You write
When there is a potential difference (voltage) of 10V between two points, it means that we are doing 10 joules of work per unit charge (electron).
This isn't correct. The statement should instead be (in more generality)
Fact. When there is a potential difference $\Delta V$ between two points, then if a charge $q$ moves from one of the points to the other, then its electrostatic potential energy will change by $\Delta U = q \Delta V$.
Therefore, if an electron moves from one point to the other, and if there are no other interactions around, then the change in its electrostatic potential energy in going from one point to the other will equal the (negative of) the change in its kinetic energy by energy conservation:
$$
U_1 + K_1 = U_2 + K_2 \implies \Delta U = -\Delta K
$$
Before thinking about circuits, let's think about two conducting spheres of charge that I connect by a wire. Before I connect the wire, sphere 1 is at voltage $V_1$ and sphere 2 at voltage $V_2$, let's say $V_1>V_2$. I find it useful, in terms of thinking about what's going on, to notice that if the spheres are the same size then saying the spheres have different potentials is equivalent to saying that the 2 spheres have different charges residing on their surfaces (you can justify this by noting that the capacitance of a sphere is determined by its radius).
Now let's connect the spheres. What will happen? Well, a current will flow in the wire. This will take positive charge off of sphere 1 and deposit it on sphere 2 [strictly speaking if you want electrons to be charge carriers, then negative charge is flowing from 2 to 1; but in terms of thinking about what's going on it's easier to imagine, and mathematically equivalent to say, that positive charges are going from 1 to 2]. This in turn changes the voltages on the two spheres; $V_1$ decreases and $V_2$ increases. The process stops when $V_1=V_2$. Again, if the spheres are the same size this condition is equivalent to the charges on both spheres being equal.
OK, now imagine a battery hooked to a resistor and a switch, the simplest circuit imaginable. Before we close the switch, terminal 1 is at $V_1$ and terminal 2 is at $V_2$. At this point, it makes perfect sense to think of each terminal of the battery as being a sphere of charge. Then we close the switch, this is like connecting our spheres with a wire. Based on our silly model of a battery, you would the voltage between the two terminals of the battery (ie, $V_1-V_2$) to decrease until eventually it reached equilibrium with $V_1=V_2$. Clearly, a battery does not behave like two spheres of charge after the circuit is closed.
The whole point of a battery is that it maintains the potential difference between its two terminals. After we close the switch, a little bit of positive charge flows from terminal 1 to terminal 2 by going through the circuit. Naively this means that terminal 1 has less positive charge and terminal 2 has more positive charge, so terminal 1's voltage decreases while terminal 2's voltage increases. Inside the battery, some process takes place to to take the excess positive charge on terminal 2 and put it back onto terminal 1. Whatever this process is, it cannot be electrostatic, because positive charges following the electric field can only ever move from terminal 1 to terminal 2 [positive charges move from high voltage to low voltage, if the only force is electrostatic].
The details of what the battery does to maintain the potential difference varies depending of the kind of battery. A conceptually simply example of a battery is a Van de Graaff generator. In a Van de Graaff generator, you have a conveyer belt that literally carries the excess positive charge on terminal 2 and deposits it back on terminal 1, undoing the naive 'equilization process.'
Most useful batteries rely on some chemical process to maintain the potential difference. For example, one can use oxidation reactions to do this. The details involve some chemistry (there's a wikipedia summary at http://en.wikipedia.org/wiki/Electrochemical_cell), but essentially you put each terminal in a bath of ions, and the chemical energy of the reactions at each terminal [balancing oxidation and reduction] forces ionized atoms to carry electrons from terminal 2 to terminal 1.
Best Answer
Actually, voltage is a difference in potential energy per unit charge. Specifically, the voltage between two points A and B is the difference between how much PE a unit charge would have at A, and how much PE it would have at B. It's important that you always have two points in mind when talking about voltage, since it's technically meaningless to talk about the voltage at a single point.
Now, you presumably know that the difference in potential energy per unit charge between two points is related to the gradient of potential energy per unit charge between those points. $$V_B - V_A \equiv \Delta V_{BA} = \int_A^B \vec{\nabla} V(\vec{x})\cdot\mathrm{d}\vec{x}$$ where $V$ is potential energy per unit charge (a.k.a. "electric potential"), $\Delta V$ is voltage, and the integral is a line integral along the path from A to B. Well, the gradient of $V$ here is physically the same thing as the electric field. And as you probably know, when a charge is placed in an electric field, the field causes it to experience a force. That force is precisely $-q\vec{\nabla} V(\vec{x})$.
Given a path, if the difference $\Delta V_{BA}$ is large, then there will be a large gradient somewhere along the path, thus a large electric field, which causes a large force pushing the charged particles from A to B, which in turn makes the particles accelerate faster and move faster, producing a larger current.