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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  2. A potential difference of 10 kV is established between parallel plates in air.
    (a) If the air becomes electrically conducting when the electric field exceeds
    4*106 V/m, what is the minimum separation of the plates?
    (b) When the separation has the minimum value calculated in part (a), what
    is the surface charge density on each plate?

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