As always, xkcd: Nerd Sniping has something right on point.
With all the side discussion, I realize that one part of the OP’s question never got fully addressed.
In the real world, there could be a path for current to flow that doesn’t go through the terminals (where the voltage is applied). In practical terms, the plate may be attached to the Earth at some point. Since we aren’t usually off the planet, the voltage relative to the Earth is almost always relevant when figuring out where current might flow aside from the obvious connections.
In the OP, the person might be touching the ceiling attached to the Earth and that provides a path. That case is the most likely to be dangerous, since there isn’t an alternate path to the Earth via the metal. Given a good conductor (as Chronos demonstrated, any metal is still quite good) and a reasonable voltage, you can be pretty far away from the “live” area and still have significant current flow through your body.
That explains a few things about electrical safety : For one, our power systems usually have one side connected to the Earth, and an additional, presumably better path is provided by the “safety ground” wire. That’s all designed to ensure that your body isn’t the best path to the Earth. Then there’s the GFCI, which actually handles a case very much like the OP describes. If the current at one terminal doesn’t equal the current at the other terminal, there’s current flow via some other path. That could very well be through a person. The GFCI will shut off the power when that happens.
GFCIs are mostly used in wet environments; as hopefully has been shown, a metal plate will get most of the current through it, but if you stick a possibly leaky bag of electrolytes in a puddle of water (or a swimming pool, per ecg), it may well be the best conducting medium available.
Thanks for addressing that. Half of the explanations are flying well above my head, so practical answers help me.
Imagine for example a mile-long metal rod instead of a plane of sheet metal. Two terminals are connected one meter apart near the center of it. If I touched one end of it (and Earth) would I get current flowing through me.?
Make all the assumptions you need. In my previous question I didn’t know enough to think of foil and sheet metal as different for the purposes of the question (although I still found the topic of how they differ interesting).
Even if a defibrillator is handy, you have about 1 chance in 4 of having a successful cardioversion.
A defibrillator actually STOPS your heart. Provided the internal pacemakers were not damaged by the wild shock, the natural pacemakers in essence reboot and a regular rhythm is reestablished.
FWIW, a current of 50 mA through your heart is enough to be fatal.
VunderBob, wearing both EE and EMT hats at the moment
Assuming that your rod has a diameter very small compared to its length (probably a good approximation, if it’s a mile long), then no, you won’t get any current in the 1D case.
are you touching the end closest to the positive or negative electrode?
!
I didn’t realize it mattered. Tell me for both then.
Also, if one mile is way too much, any interesting answer for other values are also good.
To Sapo:
Yes, in the metal rod example, significant current will flow if there’s sufficient voltage, and your body resistance is low enough to allow for it.
‘Earth’ could be replaced by any point that has a voltage relative to the terminals. It’s simply the most common example in practical experience.
I think I see what you’re asking with the metal rod question - namely, does it matter which terminal you’re closer to?
The answer depends on what the voltage from a particular terminal to the Earth is. Assume for simplicity’s sake a DC voltage is applied (which will stay the same). Let this diagram be the set-up :
…|+V-|… Voltage V between terminals (dots used only for diagram’s sake)
-----you---------------1—2--------- rod
|__________________ the Earth
The first thing to clarify: Since the rod is a conductor, the distance between you and Terminal 1 doesn’t matter much. The conductivity of the metal is high enough that if current flows, the voltage drop between Terminal 1 and you is very small. So the part of you touching the rod is, for practical purposes, at the same voltage as Terminal 1 relative to the Earth. (If preferred, this can be modeled as a very low-value resistor between you and Terminal 1).
(using T1 and T2 from here on).
The voltage at T1 is the only one that matters here. Now depending on how the problem is modeled, the voltage at T1 relative to T2 could change. But since you’re on the other side, it doesn’t matter what T2 is at. Or to put it another way, whatever T2 is relative to the Earth, T1’s voltage to the Earth is at that plus the voltage between T1 and T2. So we need only work with T1’s voltage to the Earth, since that is what your body is going to see.
