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?

Question

d=___________m

Answer 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

Answer 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?

Answer 3

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wl;w;wm
wk