Can one Voltage produce two different EMF's? Voltage is a measure of electromotive force, but it is measured by how much work a charge can do as it flows between two points.
Say I seperate some amount of positive and negative charge and in doing so create a voltage. Now let's say I seperate the same amount of positive. and neg. charge but move the two farther apart. In the second case, the voltage should be the same since I've increased the distance between the charges but decreased the force acting on them so the amount of work the charge can do has stayed the same. 
Now I have two cases where the voltage between two points is the same, but the Electromotive force is different.
How can this be? Did I make incorrect assumptions? Does this mean potential difference is not really a measure of emf.
 A: 
Say I seperate some amount of positive and negative charge and in doing so create a voltage. Now let's say I seperate the same amount of positive. and neg. charge but move the two farther apart. In the second case, the voltage should be the same since I've increased the distance between the charges but decreased the force acting on them so the amount of work the charge can do has stayed the same.

Your reasoning is faulty.
Say the first two charges are separated by distance $d$.
Consider what happens as you separate the second pair of charges. When you separate them by distance $d$, you've done the same amount of work on them as you did on the first two charges, so you the voltage between them is the same as between the first charges.
In order to separate the charges further, you must do additional work on them. Therefore once you get them further apart, the voltage between them will be greater.
The fields acting on these two particles will be lower now that they are further apart, but that doesn't relate 1-to-1 with the amount of work done to separate them. It only tells you about the differential work needed to separate them even further.
A: 
Now I have two cases where the voltage between two points is the same,
  but the Electromotive force is different.

The Photon has already addressed the misconception in the first half of the quoted statement so I'll address the misconception in the second half.


*

*The emf is the same in both cases for the simple reason that the
emf is zero.


The situation that you've described cannot 'drive' charge around a closed path but driving charge around a closed path, e.g., an electrochemical cell driving a current through a resistor, is precisely what a source of emf can do.
Moreover, it is the cell's emf that keeps the charge separated against the electrostatic attraction.  And it is the charge separation that gives rise to the voltage across the cell's terminals.  In other words, it is the cell's emf that creates the voltage rather than the other way around.
