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
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
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?
(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
wl;w;wm
wk
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