- Yes.
- You need to solve two problems.
First, given the natural radius of the spherical membrane $R_0$ (the radius with no tension in the membrane) and the current membrane radius $R$, membrane's modulus of elasticity $E$ and Poisson's ratio $\mu$, calculate tension stress in the membrane. If you consider an infinitesimally small square (with the side of $\delta l_0$) of the spherical membrane under no tension with thickness $d_0$, currently it will be a square with a side of $\delta l=\delta l_0\frac{R}{R_0}$. The tension stress $\sigma$ will expand the square in two directions. Calculation of tension stress from strain (using elasticity modulus and Poisson's ratio, and assuming the membrane material is isotropic) is a standard task, see e.g. https://en.wikipedia.org/wiki/Poisson%27s_ratio , although you may wish to find a better source. The exact value of the Poisson's ratio is not very important.
Second, given tension stress, calculate differential pressure in equilibrium. To do this, consider the condition of equilibrium for a half of the spherical membrane: $(p_i-p_o)\pi R^2=2\pi R d\sigma$ (assuming the membrane is thin: $d_0<<R_0$).
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
Yes. Though I suppose the fire danger goes up, and you certainly can't use a propane burner to warm it...
Lower density always means higher buoyancy.
Yes, and this has been proposed in various ways in science fiction literature. The engineering challenge is finding a away to confine the vacuum that is as light as a gas bag so that you don't loose the advantage to extra weight.
In general a volume $V$ of material of density $\rho$ immersed in a fluid of density $\rho_f$ experiences a buoyant force of
$$ F_b = gV\rho_f $$
and a weight of
$$W = -gV\rho $$
so the available lifting force is
$$ F_l = gV(\rho_f - \rho) .$$
Where the object is floating at the surface of a liquid the buoyant force is modified to reflect the volume of liquid displaced $F_b = g V_d \rho_f$ where $V_d$ is enough to cover the weight of the floating object.