We say that we can charge a capacitor in proportion to the potential difference we apply across its plates and the maximum potential difference depends on the dielectric strength of the medium. Now suppose there is vacuum between the plates. Is there any limit on the maximum charge we can store in it? My concern here is for atomic properties of the material, as there is a fixed number of charges in a material.

  • $\begingroup$ Fun fact -- if you deplete a metal capacitor plate out of "all its charges" (so much as removing even just "one electron per atom") it simply will explode in gaseous poof of metallic ions! (especially in a vacuum) And you will no longer have a "capacitor plate" but a gas of mutually/electrostatically-repulsive particles... $\endgroup$
    – ManRow
    Jan 30, 2022 at 19:23

2 Answers 2


The charges that accumulate on the plates of a capacitor are not provided by the material of the plates themselves but by the source that is charging them, so there is in principle no limit to the amount of charge that they can hold, if your source is strong enough.

The maximum-charge limits on actual physical capacitors are dictated by the dielectric breakdown of the medium between the plates, which typically is an insulating dielectric of some kind - common materials are paper, plastic, glass, mica and ceramics, as well as air and vacuum. In a dielectric breakdown, some of the (few) free charges in the dielectric are released by the electric field if it becomes too strong, and as they travel through the dielectric they ionize other parts of the material, setting up an avalanche reaction. The result can be a momentary spark, permanent damage to the dielectric, or even an explosion.

In air, the dielectric breakdown mode is simply avalanche sparking between the plates. If your reduce the air density between them, though, this mode becomes less and less effective, and at a good enough vacuum it will stop altogether.

The failure mode that you do get on a vacuum capacitor is leakage, which is caused by thermal electrons jumping out of the cathode and getting snatched by the electric field in the middle. This mechanism is what makes cathode rays work, but at low temperature it's still present but at much lower intensity. This effect is not a catastrophic breakdown, and it is easily modelled as the capacitor having a small component of resistive reactance.

If you really care about this (and honestly you only really care about this if you're a metrologist and you want a very steady phase reference, with low systematic errors, for AC circuits), then what you can do is cryogenically cool the electrodes, which will drive down the thermal energy the cathode electrons have available to jump out of the cathode, and therefore will limit the leakage current. For an example of this in practice, see

Capacitors with very low loss: cryogenic vacuum-gap capacitors. N.M. Zimmerman. IEEE Trans. Instrum. Meas. 45 no. 5, 841 (1996); NIST eprint.

  • $\begingroup$ Or alternatively, heat the cathodes a lot to produce diodes and tube -amplifiers! $\endgroup$ Sep 25, 2015 at 12:56
  • $\begingroup$ Taking issue with your opening paragraph: since one electrode gets excess electrons (from an external, infinite supply) and the other gets electrons taken away (finite supply), I would say there is a limit: once there is sufficient electron-deficit on the positive plate, the material stops being held together by the remaining electrons. This happens at a charge level orders of magnitude above breakdown / leakage etc. thresholds, but it should be considered an "ultimate upper limit". $\endgroup$
    – Floris
    Sep 25, 2015 at 13:08
  • $\begingroup$ @Floris fair enough, though I don't think there's a realistic experiment that will show this behaviour. $\endgroup$ Sep 25, 2015 at 13:11
  • $\begingroup$ @EmilioPisanty indeed you reach other limits first - I was just responding to the final paragraph of the question ("concern for atomic properties...fixed number of charges") $\endgroup$
    – Floris
    Sep 25, 2015 at 13:20
  • 1
    $\begingroup$ Just for kicks, I did the calculation... $\endgroup$
    – Floris
    Sep 25, 2015 at 13:48

Just to expand a little bit on Emilio's answer, and to address the specific point you made in your final paragraph about "finite charges", I decided to calculate what fraction of electrons would be "missing" from the positive plate of a capacitor when you reach breakdown.

For a vacuum dielectric, the Schwinger limit is about $1.3\cdot 10^{18}\rm{V/m}$ - ignoring all other sources of field emission (see for example the extensive description of Fowler-Nordheim analysis in this article on field emission, you cannot maintain this electric field (note - it is orders of magnitude higher than any realistic field you will ever maintain).

The relationship between surface charge density $\sigma$ of a capacitor with spacing $d$ is given by


Setting this equal to the Schwinger limit, we find the maximum surface charge

$$\sigma_{max}=2\epsilon_0 E_S = 2\cdot 8.85\cdot 10^{-12} \cdot 1.3 \cdot 10^{18} = 23 \;\rm{MC/m^2}$$

If the conductor is copper, the size of a copper atom is about 128 pm radius. This means that the number of atoms in the surface layer is

$$N_s = \frac{1}{2\sqrt{3}(128\cdot10^{-12})^2} = 1.8\cdot 10^{19}$$

If you removed one electron per atom, you would have a surface charge of

$$\sigma = N \; e = 1.8\cdot 10^{19} \cdot 1.6\cdot 10^{-19} = 2.8\;\rm{C/m^2}$$

There are about 7 orders of magnitude between these: in other words, you will never be able to create a material that maintains its mechanical integrity and generates a field even close to the Schwinger limit.

If we assume that "one electron per atom missing" is an upper bound, this would correspond to a field of $159 \;rm{GV/m}$ - still a ridiculously high value, well above what can be maintained.

  • $\begingroup$ Where did the $2\sqrt{3}$ come from? Also, shouldn't the electric field inside the capacitor simply be $E = \sigma/\varepsilon_0$, without the factor of $1/2$ ? Also, if you permit "thick-enough" plates, you can probably then extract enough electrons from the material to exceed the surface $\sigma_max=23\,MC/m^2$, although those plates will probably lose their integrity as well, as extracting one electron from every copper atom will result in some weird material comprised entirely of $\text{Cu}^+$ ions likely to simply "explode" on itself over all that electrostatic repulsion! $\endgroup$
    – ManRow
    Jan 30, 2022 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.