[Bored chemist]

a crater shaped water filled lake

That's about as good a ground connection as you could wish for. How do you plan to insulate it?

i will soon as i can't find my scientific calculator at the moment.

Even a rough calculation would do- just keep track of the zeros.

[championoftruth (OP)]

incorrect. for a 100 km plane you can go up a 100km.
https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge [nofollow]

There is quite a difference between "infinite"  and 100 km, especially if you are 100 km away from the supposed "infinte" surface. Try reading before quoting.


Anyway, suppose we can indeed spray the Sahara with aluminum. Not a big deal since it won't be carrying much current, so can be a few microns thick. What equipment are you going to use to charge the foil and the magic carpet?

Equipment is not an issue. All we are doing is giving it an excess of electron. you are correct a thin layer of aluminum and charged to a high negative voltage. Electret materials can have CHARGES frozen in place.

[Bored chemist]

Equipment is not an issue

You can not be sure of that until you do the maths.

[alancalverd]

Equipment is not an issue.

Good to hear it. So what equipment will you be using?
Please don't waste time with handwaving. I have engineers and investors standing by. We just need your specification or preferably a lead to the relevant manufacturer's catalog, and you will be credited and rewarded as the true and first inventor of the system that launches 10 kg to 100 km.
We usually start such projects with a £10,000 limited company, in which you as inventor hold 51% of the 100,000 enterprise shares. You choose the name of the company and I will do the rest when I receive your cheque  for £5100. 

Per elektron ad astra.

[championoftruth (OP)]

Equipment is not an issue.

Good to hear it. So what equipment will you be using?
Please don't waste time with handwaving. I have engineers and investors standing by. We just need your specification or preferably a lead to the relevant manufacturer's catalog, and you will be credited and rewarded as the true and first inventor of the system that launches 10 kg to 100 km.
We usually start such projects with a £10,000 limited company, in which you as inventor hold 51% of the 100,000 enterprise shares. You choose the name of the company and I will do the rest when I receive your cheque  for £5100. 

Per elektron ad astra.

Surely it would be best for you to get your engineers to do the calculations instead of relying on ME?

This way you can be confident that your MANY trusted engineers verify and confirm it independently of each other.

Look what happened at NASA to the Hubble when they outsourced it to a company which miscalculated the curvature of the mirror.

[alancalverd]

If you don't have the courage of your own calculations, you shouldn't make suggestions you can't support. But you said "Equipment is not an issue"  so you must know where to buy what we need.

It's a well-known fact that rocket science is easy (Newtonian mechanics and the spontaneous combustion of the simplest chemicals - hydrogen and oxygen) but rocket engineering is expensive and difficult. Now I'm offering to raise the money and find the personnel to do the difficult  bit, using the equipment you have stated as "not an issue". Commercial confidentiality, professional project management.....it's all there, and I've built much more complicated things in the past.

Happy to build a 1 km prototype in a field, which will increase your share value a hundredfold even before we go for the big one.

[Bored chemist]

Look what happened at NASA to the Hubble when they outsourced it to a company which miscalculated the curvature of the mirror.

They calculated just fine. They screwed up the engineering.
But calculation is cheap- so get on with it.

[championoftruth (OP)]

Look what happened at NASA to the Hubble when they outsourced it to a company which miscalculated the curvature of the mirror.

They calculated just fine. They screwed up the engineering.
But calculation is cheap- so get on with it.

The equation is
E= Sigma/2e      Newtons/coulomb..Note that distance is NOT a factor

Sigma is the charge density
 The SI unit is C/m–1. (ii) Surface charge density; σ=qA, where, q is the charge and A is the area of the surface.

e = 8.854 187 8128 x 10-12 F m-1   Vacuum permittivity

Note FORCE =ELECTRIC FIELD/Q
q = the charge

You can put in the numbers while i try to find my scientific calculator

[chiralSPO]

You can put in the numbers while i try to find my scientific calculator

you can just enter your expressions as search terms into the google search bar and it will act as a scientific calculator.

[chiralSPO]

The equation is
E= Sigma/2e      Newtons/coulomb..Note that distance is NOT a factor

distance is not a factor when an infinite plane is used, and this equation is still a reasonable approximation when the distance from the plate is much smaller than the size of the plate. If you want 100 km to be much smaller than the plate, be prepared for an expensive project!

