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?

  • $\begingroup$ Do you think your potential energy should not rise indefinitely as you move farther and farther away from a massive body like a planet? $\endgroup$ Nov 20, 2018 at 14:38
  • $\begingroup$ In practice, it stops when the voltage is large enough to ionize the air around the capacitor plates and discharge the capacitor that way. The voltage required to start the discharge from a sharp point or corner on the conductors is only a few kV. $\endgroup$
    – alephzero
    Nov 20, 2018 at 14:51
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    $\begingroup$ @AaronStevens: To be fair, it doesn't rise indefinitely, since there is such a thing as escape velocity. $\endgroup$ Nov 20, 2018 at 14:54
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    $\begingroup$ @MichaelSeifert I am not sure I am following. $-\frac{1}{r}$ is a monotonically increasing function as $r$ increases $\endgroup$ Nov 20, 2018 at 14:58
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    $\begingroup$ The field will be inhomogenous if the plates are pulled apart and the required force will diminish, similar to the gravity/planet example. $\endgroup$
    – Jasper
    Nov 20, 2018 at 17:26

2 Answers 2


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.

  • $\begingroup$ I think you could even strengthen the last sentence to "no matter how big the plates will be, the 'linear region' will be a lot smaller". $\endgroup$
    – Jasper
    Nov 20, 2018 at 19:52
  • $\begingroup$ At large distances (where the distance between the plates is much bigger than the size of the plates) the electric field around the capacitor is approximately the same as the field from two point charges, so the force follows an inverse square law. At intermediate distances, the field is something in between the "near field" and "far field" approximations. $\endgroup$
    – alephzero
    Nov 20, 2018 at 21:20

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:

enter image description here

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.

  • $\begingroup$ Ok, now I get it, that was part of the confusion because I wasn't sure how can there be current for a capacitor that is not connected to anything,Now I wonder does the same principles apply for a parallel plate capacitor when we take one plate and slide it away sideways, which means the total surface area between the two plates decreases , what happens to the charge this time? it builds up at the side of the plate that is facing the opposite plate ? but the total charge doesn't change right? $\endgroup$
    – Girts
    Nov 21, 2018 at 9:42
  • $\begingroup$ @Girts If one plate is moved sideways, the capacitance will decrease. Given a constant voltage applied to the capacitor, the charge will decrease (flow back to the power supply), similar to the case when the gap between the plates is increased: Q=CV. $\endgroup$
    – V.F.
    Nov 21, 2018 at 13:52
  • $\begingroup$ @V.F. During the second experiment how does the charge moving between the capacitor and the voltmeter register as a reading on the micro-ammeter? That would require current to flow through the micro-ammeter and so the net charge on the plates connected via your black leads would change and yet those connected to your red leads would not. $\endgroup$
    – Farcher
    Nov 21, 2018 at 14:47
  • $\begingroup$ @Farcher You are absolutely right. I've updated the circuit to reflect that. Interestingly, I made a mental note for myself regarding this, when I was analyzing the video, but then, somehow, put the ammeter on the wrong side of the tap. Thanks for your feedback. $\endgroup$
    – V.F.
    Nov 21, 2018 at 15:59

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