How can supercapacitors store $5\,\mathrm{coloumbs}$ and not implode due to the enormous force between the plates ($10^{15}\,\mathrm{N}$ if the plates are $1\,\mathrm{cm}$ apart)?
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This is a good question. It comes down to two factors: The 'plates' have dielectric material separating them, and the effective size of the plates is large, relatively speaking.
The dielectric material has positive and negative charges that align themselves with the electric field of the electrodes. Fig 2 in this link shows a very simplified view of what goes on. The charges in the electrolyte move and align themselves with the electrodes, thus the force the electrodes experience is actually just to the local charges near the electrodes, not all the way to the opposite electrodes. Also keep in mind that unlike this figure, the electrodes are surrounded on all sides by electrolyte. So In other words, it's not a net force acting on each plate, but rather just a local force on each microscopic part of the plates.
This brings us to the next part. The electrodes are made up of a highly porous matrix of carbon and other materials, like this illustration. The effective surface area is high, on the order of 1000 $\mathrm{m}^2/\mathrm{g}$. So even though the stored charge is high, the surface charge density is low, or at least low enough for the materials to handle.
We can do some rough back-of-the-envelope calculations. Assume:
- 1000 $\mathrm{m}^2/\mathrm{g}$ surface area.
- 100 F/g capacitance.
- 2.7 V breakdown voltage.
Then one obtains a charge of $q = Vc = 2.7 \times 100 = 270$ C/g. Using the naive formula for force between parallel plates:
$$ F = \frac{Q^2}{2A\epsilon_0} $$
One obtains a pressure on the order of ~1 GPa. This is by using vacuum permittivity, which doesn't exactly apply here, but we can use it anyway for a rough estimate. 1 GPa is much below what would be required to e.g. tear apart the conductors at the molecular level.
Note: Capacitors with vacuum/air dielectric do exist, however their capacitances are very low, thus the amount of charge stored on them (and the force between the plates) is low.
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9$\begingroup$ While capacitors with vacuum or air dielectric exist, supercapacitors with vacuum or air dielectric are not possible by definition. $\endgroup$– HearthCommented Dec 30, 2021 at 3:58
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$\begingroup$ @Hearth definition of what? Generally a supercapacitor is just defined as a capacitor with much higher specific energy than traditional capacitors. If somebody figured out how to make a capacitor with vacuum dielectric that has $\geq 10^4\:\mathrm{\tfrac{J}{kg}}$, then this could surely be called a supercap too. $\endgroup$ Commented Dec 30, 2021 at 15:29
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$\begingroup$ @leftaroundabout "supercapacitor" refers to an electric double layer capacitor, which is a complicated electrochemical thing that I don't fully understand. It is, however, a term for a specific type of capacitor, one which does not have a vacuum or air dielectric. $\endgroup$– HearthCommented Dec 30, 2021 at 15:40
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$\begingroup$ @Hearth no, the term for the specific type of supercapacitor that uses this complicated electrochemical thing is precisely “electric double layer capacitor”. Other types exist. $\endgroup$ Commented Dec 30, 2021 at 16:01
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1$\begingroup$ @leftaroundabout It would appear I was wrong, then. $\endgroup$– HearthCommented Dec 30, 2021 at 16:16
Kis $1 / 4\pi\epsilon_0$, so yes, it includes the dielectric constant of a vacuum (en.wikipedia.org/wiki/Vacuum_permittivity) as @physics suggested it should. Some parts of the en.wikipedia.org/wiki/Coulomb%27s_law article are careful to say "in a vacuum", but the early part of the article isn't. Clearly you should use the dielectric constant of the actual insulator, not a vacuum, although that's probably within one order of magnitude, like a factor of 4? e.g. physics.bu.edu/~duffy/semester2/c08_dielectric_constant.html 3.6 for paper $\endgroup$