# Can a balloon be inflated with electrons?

If I use a Van De Graff generator to pump electrons into a deflated balloon, eventually negative charge will start to build up inside the balloon. Assume the mouth of the balloon is sealed so air can not enter or exit the balloon. The electrons will push against each other and from my understanding build up mostly along the inner surface of the balloon.

Will these electrons push against the sides of the balloon strongly enough to inflate the balloon or will this whole setup be impossible for some reason I am not seeing like the electrons drifting through the balloon walls too quickly to build up?

Assume the balloon is as thick as necessary.

Edit 1: I was imagining that as more and more electrons are pumped in some would start colliding with the walls of the balloon. These constant collisions with the balloon walls would cause a "pressure" outward on the walls of the balloon, the same way air molecules colliding with balloon walls creates pressure. Why would the balloon not just inflate like it does with air after the relatively small amount of pressure required to inflate balloons is reached? I agree the "pressure" to add electrons would raise quickly, but that leads me to think it would inflate with fairly "few" electrons added.

Edit 2: Sorry for the follow up questions. I am by no means claiming the balloon would inflate, I just don't understand why air would work and electrons would not. If the balloon wall absorbs some electrons, won't it become negatively charged quite quickly and then directly feel a repulsive force with the other walls of the balloon and the still free electrons? Then it would inflate by a direct force instead of the indirect collisions of the electrons on its walls.

Edit 3: Thank you for all of the interest in this question. Due to some questions in comments I will provide some clarification details. After reading observations from count_to_10, Hames Large, and Marty Green regarding "hot" vs "cold" electrons, I see the experiment as the following diagram:

A) The environment outside the balloon. This could either be air or a vacuum, so long as the entire experiment could fit inside a modern vacuum chamber.

B) The interior of the balloon (marked in red). As Ulthran pointed out this would essentially be a vacuum with possibly some electrons drifting around.

C) The top ball of a Van De Graff generator. Andrea Di Biagio pointed out that we have to be careful to not let all the charge accumulate in one spot. Normally the ball at the top would be flat, but here there are points attached to help the electrons discharge from the ball in a more or less uniform pattern. This design is from my limited knowledge of Van De Graff generators. Please feel free to suggest something else if there is a better way to move "cold" electrons into the balloon.

D) Here is the mouth of the balloon. This is somehow firmly attached to the pole of the Van De Graff generator with a highly insulating material. The goal is that the electrons will not move to the pole and air will not be able to enter the balloon and electrons will not be able to exit.

I think Andrea Di Biagio is the closest to a definite answer so far. The calculation arrived at about $N = 4.633*10^{13}$ electrons required to generate enough pressure (Thanks to Rotsor for some calculation adjustments). So the question is, can we fit this many electrons inside the balloon (turns out they all end up on the inside wall of the balloon), or will something happen to stop this, for example leaking through the balloon wall, dielectric breakdown, rubber chemical bonds breaking or something else?

• Good thought experiment. But I'm not sure if its practical. – docscience Aug 24 '16 at 20:55
• For this to work, in order for a balloon to be 'inflated' by electrons, the repulsion between the electrons building up would have to be great enough to overcome the pressure of the air around. You're essentially creating a vacuum, as air cannot enter nor exit the balloon. I'm not certain, but I would guess that the balloon would begin discharging into the air before it reached that point. – Ulthran Aug 24 '16 at 20:57
• @Ulthran What would be the primary means of electrons discharging into the air? I assumed the rubber in the balloon would be an insulator and stop the electrons from conducting out of the balloon. Is a typical balloon just too thin to be a proper insulator? – Andrew Aug 24 '16 at 21:00
• Honestly I'm not sure whether or not the balloon's lack of conductivity would affect the discharge level... It's always harder to judge that kind of thing with insulators. – Ulthran Aug 24 '16 at 21:04
• Fantastic question - the sort that should be able to be answered with our theoretical knowledge, yet it appears to be a struggle to make a convincing answer! – GreenAsJade Aug 24 '16 at 23:43

Electrons in the balloon will want to get as far as possible from each other, so they will push on the surface of the balloon, which will tend to inflate it. For what I can see, the success of the experiment will depend on two things.
Firstly, will there be enough electrons in the balloon so that the forces are evenly distributed? If you put a very strong force in a limited area you will puncture the balloon.
Secondly, will the charge density be low enough so that the balloon does not discharge?

