1. (a) Will the electric field strength, E, between two parallel conducting plates exceed the breakdown V strength for air (3-106) if the plates are separated by 2.1 mm and a potential difference of 4500 V m is applied? Determine this by calculating the electric field strength between two parallel conducting plates. E = Pick a "Yes" or "No" V m (b) How close together can the plates be with this applied voltage before the air breaks down? d = mm

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1. (a) Will the electric field strength, E, between two parallel conducting plates exceed the breakdown
V
strength for air (3-106
-) if the plates are separated by 2.1 mm and a potential difference of 4500 V
m
is applied?
Determine this by calculating the electric field strength between two parallel conducting plates.
E =
Pick a "Yes" or "No"
V
m
(b) How close together can the plates be with this applied voltage before the air breaks down?
d=
mm
Transcribed Image Text:1. (a) Will the electric field strength, E, between two parallel conducting plates exceed the breakdown V strength for air (3-106 -) if the plates are separated by 2.1 mm and a potential difference of 4500 V m is applied? Determine this by calculating the electric field strength between two parallel conducting plates. E = Pick a "Yes" or "No" V m (b) How close together can the plates be with this applied voltage before the air breaks down? d= mm
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