# physics 6a

A potential difference of 4.75 kV is established between parallel plates in airf the air becomes ionized (and hence electrically conducting) when the electric field exceeds 3.00×106 V/m, what is the minimum separation the plates can have without ionizing the air?
d=___________m

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1. V/d(min) = 3*10^6 V/m, the breakdown potential

d(min) = (4.75*10^3 V)/(3*10^6 V/m)
= 1.58*10^-3 m
= 1.58 mm

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posted by drwls

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