Doesn't any massive conductor look like "ground" to an AC supply?

Question

I've been puzzling over this excellent answer to the perennial "Why don't I get shocked by a hot wire if I'm not grounded?" question. The orders of magnitude just don't seem right for two reasons:

1. The answer claims that the surface area of the "capacitive body" is the relevant metric. If the body is full of conductive mass is it not the volume that applies?
2. Household electric service is often grounded by a "ground rod" – literally a long conductor driven into the earth. The earth it contacts is not necessarily a good conductor, so I don't understand how a ground rod can dissipate many orders of magnitude more current than can, say, a person of the same height and conductive volume.

So let's look at the simple case of an adult grabbing the hot wire of a 120V, 60Hz AC household electric supply: I just measured the resistance of my skin and found it to be in the hundreds of kΩ. Let's call it 120 kΩ to make the math easy: if the RMS voltage is 120V then I can conduct 1mA of current – assuming that I have ground potential.

I touch the ground, and sure enough, no charge is transferred. I have the same electric potential as the ground. So I grab that hot wire and the question is: When do I start looking to the electric supply like I'm not the ground?

Because once that supply gets through my insulating skin I'm electrically equivalent to a 20-gallon vat of highly-conductive saline. I start at ground potential, and my body will take current until it matches the potential of the power supply. But this is AC power, so I only "charge" for 1/120s before the polarity changes and I'm giving up electrons where before I was receiving them.

At 1mA, does 20 gallons of saline acquire a significant charge in 1/120s? If not, then it shouldn't matter whether some part of my body is touching a ground or not: the shock should be the same, because my body is "as good as ground" until it is brought significantly closer to the same potential as the power supply.

Furthermore, to question #1 above, assuming that saline is in an insulated container, does it matter what shape the container is? I just can't see how that could make a difference as far as the current is concerned.

Answer 1

One main point you're missing I think is that the voltage doesn't get through your insulating skin. You have a sharp voltage drop across your skin, through which very little current flows due to the high impedance, and your highly-conductive insides are more-or-less shielded.

Another point is that the total amount of charge the circuit will give you is dependent on your capacitance. If we make the blithe assumption that you are a uniform conducting sphere of $\sim1\rm{m}$ in radius, your capacitance is $C=4\pi\epsilon_0(1\rm{m})$. The charge you can collect is given by $Q=CV\approx 10\rm{nC}$. So the average current at the very least cannot be greater than $10\rm{nA}$. (The peak current depends on the amount of resistance- lower resistance means higher peak current, as you would expect).

The earth on the other hand has effectively infinite capacitance, and so it is not limited in this regard. So if you are electrically connected to the earth, it's only the resistance of your body that is saving you. In ideal conditions, this is still usually not a lethal shock, but with something like wet skin it can be.

As far as the shape of the saline container, it can matter a lot. The charge buildup that keeps more than $10\rm{nC}$ from building up builds up around the surface of the volume (as conductors cannot build up charge inside). So if the container were instead a 20 gallon kiddie pool or something, the charge would be able to spread out a lot more, and the pool would be more capable of accepting charge (in physics terms, it would have a larger capacitance).

Answer 2

A closed conducting volume, such as a metal sphere, collects charge from a source until the net charge on the surface of the sphere is sufficient to match the driving EMF. That is, as the sphere accumulates charge, there is a voltage generated which will increase, approach, and finally be equal to that of the driving EMF.

At this point the sphere can no longer accept any further charge from that source, because the voltages are equal.

OTOH, a good ground never "fills up" because it is connected to a vast reservoir with a lower electric field, and hence a lower voltage. If your putative ground is able to accumulate charge, then it is not a "true ground". Consult your local building code for the local requirements. In many jurisdictions it is sufficient to drive a metal rod into the ground below the freeze line, because the local ground is always damp. In other locations, especially very dry ones, one must go deeper.

But a bad ground not only leads to trouble with your electronics, but can be hazardous to your well being. The key is that a current always flows from higher to lower potential, just as water always flows downhill.

Answer 3

Displacement current makes not much sense around 120v ranges. I assume there will be only very small of it will exist even if exist at all at this voltage level.

You must understand that the ground current flows from your household wiring into earth because the neutral wire has been grounded at the utility sub station. This grounding is what actually differentiate the neutral from the live on the otherwise non polar alternating current power source.

Having said that, the ground current flows because there is the closed loop through the earth although its not perfectly conductive.

Now, no matter how volumetric or huge your body is, as long as it is not properly coupled (resistively or capacitively) to the ground, the current which may flow into ground to return back to its neutral grounding at the substation will be extremely small and possibly out of the sensitivity of your body or any standard measuring equipment.