Can Impossible Vacuum balloon be possible with this idea? As it is well known, a vacuum balloon using the materials we have at our disposal is not possible, because of the sheer force they have to resist from the air outside.
I have an idea and I want to know if there is anything that makes it impossible.
1) Build a balloon out of graphene which is very light and it is conductive too.
2) Since graphene is not airtight, cover graphene (that is not yet inflated) with a good electrical insulator which is also airtight.
3) Now make the graphene store electric charge.
4) This would exert force from inside as the electric charges on the graphene repel each other.
5) This causes the balloon to inflate while having a vacuum inside.
Now my question is there any theoretical problems (not engineering problems) with this model?
 A: First, it is not "well known, a vacuum balloon using the materials we have at our disposal is not possible, because of the sheer force they have to resist from the air outside." In our patent application (Akhmeteli, Gavrilin, Layered Shell Vacuum Balloons, you can find it at USPTO site or at http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ), we show that sandwich structures made of existing materials can be both strong enough to withstand atmospheric pressure and light enough to float in air, according to our finite element analysis. As for your idea, external charge can gather on the surface of your structure due to discharge in air, say, during thunder, and ruin your design.
A: Interesting idea!  Let's check the numbers.  
For this to work at Earth's surface, the electric force has to cancel $P = 10^5 N/m^2$. 
The electric field $E = \sigma / \epsilon_0$ in the case (no field inside, all the field goes outside).
For an area $A$, the force on a conductor due to a field $E$ is $F = QE/2 = \sigma A E / 2$ so the force per unit area is 
$P' = F/A = \sigma E / 2  = \epsilon_0 E^2$
where I've expressed it in terms of $E$ so we can solve for the electric field $E$ needed to set the electric "pressure" equal to atmospheric pressure (ideally, you'd want a bit more to keep things stable):
$ \epsilon_0 E^2 = 10^5$
$ E^2 = 10^5 / 9\times10^{-12} \sim 10^{16}$
So the electric field you need is about $10^8$ volts/meter.  That's a lot.  It's 3,000 times more than the 30kV/m breakdown voltage of air.
What does that mean?  It means that the charge on your balloon is going to instantly spark off into the air and the balloon won't be held up against the atmosphere any more.
Note that insulating the balloon won't help.  The electric field has to go off to infinity to hold up the balloon against pressure. If you add insulation against sparking, several things can happen:


*

*That insulation will still have the electric field outside it, so where the insulation ends, the sparking starts, and the balloon ends up neutralized.

*Or you make the insulation so thick, without adding weight, that the surface area of the ballon is much, much, much (3000X) times bigger than the balloon itself, spreading the field out and reducing it.
So it looks like this isn't going to work at Earth's surface. Bottom line, the atmosphere pushes too hard, electrostatic forces are too weak, and the atmosphere isn't a good enough insulator.
But it's still a cool idea.
