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

Question

In case of a Van de Graaff generator, the breakdown field of air is (A) 2 × 108 V m-1 (B) 3 × 106 V m-1 (C) 2 × 10-8 V m-1 (D) 3 × 104 V m-1

Tardigrade Admin · Accepted Answer

Although air is normally an insulator but at a sufficient high voltage, it ionises and become partially conducting.
The electric field at which it occurs is called breakdown field of air and is nearly equal to

3 \times 1 0^{6} V / m

Q. In case of a Van de Graaff generator, the breakdown field of air is

$2 \times 1 0^{8} V m^{- 1}$

$3 \times 1 0^{6} V m^{- 1}$

$2 \times 1 0^{- 8} V m^{- 1}$

$3 \times 1 0^{4} V m^{- 1}$

Although air is normally an insulator but at a sufficient high voltage, it ionises and become partially conducting.
The electric field at which it occurs is called breakdown field of air and is nearly equal to
$3 \times 1 0^{6} V / m$

Q. In case of a Van de Graaff generator, the breakdown field of air is

2×108Vm−1

3×106Vm−1

2×10−8Vm−1

3×104Vm−1

Although air is normally an insulator but at a sufficient high voltage, it ionises and become partially conducting. The electric field at which it occurs is called breakdown field of air and is nearly equal to 3×106V/m

$2 \times 1 0^{8} V m^{- 1}$

$3 \times 1 0^{6} V m^{- 1}$

$2 \times 1 0^{- 8} V m^{- 1}$

$3 \times 1 0^{4} V m^{- 1}$

Although air is normally an insulator but at a sufficient high voltage, it ionises and become partially conducting.
The electric field at which it occurs is called breakdown field of air and is nearly equal to
$3 \times 1 0^{6} V / m$