Potential Due To A Point Charge At Position Of The Point Charge Electrical potential at a point P in an electric field is:
$$V_p=-\int_\infty^P\mathbf {E}\cdot d\mathbf{l}$$
Thus, for an electric field due to a point charge at the origin, the potential at P is:
$$V_p=\frac{q}{4\pi\epsilon R_p}$$
My problem is that if P is at the location of the point charge, i.e., $R_p=0$, then this equation blows up. Thus, what do I have to do if I want to calculate the potential due to a point charge at its location?
Thank you for the help.
 A: Electric potential at a point would be the work done to [in a case with a positive test charge and negative electric field] lift the particle from that point to infinitely far away. The coulomb force is defined as inversely proportional to the distance of separation, so if you were to have them both somehow "stacked" in the exact same point in space [ignoring the fact they're fermions] I'm not entirely sure how it would even be possible to separate them. I think that Equation blows up for good reason. 
Furthermore, to illustrate my point in a bit clearer of a way, if you release something in a potential and exert no other forces on it, it'll move to minimise that potential energy. Where would a charge move if you put it exactly on top of another, such that the points of their charges were exactly the same.
A: Point charges are mathematical abstractions. Real charges have a finite radius.  Even for an object as small as an electron, there is some distance from the centre at which the inverse square law breaks down.
Consider the gravitational potential of a spherical distribution of mass, which is analogous to an electrical charge. 
From outside of the sphere it can be regarded as a point mass, with all of the mass concentrated at the centre - see Shell Theorem. But even a sphere of dense matter has a finite radius, and inside that radius the gravitational potential does not continue to increase. The Shell Theorem tells us that inside the sphere the force of attraction is proportional to the mass within our current distance from the centre. For a sphere of uniform density, this force decreases linearly, resulting in a finite potential at the centre.
For a spherical shell, with all the mass concentrated at the surface, the Shell Theorem tells us that the gravitational attraction is zero inside the shell. Then the potential everywhere inside is the same as at the surface.
If the point charge is a spherical conductor or shell, this is equivalent to the gravitational shell : the potential everywhere inside is the same as at the surface. If charge is distributed uniformly throughout a sphere, then the electrical potential will continue increasing towards the centre, but not according to $1/r$. 
Potential due to uniform sphere shows that for a uniform distribution of mass or charge, the potentials outside and inside the sphere are given by 
$$V(r \gt a)=\frac{a}{r}V_0$$
$$V(r \le a)=\frac{3a^2-r^2}{2a^2}V_0$$
where $V_0$ is the potential at the surface $(r=a)$. Thus the potential at the centre is $\frac32 V_0$.
A: For a point charge located at the origin, it's potential can be written Mathematically as
$$\phi(\vec{r})=\frac{\iiint q\delta(\vec{r})dV}{4\pi\epsilon_0r}$$
When $r\to0$,
$$\phi(\vec{r})=\frac{q\delta(0)dV}{4\pi\epsilon_0r}=\frac{0}{0}$$
So, potential is no definition in $r=0$, or say potential is of Singularity. To invesigate the potential at $r=0$, you should consider the charge distribution has an finite volume such as a charged ball.
