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Since the Earth is a good conductor of electricity, is it safe to assume that any charge that flows down to the Earth must be redistributed into the Earth in and along all directions?

Does this also mean that if I release a million amperes of current into the Earth, every living entity walking barefooted should immediately die?

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    $\begingroup$ Think about lightning. $\endgroup$ – Immortal Player Mar 29 '15 at 10:02
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    $\begingroup$ Walking barefoot on a heavily charged planet would rather cause you to take up some of the charge and float away $\endgroup$ – Hagen von Eitzen Mar 29 '15 at 10:45
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    $\begingroup$ @HagenvonEitzen I think you're mixing up electrical charge and magnetic polarity. $\endgroup$ – Schwern Mar 29 '15 at 21:55
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    $\begingroup$ Do you get electrocuted every time you pick up a wire? $\endgroup$ – Mehrdad Mar 30 '15 at 1:16
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    $\begingroup$ I would say that the reason you don't get shocked is actually because the earth is a good conductor, except that it isn't actually a good conductor, at least not at the surface. $\endgroup$ – AJMansfield Mar 30 '15 at 16:34
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Electricity isn't a gas that expands out to shock anything in contact with it. Electricity is a flow from high voltage to low voltage. Touching a charged object is only dangerous if you become a current path--if it uses you to get somewhere. Even if the earth had a net charge, you aren't providing it anywhere to go, so you will not be shocked. It's somewhat like a bird on a power line.

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    $\begingroup$ The earth's dimensions are irrelevant. Currents flow due to voltage /differences/ which lead to electric fields which exert a force on charge carriers. We spend practically all our lives at earth potential. You can add ten thousand to all the voltages on the planet and it will make no difference any more than you can tumble to your death from an endless ten thousand foot high plateau. The capacity of any commonly-encountered body is way too small to lead to a dangerous shock from the one-off transfer of charge when different potentials are brought together (static discharge). $\endgroup$ – Dan Sheppard Mar 29 '15 at 16:16
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    $\begingroup$ If you release a huge amount of charge into the earth then the earth will have higher potential than bodies on it. This creates a current into those bodies that can kill them. In that sense charge is like a gas that expands into bodies standing on earth. My rep is too low to downvote but this answer is not correct. $\endgroup$ – usr Mar 29 '15 at 19:49
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    $\begingroup$ @Schwern to stay with your analogy: The human being a pipe is an empty pipe (a vaccum) and water will flow into it. The analogy is faulty, though, because water has near constant density. The electron "gas" does not. The electrons spread to fill the available space. Since we are assuming the humans are connected to the earth they are one volume and the electrons spread into the humans as well. If you assume the electrons spread in the body of the earth you must also assume they spread into the humans. Not sure how one would argue the opposite. $\endgroup$ – usr Mar 29 '15 at 21:45
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    $\begingroup$ @usr Water flowing is not a perfect analogy for electricity, don't try to stretch it. If a human is in contact with the Earth while the Earth's voltage is slowly ramping up, the human's voltage will also slowly change and there will never be a large differential. Sort of like if the ground slowly rose or fell under you, so long as it's gradual you won't notice and it doesn't matter how much it changes. If you were to change the Earth's voltage quick enough to shock a person, that would require so much energy that everything just melted. $\endgroup$ – Schwern Mar 29 '15 at 21:49
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    $\begingroup$ @Swami: If you connect the Earth to the Moon with a highly conducting wire, current will flow according to Ohm's law $V = IR$ until the two are at the same potential. The gas analogy was me making a point that, in general, you cannot shock yourself simply by touching (only) a conductive or charged object. If it's very, very charged, some of the charge will flow onto you, but in general you can't hold much charge so this won't be very much. People get electrocuted because charge goes through them, but they don't build any up in this case. I would, at this point, drop the analogy. $\endgroup$ – zeldredge Mar 30 '15 at 12:56
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I'll answer the concrete question, because it's one of those fun ones where the units are all wrong and the scales are just absurd.

Does this also mean that if I release a million amperes of current into the earth, every living entity walking barefooted should immediately die?

