We know that the electric field for a point charge is
$$
E = \frac{KQ}{R^2}.
$$
If $R$, i.e. distance from the electric field producer to the point where we want to find the electric field becomes zero, then $E$ will tend to infinity. Now take a case where you have a uniformly charged sphere and you need to find the electric field on the charged sphere. The charged sphere is not a point charge, its rather a combination of point charges. What is electric field at any point, say $P$, on the charged sphere?
Now we know that when we use Gauss's Law, we select a Gaussian surface in such a way that all the points on the Gaussian surface should experience same electric field. Many sources say that if we use Gauss's Law then on any point on the charged sphere the electric field is going to be
$$
E = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}
$$
where $R$ is the radius of the sphere and $\epsilon_0$ is the permittivity.
But now, don't consider Gauss's Law. As $P$ is at the surface of the charged sphere, then the electric field due to the small element of the charged sphere on which point $P$ lies is infinity, as the small element has charge $dq$ (let's also say we have assumed the small element of charge $dq$ on which point $P$ lies to be a point charge, and hence $R$ becomes zero, as the point $P$ lies on the small element). So electric field must be infinity due to that small element on which the point P lies. It doesn't really matter how much small the charge of the element is! When you do divide a large number and zero, you get infinity and similarly when you divide a very very small number and zero, you will again get infinity. Now for rest of the small elements which make up the whole sphere, the electric field won't be infinity; instead it would be some finite value. Say on the point $P$ for all the other small elements (except the element on which the point $P$ lies) net electric field is $X \rm\,N/C$. Still, $\text{infinite}+X = \text{infinite}$.
From here we conclude that on the point $P$ the electric field must be infinite. Then how did we get $$\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}~?$$
My second question is: Similarly how can one say that we can find the potential on the surface of the charged conductor? I saw sources saying that the potential on the surface of the charged sphere is $V=KQ/R $ if $R$ is the radius of the sphere. But how is this possible? I mean on the surface again for the above reasons we would get it to be infinite!
Please explain these doubts to me. These might be silly doubts, but still. I don't know how far my thinking is correct.
Best Answer
(Other) Rob's answer seems good to me, but let me offer another way of thinking.
As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.
If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is $\sigma=Q/4\pi R^2$, then indeed the field at the surface is $\sigma/\epsilon_0$ as in the infinite charge of plane case.
Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.