# Levitating capacitor

As the title already says, I would like to make a large capacitor levitate.

Before you dismiss this question please hear me out: In essence, this is intended to be a way to avoid the constrains of a "vacuum airship". Basically the problem with the vacuum airship is that a very rigid body is needed to withstand the atmospheric pressure, and it has been proven that there is currently no material that is rigid enough to withstand atmospheric pressure and at the same time light enough to maintain an overall density that is lower than the density of air.

In the case of a hot air balloon or a helium balloon we don't have this problem because the balloon doesn't have to be rigid. It relies on the pressure of the low density gas to counteract the atmospheric pressure, such that the net pressure on the balloon is zero.

Now my original idea was to take an evacuated (conductive) balloon and to give it a very high electric charge, such that the walls of the balloon repel each other and the balloon expands. By expanding, the balloon would occupy more space without it's mass having changed significantly, therefore the density of the balloon would reduce. At some point the balloon should start to levitate because it's density would be lower than the density of the surrounding air. A potential conductive material could be aluminium foil...

Okay, so here come some figures to see how realistic this is:

$$\text{Dry air has a density of} \;1.2754 ~\text{kg}{\text{m}^{-3}} \\\text{At}\; 6.35 ~\mu \text{m} \;\text{foil weighs}\; 17.2 ~\text{g}\text{m}^{-2} \\\Rightarrow \text{A charged cube shaped balloon with a volume of}\; 1~\text{m}^3\; \text{have an upthrust of} \\F \approx (1.2754-0.1032)~\text{kg} \cdot 9.81\text{m}\text{s}^{-2} = 11.499282 ~\text{N} \\\text{Data from various sources.}$$

The work done to charge this balloon can also be approximated easily: $$\text{Work} \approx F\cdot s = 11.5~\text{N} \cdot 1~\text{m} = 11.5 ~\text{J} \\\text{Ideally it would charge up in} \; 11.5 ~\text{s}\: \text{with a} \; 1~\text{W}\:\text{ power supply!}$$ (Assuming that all the air was displaced from the bottom of the balloon)

Edit: It would take much more than 11.5 seconds to charge up with a 1 watt supply due to the factors mentioned in one of the comments. This doesn't matter too much though, because the invested energy can be completely recovered upon deflating the balloon (ideally).

If you think that the aluminium foil is too thin to make the approximation realistic, you can scale up the entire balloon to get a much larger upthrust and a satisfying thickness. Essentially the key point of charging the aluminium foil is so than the balloon can be scaled up without us having to change the thickness of the aluminium.

Off course it's not realistic because aluminium would have to have a very high charge and this would quickly dissipate into the surroundings by ionizing the air.

What I actually tried to do in an attempt to use this effect is:

1. I took an empty, thin, airtight plastic bag.

2. I placed some graphite powder in the plastic bag.

3. I charged the graphite powder through an aluminium contact in the plastic bag, using a 1 watt negative ion generator.

Basically what I just described here is 1/2 a capacitor because the charge generator only generates negative ions (I have no clue what it does with the positive charge). If I would have a high voltage source (positive and negative terminals), I would off course take a second balloon/plastic bag for the opposite charge.

When I tried it out, I was disappointed to see that nothing much happened. There were some sparks from time to time and I did manage to observe a tiny bit of repulsion between the graphite particles, but it was nothing significant.

Finally the question is:

What am I doing wrong?

Could it have something to do with the fact that I only have one terminal and I don't know what is happening with the other charge?

Could it be that I am losing too much charge to the surroundings? (Definitely part of the reason).

Could it be that that I need much higher voltages to achieve this kind of repulsion? (I am having trouble judging the required voltage, because coulombs law get's a bit tricky with so many particles involved and I also don't know the mass of my graphite particles).

And last but not least: Could it be that I made a catastrophic mistake in my reasoning?

Off course I would be really happy if someone with access to a high voltage source is willing to try it out. The applications of such a balloon/air ship would be incredible...

Please note that I am not a physicist and so I may have made mistakes in my reasoning.

Thanks :-)

• The following link might be of interest. It describes a patent for an expandable capacitor which will break its connection if the capacitor is overloaded: patentgenius.com/patent/4283750.html. However the force of expansion is caused by gas which forms if the capacitor is overloaded. Commented Aug 23, 2015 at 0:46
• Work equals force times distance but you appear to be comparing it with just the linear expansion rather than a 3D expansion. I'm not sure that's correct. You need to consider the atmospheric pressure on six square meters, the surface of your balloon, and the amount of force it'll take to maintain that volume against the atmospheric pressure. Commented Aug 23, 2015 at 11:38
• That's a very good point, here I only calculated the work needed in order to displace 1m^3 of air. I should have estimated the work against the atmospheric pressure (which requires a lot greater).. Commented Aug 23, 2015 at 11:59
• If your balloon is so highly charged, then isn't it going to be strongly attracted to the ground by the electrical force? Commented Aug 23, 2015 at 19:35

## 1 Answer

Your description is not very complete, but I guess what happens is exactly what you expected: to get significant repulsion (to counteract the atmospheric pressure), you need very high charge, which will be necessarily limited due to air discharge (maybe that is why you observed sparks). I don't see how replacing an aluminum shell with a plastic bag containing graphite can change anything: the electrostatic field of charged graphite is not attenuated much by the plastic bag and causes air discharge.

As for the constraints on vacuum balloons that you give a link to, I derived those constraints myself (with my coauthor). The authors of the Wikipedia article just quoted our patent application. The irony is the derivation only proves that homogeneous shells cannot be used as vacuum balloons. In the same patent applications we emphasize, based on finite element analysis, that non-homogeneous spherical shells (sandwich structures) made of commercially available materials can do the job with a decent safety factor. You may wonder why I don't bother to amend the Wikipedia article myself. Maybe I should, but I have not done it so far for at least two reasons: first, the patent application is not peer-reviewed, second, I had bad experience trying to amend another Wikipedia article (https://en.wikipedia.org/wiki/Dirac_equation#As_a_differential_equation_in_one_real_component ) based on my article in a decent journal. My amendment was undone at least twice and was restored only after a rather nasty discussion:-)

• Maybe it would be better to use magnetic fields rather than electric fields.. Commented Aug 25, 2015 at 9:29
• @ChandranGoodchild en.wikipedia.org/wiki/Hairy_ball_theorem Might work with a donut balloon...>_<
– Aron
Commented Jul 5, 2016 at 5:12
• @ChandranGoodchild "less charge would be on the surface" Nope. That isn't how the skin effect works. "easier to get an airtight plastic bag than to make an airtight balloon with aluminum foil" Nope. See foil balloons. "I thought I might need a lower charge as the distance between the graphite particles is smaller than the distance between opposing sides of an aluminum balloon" Maybe, but since the density of plastic is higher than air, the it actually the added mass is more than the added volume...So NOPE.
– Aron
Commented Jul 5, 2016 at 5:19
• As already mentioned, you would need a lot more energy to make the sides repel. However, it order to get this energy in there, you would have to insulate the outside so that the air did not conduct it elsewhere. Commented Jan 6, 2021 at 14:34