It depends on how long you do it and with how much power. And (surprisingly) everything doesn't die in a fire! They'll just die in the following global environmental catastrophe.

A million amperes seems like a lot. It's how much current is used in a typical railgun. But its only about 10 lighting strikes. The Earth gets hit by lightning a lot and we're still alive. That's in part because the Earth is so huge it requires a tremendous amount of energy to change its charge, but it's also because lightning is so brief.

The original question is nonsense because the Amp is a measure of flow, not quantity. It's like asking if you could flood the Earth with a million pounds of water pressure. That's a lot of pressure (amps), but it doesn't mean anything unless you know how much water there is (volts) and how long you're applying it (time).


I'll give you another example of how electrical units can be easily misunderstood.

Amps aren't a measure of energy. They're just a measure of rate of electrical flow. You also need to know how long that flow is kept up. A Coulomb measures the potential total flow of charge over time. 1 Coulomb is 1 Amp for 1 second. 1 million Amps for 1 second would be 1 million Coulombs. 1 million Coulombs is twice the charge of the Earth. 1 million Coulombs sounds big and dangerous, right? Not so fast.

1 Amp for 1 million seconds is also 1 million Coulombs. So is 1000 Amps for 1000 seconds (about 18 minutes). What else has 1 million Coulombs? A car battery. Yes, a car battery can kill you, but it's not going to murder humanity if you drop it on the ground. A more familiar term for charge is the Amp-hour which is exactly what it says, how many hours can a thing supply 1 Amp. 1 Amp-hour is 3600 Coulombs. 300 Amp-hours is 1 million Coulombs.

It's like the difference between having 1 million liters of water dumped on you all at once, and having 1 million liters of water slowly drizzled on you. It's still a million liters of water, but one will drown you and the other will just make you soggy and annoyed.


The final piece of the puzzle is the electrical potential difference or volts. In order to know how much energy is being delivered how fast you need to know the Amps and the volts and for how long. Long story short, 1 Amp for 1 second (which is 1 Coulomb) at 1 Volt is 1 Watt-second which is 1 Joule, the standard unit of energy.

So to know how much energy is being delivered we need to know Amps and volts and seconds.


Let's take the Mythbusters approach here and give you the effect you want. What would it take to actually shock all humans? I'm going to go with the lower bound of 300 mA of DC at 110 volts required to make the heart go into fribillation (AC is deadlier, but the Earth isn't an AC generator). And let's say just 100 milliseconds (my source only says "fraction of a second"). I'm using the lower bound to give this the maximum chance of working.

In order to shock humanity to death, we need to very rapidly change the charge of the Earth by 300 mA. If we do it slowly, humanity's charge will slowly change along with it and there will be no noticeable effect.

As always when doing these sorts of large scale calculations, it's good to figure out the basic energy involved. 300 mA at 110 volts is 33 Watts. For 100 milliseconds is 3.3 Joules. We need to kill 7 billion people, that's a minimum of 21 billion Joules. That's roughly how much energy it takes to heat a home for a year. Totally doable.

UPDATE At this point I made a mistake and used volume equations instead of surface charge and generally got it all wrong. @Floris gets it right in their answer but I will leave my incorrect scribblings up here for historical reasons.

300 mA for 100 milliseconds is 0.3 Amps for 0.1 seconds or 0.03 Coulombs. In order to reach everyone, we need to raise the charge density (Coulombs per cubic meter) of the whole Earth by just 0.03. How hard can that be? The Earth's volume is is 1e21 m^3, multiply that by 0.03 C/m^3 and you get 3.25e19 C. It seems like a lot, and it is, but it's meaningless without considering the voltage.

To determine how much energy this will take, we need to plug in the voltage: 110 volts. 3.25e19 Coulombs at 110 volts is 3.5e21 Joules which is about how much energy from the Sun hits the Earth in six hours or half our estimated petroleum reserves. That's A LOT of energy being delivered in 100ms.

