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:
- Latex: Ca(NO3)2 + H2O + C2H6O, C5H8, (2-methyl-1,3-butadiene)
- Pigments are; Ultramarine Blue, (Na8-10Al6Si6O24S2-4) + (Na3CaAl3Si3O12S), Red, Hematite (Fe2O3), and Yellow Ochre, FeO(OH) nH2O.
- 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.
Best Answer
I think you'll be unsatisfied with an answer about the gravitational field of the electron--to my knowledge, no one has tested anything involving the gravitational field created by microscopic particles. The closest we've come is tests of gravitational redshift involving scattering microscopic particles in the external field of the Earth.
There, is, however, a known solution to Einstein's equation and Maxwell's equations that represents a black hole with a nonzero charge, known as the Reissner-Nordström metric, given as:
$$ds^{2} = - f dt^{2} + \frac{dr^{2}}{f} + r^{2} d \Omega^{2}$$
Where $f=1-\frac{2M}{r} + \frac{q^{2}}{r^{2}}$. In the context of this metric, there is a difference in the gravitational field induced by the charge from what it would be without the charge--you get the $r^{2}$ variation in what becomes the gravitational potential function for timelike geodesics. This is, in principle, measurable (and a failure to measure it would be a contradiction of either Maxwell's equation or General Relativity, which are both stringently tested--the former moreso than the latter).
One, however, must be careful with what one means by 'active gravitational mass' of a system like this--the ADM Mass of this system is still $M$, and is not modified by $Q$, even if particles near the black hole feel different forces due to the presence of the charged particle.
Finally, as a bit of a interesting aside, it should be noted that the presence of a charge moves the location of the horizon to the location $r_{\pm}=M \pm \sqrt{M^{2}-Q^{2}}$, so there is no horizon at all if $Q>M$. It turns out that if you put the values for the electron mass and charge into this equation, you will find that it predicts that the electron should be a naked singularity.