Energy (J) = Potential Difference(V) × Charge (C)
In this case, we have:
Energy (J) = 7 x 106V × 50C
Therefore, Energy dissipated = 350 x 106 Joules or 350 Mega Joules.
A joule is defined as “the work done or energy transferred when an object with one newton of force moves for one meter” and can be used to measure all forms of energy such as heat, light and mechanical work. It can also be used to measure electrical work which refers to how much energy has been expended when transferring electricity between two points at different voltages. In our case where there was a transfer of 50 coulombs from a thunder cloud with an electric potential difference of 7MVolts,350 Mega Joules were dissipated in total as electrical work done during this process. This amount although considerable is not even close to what happens during normal thunderstorms which typically release around 1 Gigajoule per flash – equivalent to approximately 28 Megawatt hours! So while significant power was still released here it pales in comparison to what usually takes place during storms.
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