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Q.

A potential difference of 2000 V is established between the parallel plates of a capacitor with air between the plates. If the air becomes conducting when the electric field exceeds 3 x 106 N/C, the minimum separation between the plates should be

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a

0 .01m

b

0.001m

c

0. 667 mm

d

1.34 mm

answer is C.

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Detailed Solution

Let the minimum separation should be x. Then E=Vd=2000x=3×106∴ x=20003×106=23×10−3=0.667mm
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