[championoftruth (OP)]

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.

You could say the same for any field and implies a cessation of progress...

[championoftruth (OP)]

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?

Well this guy is using 200kv and that forces looks more than 0.0026 grams.

[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

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.

force more than 0.25 grams and he only using 50000 volts!

This proves your calculations are in error or your assumptions in error or a misplaced decimal point.

[Bored chemist]

https://en.wikipedia.org/wiki/Inverse-square_law#Electrostatics

[Bored chemist]

Quote from: championoftruth on 24/01/2021 12:31:25:  "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.

You could say the same for any field and implies a cessation of progress...

There has been plenty of progress in electrostatic technology since 1770, but no  change in the laws of physics for the last 13.8 billion years..

[evan_au]

When you increase the voltage... it moves down

With 50,000 Volts in (say) 0.5 meters = 100,000 V/m, you can levitate a small aluminium foil shape.
- When you increase the voltage, it doesn't go as high.
- Extrapolating the two data points provided, you should be able to launch aluminium foil into orbit with zero volts...

[alancalverd]

Exactly so. If you put two flat plates in contact and share a negative charge between them, they are both at the same potential but repel one another. Remember the gold leaf electroscope? So you don't need a high voltage source to propel yourself into space!

Where's the tongue-in-cheek emoticon?

[charles1948]

Exactly so. If you put two flat plates in contact and share a negative charge between them, they are both at the same potential but repel one another. Remember the gold leaf electroscope? So you don't need a high voltage source to propel yourself into space!

Where's the tongue-in-cheek emoticon?

Doesn't the gold-leaf electroscope only work because it uses very thin metal foil. 

If you tried to make an electroscope with thick metal sheets, say 6-inches thick, the sheets wouldn't move at all.

They would be kept still by their inertial mass.

But - could a spaceship be built of thin metal foil?  The foil would be pressurised internally against the vacuum of outer-space.  That would keep the ship's foil-hull, rigid and taut.  It would be as light, and efficient, as a rubber balloon.

[championoftruth (OP)]

Exactly so. If you put two flat plates in contact and share a negative charge between them, they are both at the same potential but repel one another. Remember the gold leaf electroscope? So you don't need a high voltage source to propel yourself into space!

Where's the tongue-in-cheek emoticon?

Doesn't the gold-leaf electroscope only work because it uses very thin metal foil. 

If you tried to make an electroscope with thick metal sheets, say 6-inches thick, the sheets wouldn't move at all.

They would be kept still by their inertial mass.

But - could a spaceship be built of thin metal foil?  The foil would be pressurised internally against the vacuum of outer-space.  That would keep the ship's foil-hull, rigid and taut.  It would be as light, and efficient, as a rubber balloon.

Depends on how much charge you give it the higher the charge the greater the force. f=q1q2/r2 x Eo
in a flat plane the charge density is the key.

[Bored chemist]

Doesn't the gold-leaf electroscope only work because it uses very thin metal foil. 

It works because it is light.
You could just about use aluminised Mylar or something. Though the "usual" grade is heavier (per unit area) than gold leaf.

Depends on how much charge you give it

Yes, that's why I asked this

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.

You don't seem to have answered it.
Did you get distracted by nonsense where you forgot about the inverse square law?

[alancalverd]

from: charles1948 on 29/01/2021 00:24:24
Doesn't the gold-leaf electroscope only work because it uses very thin metal foil. 

It works because it is light.

But the fun bit is that, whatever material you use, both leaves are at the same potential. Therefore you do not need a high voltage source to drive your space elevator - the elements repel each other at zero potential difference!

[Bored chemist]

from: charles1948 on 29/01/2021 00:24:24
Doesn't the gold-leaf electroscope only work because it uses very thin metal foil.

It works because it is light.

But the fun bit is that, whatever material you use, both leaves are at the same potential. Therefore you do not need a high voltage source to drive your space elevator - the elements repel each other at zero potential difference!

You can use an electroscope as a crude way to measure voltage. The higher the potential, the more the leaves diverge.
With a potential of zero, you get a divergence of zero.

Sadly, it's only in championoftruth's fantasy world where we could launch a ship for free (or, indeed, where the idea could be any use at all).

[alancalverd]

The higher the potential,

with respect to what?

the more the leaves diverge.

Two electrons in deep space, with no third body reference potential, will repel each other.
F = k_eq₁q₂/r² - no mention of any potential, just charge
Indeed if there is no third body within a distance D >> r, they will fly apart indefinitely.

[Bored chemist]

with respect to what?

With respect to the potential it would have without a charge.

You can measure potential by looking at the work done bringing a small test charge from infinity to the object.
You can calculate the potential of, for example, an isolated sphere by calculating the capacitance and then combining that information with the charge (which you can usually find by counting electrons and protons).

[alancalverd]

So I have two metal plates in contact. Being conductors, they share charge so they both have the same surface electron density. So they fly apart. Which is why Champion asserted that the power source isn't a problem.

Or more to the point, they don't.

[yor_on]

I think I saw something involving a canon once?
How to get your payload to the moon I think?
=

Eh, sorry, remembering some more. It involved a cannon, not a canon.