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there are many topics on the web (for instance this, or this famous lecture) about this topic, but I do not understand some basic concepts.

The statement from which I want to start is this (taken from the Feynman Lecture):

On an ordinary day over flat desert country, or over the sea, as one goes upward from the surface of the ground the electric potential increases by about 100 volts per meter.

It is easily described by this picture.

enter image description here

My first question is:

1) why we are not crossed by a current in our body? Consider a person with 1.8m height: he should be crossed by, if for instance we consider a mean value of human body resistance equal to 10kOhm, 180V/10kOhm = 18mA of current, that is quite high for human body. We will surely perceive it, since it is similar to that we perceive if we touch the phase of our home electric network.

A first answer comes from the Feynmann Lecture, which states that, since the human body is a conductor, we have this situation:

enter image description here

So, If I have understood it correctly, no current flows because no voltage is applied between head and feet of our body. There is a voltage between air and our head, but air is a good dielectric and so no current will flow. But there are many things I do not understand:

1.1) The Human Body is not a perfect electric conductor, therefore its surface will be equipotential only after a certain transient in which all the charges have been distributed on all the body. During this transient, we shoud have perceived the current across our body.

1.2) Apart from the question about the transient, I do not understand this general explanation. In fact, a conductor keeps equipotential at equilibrium, as we have just said. But this situation is different. I see it in this way:

enter image description here

From this scheme, we conclude that the body should be crossed by 18mA of current.

1.3) Suppose our shoes are able to isolate the body from the earth. How does the situation change? I'd say that, from a circuital point of view, it is like in 1.2 (a floating node connected to the earth will not affect the current of the circuit):

enter image description here

Also in this case, the body should be crossed by 18 mA of current.

Now let's go to my second question.

Let's consider a certain component, for instance a LED, and let's put it with its anode towards sky. the physical distance between its terminals (enlarged and elongated with copper wires, for instance) may be, for instance, 5cm.

enter image description here

So the voltage on it shoud be equal to:

100V/m * 0.05m = 5V

2) Why does it keep off and is not switched on?

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  • $\begingroup$ How do you suppose the air is supposed to supply electrons to flow through the body, like the battery you depict.? $\endgroup$
    – Bob D
    Mar 28 '20 at 20:34
  • $\begingroup$ I imagine it as the situation in which the air supplies a receiving antenna. Why cannot our body be like a dipole antenna? $\endgroup$
    – Kinka-Byo
    Mar 28 '20 at 20:36
  • $\begingroup$ And the equivalent circuit of a receiving antenna is that of a load (antenna impedance) supplied by a voltage source which represents the voltage induced by the electromagnetic field in free space $\endgroup$
    – Kinka-Byo
    Mar 28 '20 at 20:37
  • $\begingroup$ But you are talking about air delivering electrons to the body. That requires the air to become ionized (break down). That, in turn, requires a voltage gradient on the order of thousands of volts per cm. How do you suppose you get that to happen when the actual voltage gradient is 100 V/m? $\endgroup$
    – Bob D
    Mar 28 '20 at 20:43
  • $\begingroup$ But which is the difference between this situation and that of a receiving antenna? $\endgroup$
    – Kinka-Byo
    Mar 28 '20 at 20:47
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I have a couple of issues with your analysis.

The first is your treatment of air like a battery supply. In order for the air to supply current to the body, it would need to become ionized so that free electrons are available. That requires the air to undergo dielectric breakdown. The breakdown voltage for air depends on a number of variables including the electrode gap and air pressure (Paschen’s Law), and the electrode configuration, the latter of which determines the degree to which the electric field is homogenous vs inhomogeneous. Generally, however, the breakdown voltage of air is on the order of 20-40 kv/cm. This is certainly much greater than the voltage gradient 100 volts/meter cited for air near the earth.

The other issue is the body impedance. You are obviously correct that the body is not a perfect electrical conductor. But relative to what? Relative to the electrical resistivity of air, which is on the order of 10$^{16}$ Ohm-meter, the 10 K ohm body impedance looks like a very low impedance conductor indeed. It is for that reason the 1.8 m height of the body can be considered at ground potential whereas the same height of air would not. That's why Feynman shows the body all at the same (zero ground) potential.

Hope this helps.

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  • $\begingroup$ Thank you for your answer. Now it is clear. When I said that the circuit of a receiving antenna contains a load and a voltage source, I did a big mistake: the voltage source is the antenna itself(thanks to the voltage induced by the time- varying electromagnetic field), and not the air. Clear, thank you very much. $\endgroup$
    – Kinka-Byo
    Mar 30 '20 at 19:41
  • $\begingroup$ @Kinka-Byo Your welcome. It was a good question. I just up voted it. $\endgroup$
    – Bob D
    Mar 30 '20 at 19:43

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