What Happens to a Charged Sphere During Field Evaporation?

AI Thread Summary
When a positively charged small sphere is subjected to high voltage, field evaporation may occur, raising questions about the dynamics involved. The rate of evaporation could potentially increase exponentially as the sphere's radius decreases, but evaporated particles might also reduce the sphere's charge, limiting further evaporation. If the sphere is charged to a voltage significantly above the threshold for field evaporation, it could lead to an explosive reaction. To analyze this, one can apply the method of images to calculate the electric field strength at the sphere's surface. This approach allows for the computation of the forces acting on the sphere's surface elements, providing insight into the behavior during high-voltage conditions.
BrandonBerchtold
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If you were to positively contact charge a small ~1 mm diameter sphere using a Van de Graaff generator, and were to charge it sufficiently high enough that field evaporation began to occur, what would happen?

Would the rate of evaporation increase exponentially as the field strength would increase with decreasing sphere radius? Or would the evaporated particles reduce the positive charge of the sphere enough to prevent further field evaporation?

Alternatively, what if you contact charged a small ~1 mm sphere to a voltage much higher than that at which significant field evaporation occurs. Would it explode?
 
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Don't know the answer but here's a suggestion for getting it. A small conducting sphere (particle) coming in contact with a high voltage electrode would be brought to the potential of the electrode. Assume a flat perfectly conducting (PEC) electrode. A charged sphere above a PEC is a canonical problem (method of images if I recall). This calculation would give the field strength at the surface of the particle. Knowing this one may compute the force on each element of the particle as ##F=dq E##.
 
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