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?
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.