Poisson's equation with a point charge source How do you derive the solution to Poisson's equation with a point charge source? Without using Coulomb's law or the electric field! To be more explicit, we have a point charge at $(0,0)$ of charge $q$ and we want to solve Poisson's equation to find the potential. 
 A: I'll combine the comments into a guiding answer with some extra details here and there.
An empty space containing a single point particle with charge $q$ is defined by two properties:

*

*The system's charge distribution $\rho(\vec{r})$ is zero everywhere except at the location $\vec{r}_0$ of the point particle.


*The integral over the entire space of the system's charge distribution is $q$, i.e.
$$\int_{V}{\rho(\vec{r})d\vec{r}} = q$$
Let's use the last property to state $\rho(r) = q\tilde\rho(\vec{r})$ where $\tilde\rho(\vec{r})$ is a 'function' (distribution) which integrates to $1$ over all space and has the property $\tilde\rho(\vec{r}) = 0$ for all $\vec{r}\neq\vec{r}_0$. Of course this is the Dirac delta $\delta(\vec{r}-\vec{r}_0)$. Now imagine we put the origin at the location of the particle: $\vec{r}_0 = \vec{0}$. Then $\rho(\vec{r}) = q\delta(\vec{r})$.
Next, take a look at the Poisson equation:
$$\Delta\phi(\vec{r}) = q\delta(\vec{r}).$$
This is a hard problem to solve. But why exactly? Well, we have a second derivative on the one side and a Dirac delta on the other side. The second derivative is not a problem per se, but if we try to integrate the equation, the Dirac delta makes life hard for us. Note that the definite integral of a Dirac delta is easy to evaluate, the indefinite integral is much harder if you don't know the answer. Intuitively it might feel natural to you that the integral of the Dirace delta is in fact the Heaviside step function, but remember that we have to integrate twice to get $\phi$, so we also have to know how to integrate this step function.
All of that is work I wouldn't want to do. Not when there's an easier way. And that way is provided by the Fourier transform. We know that it has the property of transforming differentiation into multiplication and it involves a definite integral over all space, so the Dirac delta will be easy to transform. Therefore, it's a great tool to use on differential equations like this one. It transforms a difficult problem in $\vec{r}$-space into an easy one in $\vec{k}$-space. We can then solve the easy problem and inverse Fourier transform the solution back to $\vec{r}$-space. This will yield us the correct solution of the problem in $\vec{r}$-space, due to another property of Fourier transforms which is sometimes called the Fourier inversion theorem.
The problem you are trying to solve is actually known as the Green's function problem for the Poisson equation. The method of Green's functions for solving a differential equation with a general source term $\rho(\vec{r},t)$ consists of solving the same problem, but with Dirac delta source term $\delta(\vec{r})\delta(t)$. The solution to that problem is called the Green's function for that class of problems and often it is useful to Fourier transform the equation to find this function.
The same Green's function can be used to solve any differential equation with the same structure, regardless of the source term. The underlying idea is to build up the full source term out of lots and lots of point sources and finding that the solution can also be built up in a similar way, out of lots and lots of point source solutions. The building up is done by means of convolution.
