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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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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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 $I \propto \frac{1}{L^3}$. |
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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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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. |
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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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