QUESTION: thank you, please tell me, in the case of a well designed van de graaff, would most of the current losses be into the air or into the column. for example, in a 600kV VDG, with a 60 micro amp output. with the tower/column being the usual PVC at 1.5m tall and the sphere being 1m diameter, polished steel. where does most of the 60 micro amps go to?
ANSWER: The voltage on a Van de Graaff generator isn't something that you just set at a given value. Most of them are rated to the maximum value that they will get before discharging in dry air and with a clean column. That's limited by the dielectric breakdown strength of air or the cleanliness of the column and belt. If you have a nice, clean column and clean belt with low surface conductivity characteristics, then it will generally charge until it arcs to ground. The Van de Graaff terminal on the tandem particle accelerator I worked with in grad school, for example, would go up to 6,500,000 V. That's not a typographical error, it was in the middle of a buffer gas of SF6. There are too many variables to give you a simple answer, everything from humidity and atmospheric dust to the surface properties of both the belt and the plastic column. For the numbers you mention, however, I'd estimate that the most that the air could withstand would be just under the 600 kV you mention. That's a huge sphere, too, so maybe it could...but it would probably build up voltage towards maximum and then arc to ground through the air.
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QUESTION: thank you, so the size of the terminal's curvature on a good clean VDG determines the maximum voltage that can be sustained; all other factors being equal. is that the electrical break down of air or leakage of charge that is the reason why the curvature is correlated to maximum voltage per equal amperage and if so than that is another way of saying the ionization of air, correct. so all up, would the current input from the belt start off charging the terminal and once the terminal is charged to its maximum, then the belt current would be used to equal the leaked current. the rest of the current would just be leaked out through the terminal (in a scenario where there is no grounded object near enough to discharge the terminal). ionizing the air in the process.
No, the size does not determine the maximum voltage by itself. The size helps determine that, but that's largely also dependent on the length of the insulating column. The important factor is the electric field, all dielectrics (like air, plastic, very pure water) have a dielectric breakdown strength that is given by the maximum electric field it can sustain without sparking through it. So it's not just one factor. The larger the radius, the lower the field...to a point where the sphere is so big that the radius is huge compared to the distance separating it from the ground. You're talking about running in a steady state. Van de Graaff generators without leaky columns generally build up and then that charge arcs down to ground, discharging the terminal to zero volts, then start charging again.
If you have a dirty column or humid air, then yes your belt current will equal the current leaked back to ground via these channels. But this is not normal operation unless you have a controlled Van de Graaf generator (like a particle accelerator) with a coronal discharge terminal to regulate the voltage in such a way.