Electric field calculations for a spherical grid

  • #1
Jz21
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0
TL;DR Summary
I need help calculating the radius required to achieve a electric field strength of 1 x 10/6 v/m on a spherical grid inside a vacuum chamber
Hello, I need some help regarding how to calculate the needed radius to achieve 1 x 10/6 v/m on the surface of a spherical inner grid, inside a conductive vacuum chamber. I have used various equations, however I don’t know if they are very good and would like to know other methods. Thanks!
 
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  • #2
Welcome to PF.

Jz21 said:
TL;DR Summary: I need help calculating the radius required to achieve a electric field strength of 1 x 10/6 v/m on a spherical grid inside a vacuum chamber

Hello, I need some help regarding how to calculate the needed radius to achieve 1 x 10/6 v/m on the surface of a spherical inner grid, inside a conductive vacuum chamber. I have used various equations, however I don’t know if they are very good and would like to know other methods. Thanks!
Woo-wee. 1MeV/m -- whatcha' tryn' to do?
 
  • #3
A lil bit of amateur fusion, that’s all
 
  • #4
Jz21 said:
Hello, I need some help regarding how to calculate the needed radius to achieve 1 x 10/6 v/m on the surface of a spherical inner grid, inside a conductive vacuum chamber.
If the walls of the chamber are smooth, without projections, then knowing the voltage, you must place a sphere in the chamber that has a maximum diameter consistent with that voltage gradient.

What is your working voltage?
What is the radius of the vacuum chamber?
 
  • #5
Just a lil bit of fusion, nothing else
 
  • #6
Jz21 said:
spherical inner grid
If you mean a wire mesh. I can tell you right now that this isn't going to work.

The field near the wire will be driven by the radius of the wire, not the radius of the sphere. That will be maybe two orders of magnitude greater, and the wire will not survive.
 
  • #7
Sorry, I didn’t thought I didn’t reply to the other reply, my bad. I am working with 40 kv dc, and the radius of the chamber is 5 inches. There are also no projections, with only a feed through stalk.
 
  • #8
Vanadium 50 said:
If you mean a wire mesh. I can tell you right now that this isn't going to work.

The field near the wire will be driven by the radius of the wire, not the radius of the sphere. That will be maybe two orders of magnitude greater, and the wire will not survive.
It is not wire mesh, but just standard wire loops.
 
  • #9
Vanadium 50 said:
If you mean a wire mesh. I can tell you right now that this isn't going to work.

The field near the wire will be driven by the radius of the wire, not the radius of the sphere. That will be maybe two orders of magnitude greater, and the wire will not survive.
Oh ok I see what you mean.
 
  • #10
Jz21 said:
I am working with 40 kv dc, and the radius of the chamber is 5 inches.
40 kV, radius = 5".
5" = 0.125 metre.

1 MV/m is 1 kV/mm, so there needs to be a gap of 40 mm between the chamber wall and the sphere.
The sphere radius will be (125 mm - 40 mm) = 85 mm, or 170 mm diameter = 6.75" diameter. It will need to have a smooth surface.

The support stalk for the sphere will need to be carefully constructed to prevent voltage breakdown along its surface.
 
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  • #11
Baluncore said:
40 kV, radius = 5".
5" = 0.125 metre.

1 MV/m is 1 kV/mm, so there needs to be a gap of 40 mm between the chamber wall and the sphere.
The sphere radius will be (125 mm - 40 mm) = 85 mm, or 170 mm diameter = 6.75" diameter. It will need to have a smooth surface.

The support stalk for the sphere will need to be carefully constructed to prevent voltage breakdown along its surface.
Thank you. But I now realize I was going about it all wrong. I was approximating the grid as a sphere as it forms a sphere, but is still made of wire. So now I know we’re to go.
 

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