In case of a Van de graaff generator, the breakdown field of air is

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$3 \times 10^{6} {Vm}^{- 1}$

$2 \times 10^{- 8} {Vm}^{- 1}$

$2 \times 10^{8} {Vm}^{- 1}$

$3 \times 10^{4} {Vm}^{- 1}$

answer is B.

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Detailed Solution

Air generally acts as an insulator, but when exposed to a sufficiently high voltage, it ionizes and begins to partially conduct. The electric field at which it occurs is called breakdown field of air and is nearly equal to $3 \times 10^{6} {Vm}^{- 1}$

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