It will cause an enormous explosion on the order of magnitude of the largest known eruption in the Earth's history. Like 5000 of our biggest atomic bombs going off at once. Much smaller than the impact that killed the dinosaurs, but probably enough to destroy everything in, say, Colorado and likely kill everyone in the neighboring area.

And that's for a perfect energy transfer. There's likely to be lots of inefficiencies and losses, so you'll likely need 5 to 10 times more energy. Still not enough to melt the surface of the Earth.

Congratulations! If you can get your hands on 3.5e21 Joules, turn it into an electrical charge, and somehow deliver it all to the Earth in 100 ms, you could give everyone on the planet a heart attack! And you would only destroy only one or two large US states in the process. Anyone who survives probably dies in the resulting global environmental catastrophe. Have fun!

UPDATE Again, my calculations for the whole Earth charge are wrong. @Floris gets it right in their answer. I leave my back of the envelope mistake up for historical reasons.

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    $\begingroup$ Your description of Coulombs as "total flow of charge over time" sounds off; at the least, it's an oversimplification. Coulombs measure charge, an inherent property of matter, much like mass. The only way you get "flow" is if they're moving, but you can have non-zero Coulombs with no motion. $\endgroup$ – jpmc26 Mar 30 '15 at 9:48
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    $\begingroup$ An xkcd What If? already answered; just needs Randall's stick figures inserted :-) $\endgroup$ – Mark Hurd Mar 31 '15 at 1:41
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    $\begingroup$ @MarkHurd What a compliment! :) I've thought a number of my answers here would make a good What If style blog. $\endgroup$ – Schwern Mar 31 '15 at 2:22
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    $\begingroup$ To start, let me say I like the approach you take. But I believe it is flawed. Note that you are making a significant mistake when assuming you need to raise the charge per unit volume of the earth - you have to do this as a surface charge. Either you assume the earth has enough conductivity to pull this off or it doesn't. If it does the charge will all appear at the surface. In order to generate a current in every body you need to look at the required $\frac{dE}{dt}$ at the surface - a completely different calculation. $\endgroup$ – Floris Mar 31 '15 at 12:04
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    $\begingroup$ @Schwern - you have your wish. Happy reading. $\endgroup$ – Floris Mar 31 '15 at 16:30
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Firstly we are not the best conductors, so current might be having a relatively hard time getting through us.
But I believe the real reason is that you also need a high potential difference in order to get current flowing through you.
Like lightning which needs a huge potential difference between the clouds and earth (so big that most of times a neutral earth does not give this high difference but needs some accumulation of opposite charge from that on the cloud).So without that potential difference nothing happens.
(Note: the Earth actually causes voltage difference at about 200V per meter, so we do have that big potential difference, but Earth tends to make every object that it is touching neutral--to have the same potential as its surface-- so we distort the equipotential lines that the earth produces. For more on this, check Feynman Lectures Vol2)

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  • $\begingroup$ "Current might be having a relatively hard time getting through us": Hand-to-hand impedance at 220V/50 Hz is 1.3kΩ (median). The most common cause of lightning injury or death is not bolt hitting victim's body. The current actually comes from bolt hitting the ground nearby, creating a potential gradient over the ground surface. If someone has legs apart in this gradient, a current flows through the a victim body in parallel. So not so bad conductors according to bolt preferences, and better having feet close to each other. $\endgroup$ – mins Oct 11 '16 at 9:04
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A counterpoint to Schwern's answer (which was instructive, but I believe wrong on some key points - but I will borrow a couple of numbers from it).

I think the correct way to pose the question is:

If a 300 mA current for 100 ms will kill a human, what should be the rate of change of the electric field around the body to induce that current?

Treating the earth as a conducting sphere of radius R with charge Q, the field at the surface is

$$E = \frac{Q}{4\pi \epsilon_0 R^2}$$

And so the rate of change of the electric field is

$$\frac{dE}{dt} = \frac{dQ/dt}{4\pi \epsilon_0 R^2} = \frac{I}{4 \pi \epsilon_0 R^2}\tag1$$

Now we need an estimate of the current induced when the electric field changes. We can use Maxwell:

$$J = \epsilon_0 \epsilon_r \frac{\delta E}{\delta t}\tag2$$

Combining (1) and (2) we get

$$J = \frac{\epsilon_r I}{4 \pi R^2}\tag3$$

The relative permittivity of the body varies greatly with tissue and frequency - see for example this paper. For the purpose of estimating, I will use a round number for the permittivity of water - 100 - which is much lower than many tissues, and thus gives us an easier target to hit.