[chiralSPO]

Also, quick question:

Why don't charged clouds in thunderstorms ever launch anything into space?

[Bored chemist]

Please calculate the force on a spherical craft 10 metres in diameter when it is two hundred  metres above a large flat plane  and when the potential difference is a million volts.
For extra credit, please calculate the potential to which this  10 metre diameter craft can be charged before the air round it is ionised by the field gradient.

Also calculate the up-thrust due to the density of air (per Archimedes principle)

If you want something done properly, do it yourself...
OK
The spherical craft acts as a capacitor.
We can calculate the capacitance as shown here
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html
C= 4 pi e0 R
R= 5 metres
Pi =3.141
e0 =  8.8541878128×10^−12

That's 5.5 *10^-10 farads.
At a million volts it will have a charge Q= CV so that's 5.5*10^-4 Coulombs.

And we need to see how strongly that's repelled.
The force between  a charged body and a large plane is the same as the force between two charged bodies with the same charge, and twice as far apart as the distance from the charge to the plane.
This guy explains it about 5 minutes in.

It's as if the charge sees its "mirror image" reflected in the plane.
The charge is 200 metres above the plane.
So the force is the same as having two charges 400 metres apart.

OK that's just Coulombs law.
F= K Q1 Q2 /R^2
OK, I'm lazy, so I'm going to find an on-line calculator for that.
https://www.omnicalculator.com/physics/coulombs-law
and stuff the numbers in.
It gives me a force of 0.01404305 Newtons.

About one gram's worth of force.

I admit I have cheated slightly, I used the formula for a charge and an uncharged (earthed) plane.
But you are going to charge the plane electrode. I think that increases the force by a factor of 4.
So that's 4gf

Now, for the second part of the exercise- the up thrust due to density.
I said I didn't need precision; I'm going to pretend that the sphere is a cube.
Well, the up thrust is bigger than that on a cube half as big as the sphere.
So it's bigger then the buoyancy  of a 5 metre cube.
And that's the density of air times the volume of the cube
 about 1.2 Kg/m3 times 5*5*5. That's about 150 Kgf (or 150000gf)

So the craft would work better if it was a balloon by a factor of about 150000/4

(Nobody tell Alan; you might still get £10K of investment out of him.)

[Bored chemist]

Don't forget, the force reduces as you get higher (as the inverse square of the distance).

In  space-  say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of  micrograms by the time you were reaching space.

[alancalverd]

Thank you for your concern, BC, but whilst waiting for the lad to respond I put the money on a horse instead. The great thing about horses is that they have a very good sense of things like energy, speed and distance, and make fairly reliable business partners compared with some humans. 3:1 ain't a bad return for a few minutes' work. Might do some physics tomorrow. 

[championoftruth (OP)]

Don't forget, the force reduces as you get higher (as the inverse square of the distance).

In  space-  say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of  micrograms by the time you were reaching space.

inverse square law does not apply. The calculation is wrong.
see the diagram and derivation below:-

https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge

[alancalverd]

We have been through this before. Literate correspondents will have read the introduction in that reference, my reply #19, and chiral's response # 29. I won't bother to say it again 'cos there's none so blind as them as will not look.

For numerate readers and anyone aspiring to electrostatic engineering, the near-field approximation is fairly good if the separation d is less than the radius r of the plane. From d = r to d = 10r, the force falls off roughly as 1/r, tending to 1/d^2 beyond that.  We use similar approximations for radioactive contamination.

[Bored chemist]

Don't forget, the force reduces as you get higher (as the inverse square of the distance).

In  space-  say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of  micrograms by the time you were reaching space.

inverse square law does not apply. The calculation is wrong.
see the diagram and derivation below:-

https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge

Did you see the bit where it said "This example was for an infinite plane of charge. In the physical world there is no such thing," ?

That page is great- if you know what sigma is.
It's the charge density on the plane, but how do you figure that?

Anyway, let's have a look at a slightly different way of considering it.
Imagine that we have two plane electrodes- one on the ground and one "a long way up".
If we put a voltage across the two plates the a charged object- like our ship will be repelled by one plate and attracted ot the other.