To do some calculations, assume that there are $N$ electrons inside the balloon, and that they are evenly distributed on its surface spherical surface, with radius $r$.

Due to the symmetry of the situation, each electron will experience a radial force, equal to that it would experience if a charge of $(N-1)q_e \simeq Nq_e$ was placed at the centre of the sphere.

The pressure is the force per unit area, hence $$p = \frac{N}{4\pi r^2}\frac{kNq_e q_e}{r^2} = \frac{kN^2q_e^2}{4\pi r^4}$$

Now, for this situation to even be plausible, $p$ needs to be of the order of atmospheric pressure $101~\mathrm{kPa}$, which will inflate the balloon before engaging the surface tension. Plugging in $r=2.5~\mathrm{cm}$ (a fair guess on the radius of a floppy balloon) and solving for the number of electrons we get: $$N\simeq 5\times 10^{13}$$ giving a number density of $\rho = 6\times 10^{15}~\mathrm{m^{-2}}$.

We can now check our initial assumptions. The average separation, between two electrons will be of the order of the side of a square with area $1/\rho$, that is about $10^{-8}~\mathrm{m}$.

I will need some help here to continue. This means to me that the electrons are dense enough so that the forces are not too concentrated to rip the balloon mechanically. However, several electrons so close to a molecule could disrupt their chemical properties, thus degrading the material. Unfortunately, I don't know enough chemistry to answer. Hopefully someone can contribute?

• Would all the electrons really distribute only on the surface of the balloon? If one electron was placed 1mm off the wall, wouldn't it feel a net force toward the center? A very strong force from the wall it is right next to, and a weak force from the opposite wall, because the force is proportional to $\frac{1}{r^2}$. – Andrew Aug 24 '16 at 21:52
• I think I don't understand why we can model the total force as a point charge at the center with $(N-1)q_e$ charge. If you place an electron 1 picometer from the center it will experience almost 0 force due to symmetry, but this assumption will say it will experience a massive force. – Andrew Aug 24 '16 at 22:00
• This kind of approximation is actually quite common in problems involving forces of the $1/r^2$ kind. See en.m.wikipedia.org/wiki/Shell_theorem – Andrea Aug 24 '16 at 22:07
• Hi Andrea, congrats on the math, I would still be on the first line :). I list the compounds of a balloon below in my answer, so work away with them, but I don't believe there will be any inflation, there are just too many options for the tiny hot electrons compared to air. It's a fun question though, best of luck with your answer, – user108787 Aug 24 '16 at 22:13
• @count_to_10 thanks! I will look at it in the morning (it's almost 12am here) and see what I can do. Tbh I was just thinking of an electrostatic problem, where electrons "appear" on the surface of the balloon, with no dynamics whatsoever. – Andrea Aug 24 '16 at 22:31

The electrons may interact with the material of the balloon but, if you read my reasons below, I think the electrons will travel straight through the plastic.

Balloons are made of three parts:

1. Latex: Ca(NO3)2 + H2O + C2H6O, C5H8, (2-methyl-1,3-butadiene)
2. Pigments are; Ultramarine Blue,  (Na8-10Al6Si6O24S2-4) + (Na3CaAl3Si3O12S), Red, Hematite (Fe2O3), and Yellow Ochre, FeO(OH) nH2O.
3. Coagulant : Ca(NO3).

I am by no means claiming the balloon would inflate, I just don't understand why air would work and electrons would not. If the balloon wall absorbs some electrons, won't it become negatively charged quite quickly and then directly feel a repulsive force with the other walls of the balloon and the still free electrons? Then it would inflate by a direct force instead of the indirect collisions of the electrons on its walls.

