You are swimming in the ocean when a bolt of lightning strikes the water 20 mete

Question

You are swimming in the ocean when a bolt of lightning strikes the water 20 meters away from you. What is the current density in the water where you are swimming created by the lightning? What is the potential dierence across your body? Work this problem in cgs units and then convert the final potential difference into SI (volts).

Please explain the answer step by step so I can understand it easily as that is my humble request and I would really appreciate an explained step by step solution. Thank You.

Explanation / Answer

Please see if this can help...

At the risk of appearing lazy...this answer is a modification of an answer I gave to a similar question. It boils down to the voltage drop across your body as a function of distance from the point of contact...In both fresh water and sea water, you're certaily safe at a distance of a few miles. But there are serious differences at closer range. Assume peak current of 30,000 amps and that the charge spreads out uniformly in all directions - as an expanding hemisphere. Current density will be 30,000/(2*pi*r^2)....we'll round that to 4800/radius^2 in meters. (amps/meter^2) At 100 meters that implies a current density of about .48amps/m^2. Assuming a conductance of 5mho/meter...that works out to a voltage drop of 0.096 volts per meter. You're safe a football field away. Even at 30 feet the electric field will only be about 10 volts per meter.Fresh water...completely different story...the lower conductivity of fresh water (by a factor of 1000 or more) means the electric field will be much higher at any given distance...at 100 meters you could easily be subjected to a difference of 100 volts across your body - depending on its orientation. At ten meters...about10,000 volts. Even at the longer distance you're probably still in danger in fresh water.

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