field is greater where the radius of curvature of the end is smaller.
because at a point of increasing radius of curvature, if two charges on opposite sides are at equal distances
measured along the surface, the one on the side of smaller radius of curvature will be
actually nearer …
so in equilibrium, …
ooh, that's the
wrong result, isn't it?
ah … let's start again

…
at radius of curvature r, the amount of charge in an arc between fixed distances s and s + ds (measured in 3D space,
not along the surface) will be less for greater r (by a factor cos(s/r), the radius of the arc), and its component along the surface will
also be less (by a factor cos(s/2r)) …
so for smaller r, s/r is larger and so cos(s/r) and cos(s/2r) are smaller, and so the amount of charge and its "component" (loosely speaking) at a fixed distance is smaller, so the repulsive force will be smaller on the side of decreasing r
so
in equilibrium, there must be a higher charge density at decreasing r!
(does anyone know the exact dependency on curvature?)