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I made a simple bulb-battery circuit and then I cut one of the wires and attached both ends to cemented floor, the bulb didn't glow, this means cemented floor is a poor conductor of electricity. Then how does earthing work ? This idea of activity came from when I got a shock being barefoot but got no shock from same source with slippers on.
So,

How can electrons pass through insulator like cemented floor during earthing ?

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    $\begingroup$ For comparison with your situation, in the US a household earth connection is typically made with an 8' length (2 m?) of ~15 mm diameter copper rod driven vertically into the ground. This gives a lot more contact area between the building circuits and the earth than just touching a couple of wire ends to the surface of the ground. It also generally gets the contact down where the soil is wetter than it is on the surface. $\endgroup$ – The Photon Apr 8 '14 at 18:18
  • $\begingroup$ The bulb might require a few volts at 500mA (so perhaps a few ohms,) while your shocked body was being driven at 120v at perhaps 1mA (so perhaps many tens of kilohms.) If the concrete floor is a resistor, it could be way too high a resistance to light a bulb, yet way too low to prevent the shock. Dangerous experiment: connecting a neon bulb between AC hot conductor and a piece of foil laid on concrete. Since NE-2 will easily light up with less than 1mA, it's a better model for a human body experiencing an AC shock. $\endgroup$ – wbeaty Dec 5 '14 at 7:16
  • $\begingroup$ I believe your question has to do with with this seemingly newer idea that being barefoot on Earth allows the the Earth's electrical current to "pass through" the body, with the presumed effect of healing inflammation, which these practioners posit as a primary source of most suffering. Look here for some particular poor, though representative examples of these "research" papers: ncbi.nlm.nih.gov/pmc/articles/PMC4378297 and ncbi.nlm.nih.gov/pmc/articles/PMC3265077 $\endgroup$ – Jimmy G. Jun 17 '18 at 19:26
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Earthing something means dumping the electron flow into the earth. Since the earth is so big, it can absorbe/give a practically infinite amount of charge without changing potential, this means that you can treat earth as a reservoir of ready to use electrons.

If you plug the phase of your home power line into the ground (without safety devices in the middle), you are actually dumping the electrons in the earth. (In reality -since we use AC- you are repeatedly dumping and taking back electrons 50 times per second).

Note: the other wire of the power line that gets to your home is connected to earth at the nearest distribution node.

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    $\begingroup$ I know all that but I want to know that how do they pass through insulator like cemented floor ? $\endgroup$ – Mukul Kumar Apr 8 '14 at 17:34
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    $\begingroup$ + @MukulKumar: They don't. The reason the earth is conductive is because it contains a certain amount of water, and plenty of ions in that water. $\endgroup$ – Mike Dunlavey Apr 8 '14 at 17:52
  • $\begingroup$ @MikeDunlavey, Your comment explains about the conductivity of Earth. If possible, could you comment on the conductivity of a cemented floor? Of course it must conduct and only because of that we receive electric shocks. But when we try to complete a circuit with a cement rod (without any metals), no current passes through it. I think conductivity of cement is less but not zero. Or resistance is too high but not infinite. But why does it have significant effect in one case but not in the other? I would be glad to hear your opinion on this. Thank you. $\endgroup$ – Guru Vishnu Dec 15 '19 at 6:54
  • $\begingroup$ @M.GuruVishnu: I think you're right, but that's beyond what I know about materials. If you learn anything about conductivity of cement, please share it. Thanks. $\endgroup$ – Mike Dunlavey Dec 15 '19 at 23:43
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Actually according to my opinion it give you a shock but it is negligible. If you give a high energy and try to make contact with earth it will neutralize by giving a little to you. The reason is, if there is a good conductor like body it goes through the body other than only earth.
Just take a lightning application and think of that.

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    $\begingroup$ The most useful answers will have reference to the facts you present. Consider what you are about to say and present it in a way that makes it easy to understand. $\endgroup$ – LDC3 Apr 18 '14 at 3:41
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There are two kinds of "earth" being talked about here. There is:

  1. The kind that aims to use the ground itself as a "return path"; and
  2. A protective "earth", which is actually a separate conducting line laid throughout a building.

For the "return path" earth of 1. there are several ways wherein this will work:

  1. There is actual conduction (i.e. drift of charge like ions) through the water and dissolved ions in the ground and this effect is sometimes (albeit seldom) used to lessen the amount of infrastructure one needs in remote areas to deliver mains power to sparsely spread buildings. As you can imagine, its efficiency depends greatly on groundwater levels and so this is far from ideal. See the Wiki page on Single Wire earth Return for details.

  2. There is "displacement current return path" as described by Kornut's Answer, which I'll add a little more to. As I talk about in detail here, one actually doesn't need a return path to make an electric "circuit" work. We can put some numbers in to flesh out Kornut's Answer: imagine that your "earth" connexions are instead big conductive spheres that store charge as in my drawing below.

Single Wire System

You can easily do the Gauss's law calculations for widely separated spheres to show that the voltage difference between them when one has been "pumped up" with positive charge $+Q$ and the other with negative charge $-Q$ in the way described by Kornut in his answer is:

$$V = \frac{Q}{2\,\pi\,\epsilon_0\,R}$$

where $R$ is the spheres' radius (assumed the same), so that the above electrically looks like a "circuit" with the spheres replaced by a capacitor of value $2\,\pi\,\epsilon_0\,R$. So, at $50\mathrm{Hz}$, say, with one metre radius spheres, the impedance $(i\,\omega\,C)^{-1}$ will have a magnitude of $(100\,\pi\,\epsilon_0)^{-1} = 360\mathrm{M\Omega}$ so you're going to need awfully big spheres or high frequencies to make the above scheme light the lightbuld. Even at $1000\mathrm{km}$ radius, the figure is $360\Omega$.

There have been some really bizarre truly one-line power transmission systems thought of in the past, where the one line works as a waveguide and does not need a return path. See the Wiki Page for the Goubau Line for details.

Now to discuss the notion of protective earth. In home electricity supplies, one side of the supply ("the neutral") is "tethered" to the same potential as a protective earth circuit. This latter is simply a system of conductors, going through the third "Earth" pin on the socket outlet, that tethers any conducting surface of an electrical appliance to the one side of the supply. The other side of the supply is called the "active". If a fault happens in an appliance such that the active touches the conductive housing of an appliance (say of a toaster), we have a dangerous situation, since anyone touching the appliance can then get an electric shock. However, if the housing is connected to the protective earth circuit, there is a redundant path back to the supply's neutral. This leads to a high current in the redundant path and hopefully a blown fuse (the redundant protective earth path has a fuse in it that blows with a very low current).

A more modern and safer way to achieve this protection is earth leakage protection, which I talk about in this answer here.

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