We have the technology for pumping cold air against a pressure gradient, but not for pumping cold electron gas, so we cannot equate the two as regards filling the ballon. So using hot electrons is our only option and this will destroy the balloon.

My sincere thanks to James Large for pointing this out to me and my apologies to the OP for not grasping this point earlier, if that is what is being referred to in his question above.

I contend that most of the hot, tiny, fast moving electrons will either combine with one of these many compounds in the wall of the balloon or, far more likely, just pass straight through it. In other words, the balloon walls may as well not be there in the first place.

Image from Electron Gun Wikipedia

An electron gun, from an old TV set. The screen of these tvs incorporated lead-oxide glass since fast electrons are dangerous, and the k.e. of the electrons is probably high to easily burn through plastic. (Correction to original text thanks to James Large)

What happens if we keep pushing electrons into the balloon, (even if we did have a cold electron gas system)? The kinetic energy of the electrons inside increases, but the force necessary to push more and more electrons into a stronger and stronger sphere of negative charge would be considerable. It may well be that the heat generated by the k.e. of the electrons and the system needed to pump them into the balloon would create enough heat to melt the plastic in a very short time.

• Comments added to question for clarity – Andrew Aug 24 '16 at 21:37
• Surely if he kinetic energy of the electrons is high enough to make it hard to inject more, it's also enough to inflate the balloon? – GreenAsJade Aug 24 '16 at 23:43
• Re, "The screen of these tvs was lined with lead (that's why they are so heavy)" No, not lined with lead: They were made with lead-oxide glass--the same stuff that fine crystal is made of. It's not that much heavier than regular glass. The main reason big CRTs weighed so much is that the glass had to be thick to withsatand atmospheric pressure. (Nothing inside but hard vacuum!) The face of a big color TV screen might be 450 square inches or more. That would have to support more than three tons of atmospheric pressure at sea level. – Solomon Slow Aug 25 '16 at 0:53
• Re, "hot, tiny, tiny fast moving electrons," Yes. A high-energy electron beam would burn right through the balloon, but I think the OP was imagining that we could somehow fill it with cold electrons. I don't know of any technology that could do that, but if there was such... I'm guessing that the ability of the balloon fabric to contain the electrons would be limited by its dielectric strength. It's the strength of electric field (measured in Volts per meter) that will cause a material to conduct electricity. – Solomon Slow Aug 25 '16 at 1:01
• Re, "what happens if we keep pushing electrons into the balloon? The kinetic energy of the electrons inside increases, but the force necessary to force more electrons into a stronger and stronger sphere of negative charge would be considerable." That also sounds a lot like what would happen if we tried to keep pushing more and more air into a balloon. The end result being, that some of the energy goes into stretching the balloon fabric. The main difference is, that we have technology for pumping cold air against a pressure gradient, but not for pumping cold electron gas. – Solomon Slow Aug 25 '16 at 1:07

One thing you CAN'T do is inflate a balloon by rubbing it on your sweater. You can get quite a bit of charge on the balloon in terms of voltage (maybe $10$kV) but the balloon doesn't get any bigger.

I'm doing some ballparking here and for a pressure of 1 atmosphere, I get $10^{15}$ electrons at a $10$cm radius, which is close to @Rotsor in the comments field of Di Biagio's answers. That gives me a voltage of around $10$MV, or $1000$ times what you get by rubbing the balloon on your sweater - and in the ballpark of what you might get from a very efficient Van de Graaff generator. Which is (significantly) cold. I guess if you could get those electrons inside a balloon so they couldn't discharge, you would inflate it.

(I love this question, physics whatif's are a lot of fun. In the absence of being able to comment, I'll just plow ahead with a related answer, sorry.)

