A potential difference of 4.50 kV is established between parallel plates in air. If the air becomes...

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Physics

A potential difference of 4.50 kV is established betweenparallel plates in air.
If the air becomes ionized (and hence electricallyconducting) when the electric field exceeds 3.05Ã—10⁶ V/m, what is the minimum separation the plates can have withoutionizing the air?
Answer in mm.

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3.9 Ratings (464 Votes)

Potential difference (Î”V) between the plates of the capacitor = 4.50 kV

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.50 x 10³ V

Electric field (E) between the plates of the capacitor = 3.05 x 10⁶ V/m

Then the minimum separation (d) between the plates of the capacitor is calculated as

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â E = Î”V / d

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â d = Î”V / E

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (4.50 x 10³ V) / (3.05 x 10⁶ V/m)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.475 x 10^-3 m

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.475 mm

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