Potential Energies of Charges Do particles have potential energy because they experience Coulombic forces from other particles? If a particle is placed in a region where there is no charges present around, then is the potential energy zero?
 A: In a word, yes. The important thing to remember here is that charges do not feel the effects of their own fields.
A: Charged particles or objects acquire potential energy when work is done on them in order to move them in a region where an electric field is present. The change in potential energy is, then, equal to the opposite of the work done (assuming only electrostatic field is present in the region):
for example if a charge $q = 1\,\, C$ is moved by $1$ m in a region of constant electric field of magnitude $E =1 $ $\frac{V}{m}$ parallel to the displacement, the change in potential energy of the charge is $\Delta U = - W =  q E \Delta x = 1 \,\,J$
As you can see, I made no explicit reference to the presence of source charges: what is important is the presence of an electric field in the region of interest. Let's say you have a region $\Sigma$ somewhere in the Universe where there are absolutely no charges except for the one you put there (your test charge), but an electric field is present. Your particle will then experience a force due to the electric field and, therefore, change its potential energy, even if no other charge is present inside $\Sigma$. Of course, since charges are the ultimate sources of every electric field, there will be, in some other region of the Universe, a distribution of charge that creates the electric field in $\Sigma$.
So to summarise the answer, the presence of a distribution of charge in a given region is not strictly necessary in order for a charged test particle to change its potential energy; all there needs to be is an electric field in the region. This implies, of course, that a distribution of charge is present somewhere to generate the electric field of interest. It just doesn't have to be in the region of interest.
EDIT: Mathematically, what I am saying is that even if the charge in a given region $\Sigma$ is zero, the potential $\phi$ is not necessarily zero: in fact it satisfies the Laplace equation $\nabla ^2 \phi = 0$ that, with appropriate boundary conditions, has an unique solution for the potential in the region of interest.
