I'm swimming in the ocean and there's a thunderstorm. Lightning bolts hit ships around me. Should I get out of the water?
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$\begingroup$ great question! $\endgroup$– Pratik DeoghareCommented Nov 16, 2010 at 14:53
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1$\begingroup$ Should go on Mythbusters! $\endgroup$– SklivvzCommented Nov 17, 2010 at 20:42
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6 Answers
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In fresh water what makes lightening so dangerous to a swimmer is that most of the current travels on the surface of the water, so rather then getting a $1/r^2$ falloff in current density, you see a $1/r$ falloff. Obviously eventually it will be conducted down into the mass of the water, but this takes a many meters. In salt water, this should happen much quicker. I'm not sure how the conductivity of the inside of your body compares to seawater. Even if it is less, some current would still flow through you.
For normal dry skin, it takes considerable voltage to penetrate the skin (maybe a hundred volts), wet your skin with saltwater and you'll conduct electricity quite well! As a teenager playing with chemistry and water, that happened to me once, 12 volts AC and ionic solutions made for a pretty nasty shock. Normally 12 volts won't penetrate the skin, so I was unrealistically confident!
I have a spark generator that makes roughly 20KV sparks (from a capacitor), discharge it into water, and you see surface sparks spread from the point of entry in all directions.
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$\begingroup$ I'm choosing this answer as the other more detailed answer is too detailed. :) Upvoted both though. $\endgroup$– scrrrCommented Nov 24, 2010 at 14:09
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$\begingroup$ @scrr but neither this nor the answerby @machine answers your question, while other answers do. I would have chosen the answer by @matt or @cem. $\endgroup$ Commented May 15, 2011 at 20:20
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Here's a crude way to look at the problem:
Suppose there are $N$ wires. Each has resistance $R$, common potential difference $V$ and are connected in parallel. So the current through each wire is $I = \frac{V}{NR}$.
Let's imagine a hypothetical wire formed by sea water which has a length, $L$ and cross sectional area, $S$. There are approximately $\frac{2\pi L^2}{S}$ of those wires in a hemisphere of radius $L$. The resistance of such a wire would be $\frac{\rho L}{S} $, where $\rho$ is resistivity of sea water.
The number of such wires that can be connected to your body (with area $A$) is $\frac{A}{S}$.
So the approximate current that will flow through your body is:
$I = \frac{\frac{A}{S}V}{\frac{2\pi L^2}{S} \frac{\rho L}{S}} = \frac{AVS}{2\pi\rho L^3}$
Now assuming $L=100, \rho=0.25, A=1, V=100M, S=10^{-2}, \rho=0.25$
$I = 0.6 Amp$
The most important things to note are that $I \propto A$ and $I \propto \frac{1}{L^3}$.
answered Nov 16, 2010 at 17:01
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$\begingroup$ If the potential difference is $V$, the current through each wire will be $I = \frac{V}{R}$, not $I = \frac{V}{NR}$. $\endgroup$ Commented Nov 17, 2010 at 6:41
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$\begingroup$ Changed it. Please suggest if there are any more problems.:) Thanks! Feel free to edit the answer. $\endgroup$ Commented Nov 17, 2010 at 7:40
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$\begingroup$ You never did fix what Mark was taking about. If $R = \rho L/S$ is the resistance of one such wire, then the current through that one wire is $I_\mathrm{wire} = V/R = VS/\rho L$. Then the current through your body is $I_\mathrm{body} = (A/S) I_\mathrm{wire} = AV/\rho L$. In particular, $S$ must drop out of the final formula, since the current in a continuous medium shouldn't depend on how you imagine chopping it up. Moreover, the proper calculation involves integrating $\rho/r^2$, which diverges unless the lightning strikes an extended area. $\endgroup$– user10851Commented Dec 19, 2014 at 19:33
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either: it comes down very close to you, so the lightning will probably go through your head and fry you
or: it comes down some distance away and dissipates rather quickly. salt water is probably even more conductive than your body, so the current might even flow right around you. I doubt it would make much of a difference.
the danger from thunderstorms while you're in the water is really that your head is the most elevated thing in a large area, not the conductivity of water.
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$\begingroup$ sea is rarely flat, so a 20cm extra elevation is mostly non relevant. $\endgroup$– njzk2Commented Aug 16, 2015 at 16:58
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While moving across a conductance, electricity tends to follow the more conductive(i.e. less resistive) path. In the salt-water case, it will, depending on the distance from you, mostly ignore you. However, if you are sufficiently close, since a lightning bolt has such a huge energy output, it can still be enough to fry you even though it mostly ignores you.
I would feel uncomfortable swimming in a thunderstorm and you should too.
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My girlfriend and I were swimming in the ocean about 30 feet from a beach in Costa Rica as a storm was approaching. We saw occasional flashes of lightning, and I was counting the time it took to hear the thunder, which was at least 7-8 seconds at the fastest, and often 10 or more seconds. I figured as it was at least a mile or more away we were fine, but after a delay with no lightning or thunder, we had a sudden flash, that appeared closer, with thunder a split second later. I felt a slight buzz in my body and I almost didn't mention it to her as we scrambled to shore, but then she said "I just felt zapped!" I would describe it as similar to static shock of touching a door knob after walking across a dry rug, but more diffuse and softer.
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Most probably the current just spreads in all directions and weakens quite fast (at least like $r^{-2}$, not counting resistance), so I don't think the hazard is much (in magnitude) larger than on land in similar conditions.
EDIT: In what I found in Internet salty water has only 10 times better conductivity than wet soil; yet on land the wet soil layer lays on insulating layer of dry soil, so the current is directed to rather "flood" than penetrate.