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# Physics/a van de graaff

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QUESTION: hello, thank you.
please tell me. how would a ~1MV van de graaff negative top terminal compare to a ~10kV needle ionizer when it comes to ionizing the air (negatively) in a room. being that the grounded parts of the VDG would be sealed off from the air. would the greater surface area and much higher voltage of the ~1MV VDG compensate for the smaller radius of curvature of the needle in releasing electrons?

ANSWER: You didn't tell me how big the terminal is on top of the Van de Graaf.  That means everything, you need to account for the electric field and the surface area both.  And if it's spherical, and if it's nearly complete.  There's some missing information here.  Sizes and curvatures of the needle, too!

---------- FOLLOW-UP ----------

QUESTION: thank you, please tell me, regarding the comparison of the needle and the van de graaff terminal sphere as to the generation of electrons into surrounding air. if the VDG terminal sphere was ~30" with 80 micro-amps, ~1MV and the needle was 100 microns by 1cm at 25kV, 5 milliamp. is it somehow calculable?
thank you,
Gene

Answer
The number of micro and milliamps is useless, because it largely depends on leakage current.  If you're dealing with the same air and a hemispherical needle end, a perfect one...and if you're dealing with a nearly spherical VDG terminal then it's possible to figure out the field and the area.  However, there's a caveat.  Air is a non-ohmic material.

So no, it's still not calculable. It's measureable, if you have something to receive charge in the area and a voltmeter to measure it on, like a sphere of aluminum foil, but it's not simply calculable.

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