Asked by Sally

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

Answers

Answered by drwls
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
Answered by Amr adel
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?
Answered by Bishnu
qklw;w'
wl;w;wm
wk
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions