Electric potential just outside a spherical shell We know that as we get closer and closer to a point charge, the electric potential approaches infinity.
Since electric potential at the surface of a spherical shell is finite (Gauss law) , so on moving away from the surface it would fall. In other words, it would be finite as well.
But if I think this way: Take an infinitesimal charge element on the surface so that we can take it as a point charge, then potential due to this charge would be similar due to a point charge. So as we approach this element, shouldn't potential go to infinity.
What's wrong with this reasoning? Please help. 
 A: You may be somewhat confusing electrical field and electrical potential. But in both cases, the flaw in your reasoning is the same: the distance from the infinitesimal charge may be infinitisimal small but so is the charge (infinitesimal small). You cannot avoid a proper integral or using Gauss law.
A: The problem here is symmetry. Use Gauss's law on an infinite sheet of charge. You get a finite (and constant field):
$$ E = \frac{\sigma}{2\epsilon_0} $$.
($\sigma$ is the charge density). It does not matter how close you get, or how far you get: the problem is scale independent, because at any point, you always see the same thing: and infinite sheet of charge occupying $2\pi$ str. Even if your get infinitely close, it's the same--and all the charge matters, even the infinite amount of charge infinitely far away. You can not ignore it.
Now consider your situation: infinitely close to a shell of charge. The field is now:
$$ E=\frac{\sigma}{\epsilon_0}$$
It doubled because there is no field inside the shell of charge, so the inside flux has to come out (doubling the outside flux). Other than that nuance, it looks like the infinite sheet of charge as you get infinitesimally close, and the field does not diverge.