So one side of your body is at the voltage of T1 (to Earth). If the other side is connected to the Earth, it is therefore at 0V relative to Earth. Then the voltage across your body equals the voltage at T1. Following Ohm’s Law, this voltage divided by your body resistance will be the amount of current flow. More or less; the impedance[sup]*[/sup] of your body is not necessarily a single resistive value, but it’s a decent enough model.
One final thing of note is that current flows whether T1 is positive or negative relative to the Earth. If T1 is +100V to the Earth, current flow would be from T1 through your body to Earth. If T1 is -100V, it’d flow the other way. If you’re like UncleFred, you can reverse those directions; it doesn’t make a difference. Hopefully you can see how this covers the AC case. (There are a few other complications of AC and your body as a conductor, which brings in impedance again.)
Side discussion, skip if you want:
As I said the model makes some difference in what is seen at Terminal 1 relative to Terminal 2. Using an ideal voltage source (and a non-ideal conductor), the voltage stays fixed. So if T2 is connected to Earth and there’s +100V at T1, your body will be at +100V relative to the Earth. In a practical situation, you don’t have an ideal voltage source; the terminals will end up close to the same voltage.
If you connect it to T1 and T2 is 0V relative to Earth, the voltage at T1 will be near 0V (and in this case the assumption is that this is the Earth or a large sink that the power supply is also connected to, to explain why it heads that direction instead of to what Terminal 1 is at). But if T2 is at 100V relative to Earth, you’d get that, or nearly that, at T1 as well, and thus at your body.
*Impedance - if you’re unfamiliar with this term, it’s simply a more generic way of saying ‘resistance’. It takes into account some time-varying aspects as well. A resistance is an impedance, it simply doesn’t vary over time.
Somehow I missed the ‘mile’ part of the rod case; that is significant. Assume my diagram is more to scale, and you’re only a few meters away. That should cover the ‘interesting answer’ portion.
About 1/4 the length, last time I measured.
Great. Now hook it up to a power source and tell us what happens.
Your simple rod question brings up a lot of different issues (in addition to Uncertain’s personal issues, which I think I’ll ignore 
)
There are two main cases you need to consider here. One is that the power supply is isolated from earth ground. The other case is that it isn’t.
Case 1: Isolated
This is your typical case where your voltage supply is something like a battery. Your example here has essentially the same problem as your metal plate example in that you are using a very conductive metal rod, and are in essence shorting out the battery. Most real world batteries will overheat and possibly explode when you do this. Ignoring that and just going with the concepts here, you are going to get a lot of current flowing between T1 and T2 (using panamajack’s terminal notation). Technically, since electricity does flow through all paths, some teeny tiny amount of current will go from T1, all the way around to the left side of the rod, then all the way along the bottom of the rod to the right side, then back up around to T2, but that current is going to be so small that you can pretty much ignore it.
If you touch the end of the rod and touch earth ground, nothing will happen to you. You will not feel a shock.
Case 2: Not Isolated
In this case, you’ve got some sort of power supply that plugs into a wall outlet, or something along those lines. This is “grounded”, meaning that the negative rail of the power supply is connected to earth ground. Now, the earth is part of the circuit.
Lets say T1 is the “ground” connection and T2 is the “hot” connection. Since T1 is closer to you, you are essentially touching a connection that is already grounded. You are at ground potential (since you are on the ground), and T1 is at ground potential. You don’t get shocked.
Now, let’s say T1 is the “hot” and T2 is the “ground”. This one isn’t so pleasant for you. There are two circuit paths, one from T1 to T2 directly, and one from T1, down the rod, through you, through the ground, and back up to T2 since it is grounded. Since the rod is a good conductor (better than you at least), and you’re not, your “electrical load” on the power supply is going to be fairly negligible compared to the load it already sees due to the metal rod. so you aren’t going to change the voltage from T1 to T2 much. Your part of the circuit is T1, down the metal rod which has a fairly low impedance, through you, then through earth which also has a low impedance. Because the impedances of the earth and the rod are so low, you can almost ignore them, and basically you are going to receive a full shock close to the T1-T2 voltage (the voltage of your power supply). A thinner rod will have a higher impedance than a thicker rod, so the thinner your rod, the more it is going to protect you from the shock (keep your thick and thin rod jokes to yourself, please ).