We need to convert the displacement current to a mean current in the body by multiplying by the body's cross section. If the average person has a waist circumference of 38 inches (source - www.cc.gov/nchs/fastats/body-measurements.htm) this is an area of about 1/7 m$^2$ so

$$J_{crit} = 7 I_{crit} \approx 2 A/m^2$$

Solving (3) for $I$ (the current we would need to supply the "earth sphere") we get

$$I =\frac{4\pi R^2 J}{\epsilon_r}\approx 10^{13} A$$

That's a lot of current to source... remember that this represents net charge we have to supply to the earth (it has to come from somewhere "not earth"). In order to put such net charge on the earth we would have to put an equivalent amount of net charge on a rocket and send it into space. But just for kicks we will accept that we can do this - perhaps there's a conductor between Earth and the Moon, and we put a big battery between. So how much power is needed to move that much charge?

For this we need the potential of the earth's surface as a function of the charge. We know this is

$$V = \frac{Q}{4\pi \epsilon_0 R}$$

With the current flowing for 0.1 second we would have a total $Q=10^{12}C$ and the potential is

$$V = 1.4\cdot 10^{15}V$$

The actual potential difference from head to toes would be 400 MV (remember - this is not "you are in contact with conductors" but "while the field changes around you, current is induced in your body").

Of course the voltage increases as the charge does, and the total amount of work done would be

$$E = \frac12 QV = 7\cdot 10^{26} J$$

Doing all this in 1/10th of a second requires an instantaneous power of $7 \cdot 10^{27} W$ which is a bit larger than the power output of the sun (which is $4\cdot 10^{26}W$ according to Wolfram alpha

This being the case, I think we're pretty safe. The only way Dr Evil could get away with this plan is to do it in reverse: first pump the charge off the earth to the moon (slowly), then let it all flow back in a cosmic lightning strike. I am not absolutely sure that the moon would stay in orbit while we charge it up... electrostatic attraction would get pretty strong. But that might be the topic for another post.

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    $\begingroup$ Thanks! Your answer has bigger numbers and more disastrous consequences for the Earth, so it's intuitively more correct. ;) I like your speculation at the end there about Moon lightning. A topic for another post you say? $\endgroup$ – Schwern Mar 31 '15 at 17:02
  • $\begingroup$ @Schwern - challenge accepted! *grin*. Later today... $\endgroup$ – Floris Mar 31 '15 at 17:34
  • $\begingroup$ @AOrtiz I don't think so. There is an $\epsilon_0$ in the numerator of one and the denominator of the other... Shouldn't they cancel? $\endgroup$ – Floris Nov 1 '16 at 12:55
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There is a concept of "voltage of a step"* in energy industry - if a high voltage power line is leaking into the ground and isn't shut down, then near that point the ground voltage difference over a single human step (when one feet is closer than the other) can be enough to kill a person; that's why it may be dangerous to approach fallen wires after a storm or something like that. Nowadays it's less of an issue due to more automated detection and cutoff systems, but a few decades ago it was an important hazard.

The distance is meaningful for high-voltage lines, e.g. 30 kV - 330 kV range, but even for such amounts it's not long range - the voltage dissipates rather quickly; and even rather close you wouldn't get electrocuted if just standing there without making a long step.

[*] possibly a different term should be used in English, this comes from other languges.

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    $\begingroup$ That's a better answer than the one selected. To support your demonstration, it's good to remember lightning mostly (40%) kills/hurt because the bolt hits the ground, and creates a potential gradient over the ground. The gradient is strong enough to kill at short distance by flowing current through the legs. $\endgroup$ – mins Oct 11 '16 at 9:12

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