I came across the following design of a nuclear fusion reactor (electrostatic configuration):

It works as follows:

  1. A vacuum is created in the spherical stainless steel chamber.
  2. A small amount of deuterium gas is let into the vacuum chamber.
  3. A DC voltage of 150kV is applied across the tungsten ring inside the vacuum chamber to start fusion.

Experimental setup of the electrostatic fusion reactor: Experimental Setup

Vacuum chamber: Vacuum Chamber

Tungsten ring carrying the current inside the vacuum chamber: Tungsten Ring

My question is, given the high voltage across the tungsten ring inside the vacuum chamber, and its close proximity to the conducting stainless steel shell, how does this design avoid current flowing across the stainless steel chamber?

Is the partial vacuum sufficient to prevent an arc between the tungsten ring and the shell, or would this design rely on the shell not being located near a ground?

If this is due to the low gas density inside the vacuum chamber, which laws/calculations could be used to verify that the current cannot cross the gap between the tungsten ring and the stainless steel chamber?

  • \$\begingroup\$ @TimWilliams Here is the source: instructables.com/Build-A-Fusion-Reactor I might be classifying it incorrectly \$\endgroup\$ Feb 19 at 20:13
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    \$\begingroup\$ @jcaron Why? 150kV should be enough for deuterium. It's basically just a gas discharge tube. \$\endgroup\$ Feb 19 at 23:14
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    \$\begingroup\$ Looks like a farnsworth fusor to me. en.m.wikipedia.org/wiki/Fusor \$\endgroup\$
    – Bryan
    Feb 20 at 0:04
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    \$\begingroup\$ The premise of the question is not correct. Current does cross the gap between the tungsten ring and the stainless steel chamber. Also, I would think that for safety purposes, the stainless steel shell must be connected to ground. \$\endgroup\$ Feb 20 at 1:46

1 Answer 1


As I understand this apparatus, high voltage is fed through an insulated feed-through into the vacuum chamber and a gas discharge through mostly rarefied deuterium gas takes place between the (grounded) chamber and the electrode at the center.

You can look up dielectric breakdown of various materials and see what thickness is required (plus you want a healthy safety factor since insulators are not perfect). Something like a ceramic or glass feed-through should have a voltage rating if it is a commercial product.

If it does arc over, the current and energy may be rather limited, depending on the supply, so it may not be all that dramatic. Not like a short-circuit near an industrial sub-station which can vaporize quite a bit of material. On the other hand, the voltage and available current are within the potentially lethal range and 150kV can jump a fair distance in air. Keep in mind also that ideal breakdown voltages in air typically assume smooth spherical electrodes, but pointy electrodes will cause the air to break down at a considerably shorted distance and/or lower voltage.

A hard vacuum is a good insulator, but low pressure gas actually breaks down at a lower voltage than atmospheric pressure (due to the free path being longer in a rarefied gas).

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    \$\begingroup\$ Put another way: at a low enough pressure, the gas might be ionized (glow discharge), but there isn't enough stuff flowing around to short out the supply. The design (or at least, intent) of a Farnsworth fusor is to channel exactly those ions into the center (something of an ion pump effect), so the insulator will do its job without excessive current flowing along/near its surface. \$\endgroup\$ Feb 20 at 1:59

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