Capacitor demo explanation I know that for a charged capacitor as one separates the plates further apart the voltage increases while the capacitance decreases.
But surely as the plates are pulled further and further apart the potential difference across the plates or voltage cannot rise indefinitely? Where does it stop?
also can someone please explain more in detail perhaps with a schematic the setup seen in this video? 
https://www.youtube.com/watch?v=e0n6xLdwaT0
Especially if he charges the capacitor with a power supply ad then  disconnects the power supply where is then the current measured as the plates are moved apart? I assume the plates aren't electrically connected otherwise the capacitor would discharge itself?
 A: 
to take the gravitational potential energy as comparison feels weird because the further a mass gets from another mass the less force it experiences until a point where the force experienced is so negligible that it counts only theoretically

The same thing is happening with the charge plates.
At a close distance (when the separation is much less than the size of the plates), the field between the plates is uniform and the potential increases linearly with distance.  This is analogous to how we treat gravitational energy near the earth.  The field is nearly uniform, so we assume energy and potential increase linearly with height.
At larger distances, we can no longer assume the field is uniform and the change in energy or potential with increase in distance starts to decrease rapidly.  At large distances, the forces/gravitational/electric fields tend to zero.
When the capacitor plates are small, the linear region for separating the plates will also be small. 
A: 
But surely as the plates are pulled further and further apart the
  potential difference across the plates or voltage cannot rise
  indefinitely? Where does it stop?

Each plate of a charged capacitor will have some charge on it. Each plate will also have some self-capacitance. 
Given that self-capacitance and that charge, we can find the potential of an isolated plate relative to infinity. Since the capacitance of an isolated plate could be 
 much smaller than the capacitance of the capacitor the plate was a part of, the potential of the isolated plate could be very high relative to the voltage on the original capacitor, but it would still be finite as would be a potential of an isolated sphere of a similar size. 
So, if we keep separating two plates of a charged capacitor, at infinite distance, the difference of potentials between them will be just $2$x of the potential of one isolated plate - not infinity.

also can someone please explain more in detail perhaps with a
  schematic the setup seen in this video?
  https://www.youtube.com/watch?v=e0n6xLdwaT0

The purpose of this setup is to demonstrate the relationship between the capacitance, the charge on a capacitor and the voltage on the capacitor, i.e., $C=\frac Q V$.
The circuit of the setup is shown below:

The electrosope (on the top) is used to measure the voltage on the capacitor. The electrometer (on the right) is used to measure charge and discharge current. 
In the first part of the experiment (left circuit), the power supply, set to a $1$V, is always connected to the capacitor, which means that the voltage on the capacitor does not change. So, when the distance between the plates of the capacitor changes and, therefore, the capacitance changes, the capacitor gets charged or discharged, according to the formula. You might notice that, for the same distance adjustments, the charge and discharge currents are more significant when the plates are closer together, since this makes relative changes of the capacitance greater.
In the second part of the experiment (right circuit), the capacitor is charged from the power supply to $1.5$V, after which the power supply is disconnected, which means that the charge on the capacitor will remain roughly the same through the experiment.    
When the distance between the plates is increased, the capacitance decreases and, therefore, according to the formula, the voltage increases, which is indicated by the elecroscope. 
Then, when a sheet of a dielectric material is inserted between the plates, the capacitance increases, which causes the voltage to decrease. 
As mentioned, the charge on the capacitor in the second part of the experiment was supposed to be constant, but we can see that the electrometer still registers significant charge and discharge currents. This happens because the capacitance of the electroscope is not negligible in comparison with the capacitance of the capacitor, so when the capacitance of the capacitor changes, some charge redistribution takes place. So, we can say, that this setup is not perfect, but it still demonstrates basic capacitor relationships.