Although the example isn’t really practical due to the basic short circuit between the power supply terminals, it does hint at some important aspects of electrical safety. One issue is that electricity in a non-isolated system you only need to touch the “hot” part of the circuit and you can get shocked due to other parts of your body being in electrical contact with earth ground. Isolated systems are inherently safer, because you have to somehow make two electrical connections before you get shocked. One connection alone (like touching one wire) won’t do it.
So of course you are going to ask why don’t we run our electrical systems in the real world isolated? They’d be safer, wouldn’t they? Well, yes, they would, and in fact sometimes we do run them that way. The next time you happen to be in a hospital, look for the red outlets. Those are fed from isolation transformers. Both the hot and neutral wire are isolated from earth ground. You can touch either wire and the earth and not get shocked. You have to touch both wires to get shocked. We don’t run electrical distribution systems (like the electrical system that feeds your house) that way, because mother nature tends to randomly insert ground connections through tree branches and similar things, which makes it very difficult to keep the entire system isolated. Hospitals have to go through yearly testing and do a lot of work to make sure their isolated systems stay isolated. That’s not practical to do on something the size of the U.S. power grid.
Wouldn’t you know it, the most interesting question was the one I didn’t know enough to ask.
What do hospitals have to gain by having isolated systems? You would think that the world is already used to them not being so and not shocking ourselves to death. What’s different in a hospital that makes the extra effort worth the while?
when your exposure to voltage is through dry healthy skin the protection methods that are in common use work well.
in a hospital you are making contact with body fluid system and body interiors which are much more conductive and sensitive to damage.
Isolated power in hospitals grew out of two things, open heart surgery and anesthetics.
In the early days of open heart surgery a lot of patients were dying, and no one really knew why. One of the theories was that tiny currents from the monitoring equipment was shocking the patient to death. The theory wasn’t well received, and there was no way to prove it since it left no trace of what happened behind. There was a lot of debate on both sides of the argument, but eventually, they started to electrically isolate their equipment. To the surprise of many, they found that fewer patients died.
This eventually led to the standardization of using isolated power in what they called “wet” locations, which has been expanded over the years to include a lot more than just hospital operating rooms.
The second issue is anesthetics, many of which have historically been flammable. If you have an electrical arc near a flammable anesthetic due to equipment malfunction or whatever, you can easily end up with a fire or explosion. Using isolated power reduces the risk of an arc, and therefore reduces the risk of fire or explosion along with it. These days, they have also tended to move away from using flammable anesthetics.
in an oxygen rich environment, which is part of using anesthetics, many things including patients are highly flammable.
This is not true. The potential, and thus the resistance, diverges logarithmically with distance. This is evident from the 3D potential plot shown earlier. The only way you will get a finite resistance between the two electrodes is to take account of the finite size of those electrodes. The resistance will then involve the logarithm of the ratio of the distance between the electrodes and the size of the electrodes.
Of course, for the electrocution problem, where your feet are distant from the electrodes by many times their diameter, the voltages will all be finite.
Your equation seems to be for the case where you have four electrodes, call them a,b,c, and d that are equally spaced and colinear. If you inject current at a, extract it at d, and measure the potential difference between b and c and take the ratio of voltage to current, you get a resistance that is independent of the distance between the electrodes and given by your expression.
It is very simple to calculate the potential pattern in the following way. First calculate the potential for current injection from a point source at the origin and a sink at infinity. The potential is radially symmetric, proportional to the current and logarithmic in the radius (diverges for both r=0 and r=infinity).
When you inject current at one point and sink it at another, simply superimpose the two solutions, getting the difference of two terms logarithmic in the distance from one or the other electrode.
Curses! Someone sent me that cartoon a few weeks ago and it worked perfectly. I spent several evenings looking for a simple trick that would enable me to find the resistance across a diagonal in an infinite square grid of identical resistors. The resistance between adjacent nodes is trivial (half the value of each resistor), there is lots of symmetry, and I was convinced there must be an easy way to find the answer. Unfortunately, I am not aware that there is such a simple trick. I finally looked up the answer, which was obtained by Fourier transforming the entire problem and calculating an integral of the continuous variable in the transform domain.