# Levitation using directed electric fields

1. Mar 27, 2013

### taylaron

Greeting PF’rs
Subject: Levitation using directed electric fields

If someone had a way to take a spherical mass and pump electrons into the mass and fill many of the valence electron shells in the atoms, there would be an enormous electric field emitted by the charged mass. (Yellow sphere in illustration). Because the electrons experience the coulomb force against one-another, the electrons would be driven to be equally spaced throughout the charged mass. Coulombs law would obviously be valid because the area in the exact center of the mass would be uncharged.

To prevent electrical discharge, the mass would need to be surrounded by a strong dielectric, say- ultra pure water of PTFE.

The electric field emitted by the object would induce charge separation in objects nearby. If the E field was strong enough and the object was above the charged mass, the object would experience upwards force exerted by the E field on the electrons in the object.

Because a charged mass the repels nearly everything nearby wouldn't be very useful, a Faraday cage would be placed around the ultra pure water, acting to block the E field lines emitted by the charged object and a hole in the Faraday cage would allow for E field lines to be emitted in only a select direction.

Ignoring the amount of electrons necessary to induce any significant repulsive force, and my conceptual understanding is correct:

1. Would a hole in the circular Faraday cage (bottom of my illustration) allow the electric field lines to extend into infinity as shown? Obviously the repulsive force decreases with the square of the distance.

2. I’m not totally familiar with the effects of a Faraday cage. Would most of the electric field lines not naturally exiting the hole in the cage be redirected out of the hole? In other words, is the field line density only proportional to the cross section of the exposed charged mass?

Cheers,
-Tay

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2. Mar 27, 2013

### Staff: Mentor

They are just on the surface (assuming some conductivity in the material), and the spherical symmetry can be broken by external charge distributions.
"Most of them" (counting them is not a good idea) would still end on the cage, some of them at the outside.

I don't understand question 2.

Note that this setup would be unstable - any deviation from an ideal position will destroy the balance.

3. Mar 27, 2013

### DrZoidberg

If all you want is electrostatic levitation, there is a simpler way to do that.

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4. Mar 27, 2013

### taylaron

I don't understand why excess electrons in a mass only exist on the surface of the object. The electrons would feel less repulsion between themselves if they were dispersed throughout the entire body. If the mass is conductive, I don't see why it wouldn't do that. Yes, the electric field in the very center of a spherical mass would be zero per columbs law, but that dosen't mean they have to be on the surface for it to be true.

5. Mar 27, 2013

### Staff: Mentor

Every net charge in the volume would generate an electric field outwards, repelling those charges towards the surface of the object. Inside, zero net charge density is the only solution to a volume free of electric fields.

6. Mar 27, 2013

### taylaron

Would you not agree that when electrons are deposited onto a sheet of foil, the charge is equally dispersed over the surface? i.e. the same number of charges per cm^2

7. Mar 27, 2013

### Drakkith

Staff Emeritus
I think he's just talking about when you form a conductive material into a sphere. The repulsive force would cause the electrons to bunch up towards the outside of the sphere somewhat.

8. Mar 27, 2013

### taylaron

I understand what he's saying, im arguing that the concentration of electrons on the surface would be consistent with the number of electrons per cm^3

9. Mar 27, 2013

### Khashishi

A Faraday cage won't direct the electric field downward like that. I'm not sure how you would direct the electric field preferentially through a hole like that. Remember that the electric field is - gradient of the electric potential, so the integral of the electric field along the path from point A to point B is independent of path.

10. Mar 27, 2013

### taylaron

So you're saying the electric field would look like this: (Ignore the straight field lines)

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11. Mar 28, 2013

### DrZoidberg

When the electrons are all on the surface they may experience a stronger force however that force points outwards and they can not move in that direction since air is a good insulator. If they were all distributed throughout the volume of the object the force they experience would move them.

12. Mar 28, 2013

### taylaron

What do you mean?

13. Mar 29, 2013

### DrZoidberg

I mean it's impossible to arrange the excess electrons in a negatively charged metal object such that none of them experience any force. So the only stable arrangement is for all the excess electrons to be at the surface because there the insulating air can stop them from moving any further.

14. Mar 29, 2013

### taylaron

I see what you're getting at.
I drew a simple diagram with 5 electrons in a row and compared the net vector sums of the electrons and found that the electrons not on the ends or center experienced a net outwards repulsive force. This is because there are more electrons on once side than the other.... Therefore, all of the charges would be located on the surface of a solid spherical conductive mass.