During lighting, a current pulse, shown in figure, flows from the cloud at a height $1.5\ km$ to the ground. If the breakdown electric field of humid air is about $400\ kVm^{-1}$ , the energy released during lighting would be (in unit of $10^9\ J$ )
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Energy released $=\mathrm{q} . \mathrm{v}.$
$\mathrm{q}=$ area under curve $=\frac{1}{2} \times 150 \times 0.2=\frac{30}{2} \mathrm{\,C}$
$\mathrm{v}=\mathrm{E} \cdot \mathrm{d}=400 \times 10^{3} \times 1500=60 \times 10^{7}$
energy released $=60 \times 10^{7} \times 15=900 \times 10^{7}$
$=9 \times 10^{9} \mathrm{\,J}$
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