What would be the potential due to a charge $Q$ at $Q$ itself? We know that the potential due to a point charge $Q$ at any distance is given by $V=\frac{kQ}{r}$, but what would the potential be at the charge itself?
 A: It should be undefined (here undefined does not mean that it is infinite,but it means that it is not having any defined or meaningful value.) Reason:
If you put a=0 in the equation the V would be kq /0,but division by 0 is undefined.
A: There are two concepts:
1) potential $\varphi_1(\mathbf x)$ due to charge 1 at any point of space $\mathbf x$;
2) potential $\varphi_{-1}$ experienced by charge 1 at the point it is located $\mathbf r_1$.
In case there are some other charges 2,3,...N,
$$
\varphi_1(\mathbf x) = Kq_1\frac{1}{|\mathbf x - \mathbf r_1|}
$$
$$
\varphi_{-1} = \sum_{k=2}^N\varphi_k(\mathbf r_1) =\sum_{k=2}^N Kq_k\frac{1}{|\mathbf r_1 - \mathbf r_k|}
$$
The first kind of potential is the potential energy a unit positive charge would have due to particle $1$ if placed at point of space $\mathbf x$. Notice that two point particles are needed to make sense of this concept. It makes no sense, for a point particle, to ask what is the potential energy of point particle 1 due to being in electric field of 1; there is no self-interaction in the Coulomb formula, the case $i=j$ is intentionally skipped. When we go beyond Coulomb formula to relativistic EM theory, there is no consistent theory of how point particle could act on itself, but there are consistent theories of charged particles without self-interaction.
The potential energy is infinite for $\mathbf x = \mathbf r_1$, meaning if the two particles get to the same point of space, the energy needed to overcome their repulsion (or energy extracted from their attraction) is infinite. This is a theoretical limiting case that practically doesn't happen.
So "potential at the point where the particle 1 is located", if it is to be a meaningful finite quantity, is usually meant to be the second kind of potential, and its value depends on positions and charges of all other charged particles.
A: *

*Physically, the answer is "infinite", meaning that it won't happen in reality.

*Mathematically, the answer is undefined (or there is no answer).


The formula you show:
$$V=\frac{kQ}{r}$$
is only valid for $r\in \mathbb R/0$, in other words: It is valid for all values of $r$ except for $r=0$. Mathematically, if something is not valid - is not defined - for certain values, then it is not allowed to be used for that value. Then it has no result mathematically. The result is not infinite, or zero, or something else. The result is not defined at all. It doesn't exist mathematically.
But the fact that this particular mathematical formula cannot give a result for $r=0$, does not mean that a result doesn't exist. It is just a limitation of this particular formula.
Physically, we expect our surroundings to be a smooth continuum. At least until we know better. That a result suddenly doesn't exist, as if reality suddenly doesn't exist, is non-physical. Also, this means that we won't expect a sudden "hole" in our results. Meaning, every change is gradual. If the value leading up to $r=0$ shows a trend, then we wouldn't expect the actual value at $r=0$ to suddenly be enormously different. So, physically, since we can't use the mathematical formula you showed, we might instead trust the limit of that formula:
$$\lim_{r\rightarrow 0}\left(V\right)=\lim_{r\rightarrow 0}\left(\frac{kQ}{r}\right)$$
And that limit must be infinity, because that is the result the values of $r$ lead up to as they get closer to $r=0$:
$$\lim_{r\rightarrow 0}\left(\frac{kQ}{r}\right)=\infty\quad\text{, meaning }\frac{kQ}{r}\rightarrow \infty \text{ when }r\rightarrow 0$$
(When dealing with infinitey, I don't think mathematicians like the writing style: $\lim_{r\rightarrow 0}(kQ/r)=\infty$, because $\infty$ shouldn't be treated like a "number". Instead, they will always use the "going towards" notation above with the $\rightarrow$ arrows.)
What this actually means physically is of course that this will never happen. When the potential must be infinite for something to happen, then the electric repulsion is so large that it in reality never will happen (assuming two equal-sign charges). In other words, you cannot place another charge exactly on top of the charge $Q$. If you tried, then they would skew aside immediately, since the electric force would be basically infinitely large. Having two equal charges placed at exactly the same position is physically impossible. If the charges are opposite, then something else will happen. A negative and a positive charge merging may result in some odd and very special phenomena that causes them to neutralise each other.
Overall, such infinity value tells us that something very "special" will happen or that it is an impossible situation.
A: Potentials and charge distributions are not mathematical, they are physical observation that lead to mathematical models, which have to describe the data.
The classical Maxwell equation solutions fit a multitude of observations with great accuracy and also are predictive for new systems. BUT note the term classical. It is in contrast with quantum mechanical.  When distances approach zero one is in the quantum mechanical regime, the one that is given by  the Heisenberg uncertainty principle. $Δx*Δp>h/{2π}$ . 
The classical potential function of 1/r cannot be tested for small distances because a test charge that would measure the potential can never reach the other charge. This is one of the reasons that quantum mechanics was invented. Together with the black body radiation, the periodic table, and the spectrum of light from atoms it was necessary to forbid two point charges to fall on each other, because atoms would not exist, the electrons would fall to the nucleus and neutralize it.
In quantum mechanics, the potential of the charged nucleus enters into the Schrodinger ( more correctly the Dirac) equation and the solutions for the electron are quantized, and there is always a ground state which allows to form stable states. ( see the hydrogen atom).
So it is the quantum mechanics from which all classical physics emerges that provides the mathematical  models for r=0 .
The same problem exists with the 1/r of the gravitational potential, that is why quantization of gravity is a holy grail of theoretical physics. An effective quantization is used at the moment in cosmological models like the Big Bang, and they are looking for definitive models.
A: This question goes to the centre of the problem of self interaction, which is unsolved. So it is an excellent question. Often self interaction is ignored, but in quantum field theory it is prominent. There it leads to the infamous infinities or divergences. In QFT self energy is assumed to be incorporated into mass. However even this approach is not satisfactory.
Let's take the example of the electron. Thomson argued that if the rest mass was made up entirely of self energy, which is positive as charge repels itself, then the so called classical electron radius would be about $2.8 \cdot 10^{-15}$. However experimentally we now know that the electron behaves as a point charge down to a scale of at least $10^{-22}$ m. This means that its potential does not deviate from $1/r$ down to this scale. If we then apply Thomson's approach we end up with a mass of 2 $GeV/c^2$, interesting not unlike that of a proton. This in my judgement is an important unsolved problem of physics.
