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# Physics/maximum voltage

Question
thank you, so in a hypothetical situation where the column of the van da graaff is too long & clean for a discharge. why does the curvature of the terminal determine the maximum voltage in dry clean normal air. is it because of the leakage of current into the air? if so than how could one calculate this leakage of current? would it not simply be approximately equal to the amps incoming on the belt? as the equilibrium between input(belt) and output(leakage) is reached and the terminal reaches the maximum steady voltage but the Faraday ice pail effect will force more extra charge onto the outside of the terminal no mater how full it is resulting in leakage of current into the air.
thank you,
gene

If you have a ridiculously long and high-resistance column, like 100 meters, so that everything else is essentially at infinity relative to the radius of the sphere, that radius determines the electric field at the surface.  The field of a sphere of uniform charge is given by kQ/r^2, where k is a constant, Q is the charge, and r is the radius.  The voltage is given by kQ/r (ignore the sign for not, not important).  How much charge is carried away in the air depends on detailed properties of the air, like pressure/dust/humidity/CO2 levels, etc.  You can't calculate it with any reliability, no.  If you reach an equilibrium state where all the discharge is through the air, then that discharge will equal the net current up the belt (though the return of the belt will probably attract some of the discharge ions/electrons, depending on what kind of belt you use and whether your terminal is positive or negative).    That's a rather extreme circumstance.  In my experience, leakage down the column is almost always a factor (the column will attract conductive dust as the generator runs), and if there's no breakdown discharge then either the column is dirty or the air is humid.  But in cases where the air is humid, the belt charge also roughly equals the discharge through the humid air.

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