Riffing off some of the answers above, I suspect that the answer is no, it won't inflate a balloon at one ATM pressure because of the materials science aspect of things. If you have to raise the voltage to 10 MV to provide inflation pressures (Analysis of voltages by Marty, great constraint on physical reality, dielectric strength is a good way to simplify the material properties.)) then you have "containment problems" Electrons don't bounce, they just follow potential, and at 10MV, you have serious dielectric breakdown issues. 10MV is enough to create an arc 3 meters long in air (see Voltages involved in Lightning) So trying to contain that potential without electron flow in a "Balloon" will turn the shape into a globe with thick walls, instead of a balloon. Is there a way to require inflation force to maintain shape, yet be dielectrically sound? Probably not at 1Atm pressure.

If you had an experiment like this in a vacuum (space), you could probably "inflate" a Mylar balloon this way because there is no inward pressure of an atmosphere to balance, though allowing the electron beam to go through an "opening" probably leaks electrons too, so you end up having to generate the electrons inside of the envelope. I base this on the fact that static electricity spreads out across an insulator (the outside of a balloon) uniformly; with an area that will flow electrons, the inside and outside of the balloon become one surface to spread electrons across. This is OK, except for the fact that electrons flow as a stream from high potential to low potential in a vacuum...

EDIT: In an edit to match up with OP's third edit I will say that it's much closer to an isolated experiment that would "inflate" the balloon.

The experiment design would go as follows, and at that point, I think you have a decent chance at seeing the balloon inflate, at least some:

1. With the balloon (loosely fastened to let the air pressures equalize) and the van de graff generator head sealed inside of a bell jar, pump out all of the air you can.
2. Flick some switch, and seal the neck of the balloon tightly against the generator neck.
3. Turn on the Generator.

You should at least get enough repulsion to lift the balloon off the generator against gravity and potentially to the point where it begins to stretch. It's seems like a 50/50 shot whether a balloon can contain enough voltage to significantly stretch the balloon without dielectric breakdown, but just the initial inflation would be awesome to see.

• I do admit that the "static electricity spreads out across an insulator (the outside of a balloon) uniformly" may well depend on having another insulator to prevent flow off of the surface, which is untrue in a vacuum situation. In the air case, can you even have an "electron beam?" That seems like a problem because electrons hit atoms before getting to the balloon... – BenPen Aug 25 '16 at 15:34
• (And yes, my background is a mix of physics/engineering, so materials science is the way an Engineer says, "Experience indicates that this is true somehow", so, perhaps my answer is lacking a particular rigorousness...) – BenPen Aug 25 '16 at 15:47

It is certainly possible to move objects by electrostatic means. See for example the gold leaf electroscope here:

https://en.wikipedia.org/wiki/Electroscope

If your balloon was made of a suitable material (ultra thin and conductive to spread charge, such as gold leaf) and open to allow air through, rather than sealed, a van de graaf generator might be able to inflate it. The distances (and forces) would be similar to those in the "van de graaf hairstyle" photos here

https://en.wikipedia.org/wiki/Van_de_Graaff_generator

If you seal it, however, it will have to inflate against atmospheric pressure. If it reaches 10 times its original size, the internal pressure would be 1/10 atmosphere. That's nearly an atmosphere of pressure differential: 14.7lbf per square inch or 10N (about 1kgf) per square centimeter. Multiplied over the whole surface the forces are tremendous, as can be seen in this video of an imploding rail tank car https://www.youtube.com/watch?v=UpWeU2fvFGs. Clearly the forces involved are much larger than those required to make someone's hair stand on end.

Conclusion: in theory it could be done. In practice you'd need astronomically high voltages to beat atmospheric pressure. Containing such voltages without them leaking or arcing away is probably impossible.

I think it doesn't work. All the electrons are to be pushed towards the inner surface of the balloon. The potential of the balloon keep increasing eventually becomes a big negatively charge balloon. You may infer what happens.

• This should be OK. all of the negatively charged balloon walls should repel each other electrostatically. But, does the potential go up beyond what can be held without dielectric breakdown or does it manage to inflate it. That's what my answer attempts to address... – BenPen Nov 30 '16 at 17:22