In that case the force is independent of the height.

It's awkward to build the top electrode, but there's an easy solution; we can use the ionosphere.
It's (very roughly) 100 KM up so that's close enough to space.

Again let's imagine that we can put a million volts across the two plates.
And let's assume that our craft starts near the bottom plate and is attracted to the top one.

How much force is on the craft?

Well, if we move the ship from the bottom to the top then we alter the electrical potential by a million volts.
And the ship has (as before) a charge of about 5.5*10^-4 Coulombs.

So the energy transferred to the ship is the product of those
That's 5.5 * 10^2 Joules. (About the energy you would get from a ounce  of low Calorie cola).

OK, we need to convert that to a force.
Well that force acts over a distance of 100 Km and in doing so it transfers 550J of energy
Energy is force times distance so we can divide the energy (550J) By the distance (100,000 M) to get the force.
That's a force of about 0.0055 Newtons
 
Which is close enough to the answer I gave before (0.014N).
I could get the "right" answer by fiddling with the distance if I wanted.

It's an interesting thought- this pair of plates looks a lot like a capacitor.
Let's consider a single square metre of it and work out the capacitance
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

That will have a capacitance of E0 times the area divided by the distance.

So that's 1/100000 times  8.8541878128×10^−12
And that's not very big.
 8.8541878128×10^−17 Farads
OK that's the capacitance of each square metre of the pair of plates.
And, if you charge that to a million volts each square metre will acquire a charge (Q=CV)  of
 8.8541878128×10^−11 Coulombs.

Some of you might be wondering why I did that.
Well, as far as I can tell, that's the mystical "sigma" in the equation in the Kahn academy page.


"The total charge on the plane is of course infinity, but the useful parameter is the amount of charge per area, the charge density: ­sigma "

And what I just calculated was the charge density.
So let's bung the numbers into the equation.
E=sigma/2 e0
Well sigma is 8.8541878128×10^−11 Coulombs per square metre.
Divide that by 2 times e0

8.8541878128×10^−11 / 8.8541878128×10^−12 gives us 10.   It's interesting to consider the units here. 10 volts per metre because we have a million volts and 100,000 metres of spacing- I could have taken the short cut, but the OP wouldn't have believed me.

And the formula says to halve that.
So the electric field near this plane is 5 volts per metre (Which is obviously different from 10; the difference is that using a second electrode doubles the field strength).

OK, so what can we do with that?
Well, it's a field gradient.
It has units of Newtons per Coulomb.

So, we should be able to multiply it by the charge on our ship, and get the force on the ship.
We know the charge; it's 5.5*10^-4 Coulombs
And so we can multiply that by 5 and get 0.00275 Newtons. (as I said, adding a second plate doubles the force)

About a quarter of a gram.
My apologies; when I said

I think that increases the force by a factor of 4.

it looks like I should  have divided, rather than multiplying.

Alan's investment in a horse is 16 times more sound than we thought it was.

[championoftruth (OP)]

So, we should be able to multiply it by the charge on our ship, and get the force on the ship.
We know the charge; it's 5.5*10^-4 Coulombs
And so we can multiply that by 5 and get 0.00275 Newtons. (as I said, adding a second plate doubles the force)

About a quarter of a gram.
My apologies; when I said

I think that increases the force by a factor of 4.

it looks like I should  have divided, rather than multiplying.

Alan's investment in a horse is 16 times more sound than we thought it was.

I think your calculations need to be reviewed as your figures are far too low.
I have an ionizor in the house and yesterday i connected  a 3 gram metal plate to one of the 4 outputs and when i put hand near it it moved towards my hand quite strongly. The force was more than a a gram.

This ionizer only uses 6 kilovolts!!!

[Bored chemist]

I think your calculations need to be reviewed as your figures are far too low.

Feel free to try.
But I did three different calculations and got practically the same response.

The force was more than a a gram.

Did you measure this or are you saying I'm wrong, based on your ignorant guess?

[alancalverd]

It all depends on "near".  From 100 km distance you'd have difficulty even seeing a 5 kV ioniser, never mind detecting its (far) field.

The quantitative study of electrostatics began in 1770 and was pretty much completed and verified by experiment  within 20 years.