Charging a magnetically levitating sphere to 1 gigavolt in a high vacuum

AI Thread Summary
Charging a magnetically levitating sphere to 1 gigavolt in an ultra-high vacuum using an electron beam presents significant challenges. Key limitations include the energy of the electron beam, the sphere's dimensions, and surface smoothness, which influence charge retention and field emission. If the sphere's surface is not perfectly smooth, small irregularities can lead to excessive field emission and discharge, potentially damaging the equipment. Additionally, the proximity of the vacuum chamber walls affects the electric field gradient, further complicating the charging process. Overall, achieving such high voltage safely requires addressing these practical and technological constraints.
BrandonBerchtold
Messages
46
Reaction score
6
What sort of limits would be encountered if you tried to charge a magnetically levitating sphere to as high a voltage as possible in an ultra high vacuum by using an electron beam aimed at the sphere? Assume the sphere is highly spherical and polished.

If electrons have sufficient energy to impact the sphere, would they stick or splash off? What if there was a hole drilled into the sphere such that the electrons impact the inside surface of the hollow sphere. Would the maximum possible charge on the sphere be higher then?
 
Physics news on Phys.org
The potential relative to infinity is Q/(4 pi eps_0 r), the field strength at the surface is Q/(4 pi eps_0 r^(2)), let's use 10 V/nm as limit, then we get r = 1GV/(10V/nm) = 10 cm. This doesn't sound bad, but direct field emission wouldn't be the limit. An electron leaving the sphere, just a stray electron from the beam, or even a random electron released from cosmic rays would hit the vacuum chamber at an energy of 1 GeV. More than enough energy to produce a couple of ions, which then accelerate towards the sphere, hitting it at an energy of 1 GeV or more. They will kick out many electrons, accelerating towards the vacuum chamber again and so on. You'll get a gigantic discharge, destroying the equipment in the process (stored energy is 5.5 MJ).

MV equipment is usually operated with SF6 as gas. It prevents this kind of discharge. You can reach something like 30 kV/mm. The same calculation as above leads to r=33 m. Good luck levitating that (electrostatic levitation sounds interesting).
In the calculation above I assumed that the other electrode is "at infinity" - which means the room would need a radius that is very large compared to 33 m. If that is impractical then the radius of the sphere must increase even more.
Oh, and did I mention that you need pressurized SF6? This giant building is like a bomb in case the pressure vessel fails.
 
  • Like
  • Informative
Likes Vanadium 50, vanhees71, BrandonBerchtold and 4 others
BrandonBerchtold said:
Summary:: What sort of limits would be encountered if you tried to charge a magnetically levitating sphere in an ultra high vacuum by using an electron beam aimed at the sphere? Could you charge the sphere to a potential in the range of a gigavolt or more?

What sort of limits would be encountered if you tried to charge a magnetically levitating sphere to as high a voltage as possible in an ultra high vacuum by using an electron beam aimed at the sphere? Assume the sphere is highly spherical and polished.

If electrons have sufficient energy to impact the sphere, would they stick or splash off? What if there was a hole drilled into the sphere such that the electrons impact the inside surface of the hollow sphere. Would the maximum possible charge on the sphere be higher then?

This description is full of holes.

1. What is the energy of the electron beam? If it is just the order of eV, then at some point charging effects on the sphere can easily repel the beam from hitting it.

2. What is the dimension of the sphere? This tells you how much charge it can hold before it will start spewing out its own charge.

3. What is the nature of the surface of the charge? Again, at some point, field-emission will take over. The degree of smoothness (nothing is perfectly smooth, and even grain boundaries has been shown to be center of emitters) will dictate how quickly the sphere will start leaking charges.

4. What is the UHV vessel and how big is it? There is a difference between estimating the boundary where V=0 to be far away versus something that is only 10 cm away. The distance between the sphere and the grounded walls of the UHV vessel WILL dictate the strength of the electric field gradient.

etc... Like I said, there are major holes in the details here.

Zz.
 
mfb said:
The potential relative to infinity is Q/(4 pi eps_0 r), the field strength at the surface is Q/(4 pi eps_0 r^(2)), let's use 10 V/nm as limit, then we get r = 1GV/(10V/nm) = 10 cm. This doesn't sound bad, but direct field emission wouldn't be the limit.

There are other practical/technological limits related to the direct field emission mechanism as well. In order to be charged to 1 GV, the surface of R=1 m sphere placed in ultrahigh vacuum must be incredibly smooth. Small irregularities (on nanometer scale) would cause powerful emission and high density currents. This kind of problems are met in medium voltage vacuum circuit breakers where the allowed max field is less than 1MV/cm
 
  • Like
Likes BrandonBerchtold
I was using the Smith chart to determine the input impedance of a transmission line that has a reflection from the load. One can do this if one knows the characteristic impedance Zo, the degree of mismatch of the load ZL and the length of the transmission line in wavelengths. However, my question is: Consider the input impedance of a wave which appears back at the source after reflection from the load and has traveled for some fraction of a wavelength. The impedance of this wave as it...

Similar threads

Back
Top