Why is the voltage in inhomogeneous electric fields undefined? For example I'm thinking about a radial symmetric field like a point charge. The voltage $U$ can be determined by $$\int_{r_1}^{r_2} E(r)\mathrm{ds} = \frac{Q}{4\,\pi\varepsilon_0}\,\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$ My questions are: why can't the voltage between the origin $r_1 = 0$ and any other point $r_2$ not be calculated? For instance in a homogeneous field like the one inside of a plate capacitor $U$ is given by $$\int_{r_1}^{r_2} E \,\mathrm{ds} = E\,(r_2-r_1)$$ $E$ doesn't depend on $r$, right? That means, $r=0$ seems totally legitimate. But where does this huge difference come from?
 A: I presume (perhaps incorrectly) that you are referring to the singular behavior of the electric fields (and potentials) of point charges. The origin of this singularity is a deep question which requires quantum field theory to adequately answer.
Computationally, this issue is brushed under the carpet by setting $V=0$ at infinity, in which case the potential of a point charge a distance $r$ away from the charge is given by
$$\int_{\infty}^r \text{d}r ~E(r) = \frac{Q}{4\pi \epsilon_0 r},$$
which is finite for any $r > 0$. However, the singularity you describe still occurs in the $r\to 0$ limit. The classical explanation of this points to the fact that the charge distribution of a point charge is given by
$$\rho(\mathbf{r}) = Q\delta^{(3)}(\mathbf{r}-\mathbf{r}_0),$$
which is infinite as $\mathbf{r} \to \mathbf{r}_0$. In the case of charged parallel plates the charge distribution is "smeared out" and defined by
$$\rho(\mathbf{r}) = \sigma\delta(z-z_0).$$
Notice that although the latter also has the delta function, it's a "weaker" singularity in the sense that
$$E \sim \int \rho ~ \text{d}z = \sigma \int \delta(z-z_0)~\text{d}z = \sigma  \quad \text{(finite)},$$
but an analogous integral for the point particle will leave a "two dimensional" delta function behind, which exhibits singular behavior.
A: The electric field for a point charge is divergent as $r \to 0$, which means it cannot be integrated with $r = 0$ as one of the endpoints (the integral does not exist).
As to why the electric field is divergent, it's best to think about it in terms of Gauss's Law.  We know that (roughly) field lines begin at positive charges and end at negative charges.  We also know that the number of field lines per area is (roughly) proportional to the electric field in a region.  So let's consider a sphere of radius $r$ surrounding a point charge.  The number of field lines poking through that sphere is independent of $r$, while the area of the sphere is proportional to $r^2$.  Thus, the electric field is proportional to the number of field lines per area, which is proportional to $1/r^2$ according to the above logic.
