Research website of Vyacheslav Gorchilin
2019-06-29
The distribution of charge and potential in the Earth's mantle. Experience
In this article will be considered a possible distribution of charge and potential in the Earth's mantle on the basis of experience. After more serious experiments on its basis it will be possible to design stations for the free energy potential of our planet. Its electric charge is the same natural resource as oil or gas, only much more economical and environmentally friendly. In addition, he quickly recovers.
The fact is that the famous classical model of potential Land does not hold water. According to this model, the charge of our planet is 500 thousand Pendant ($$q_E=5\cdot 10^5$$), and the planet itself is considered as a uniformly charged sphere. While never considered to be a potential on its surface, which under this model should be this: $\varphi_E = {q_E \over 4\pi \varepsilon_0 R} \qquad (1)$ where: $$R=6.37\cdot 10^6$$ is the average radius of Earth in meters [1], $$\varepsilon_0 = 8.85\cdot 10^{-12}$$ is the absolute dielectric permittivity [2]. Counting, we find that the potential on the surface of our planet should be ... 705 million Volts, about a billion! In this case, all neutrally charged meteorites falling into our atmosphere would glow and sparkle, and in the calculations of the attraction between the planets would have included the Coulomb component [3]. In addition, if all the electrons were collected in the surface layer, then there would be a strong attraction between them and positively charged ions of gas in the atmosphere, and thus we would observe a constant air flow from top to bottom and the concentration of these ions directly at the surface of the earth. On the tops of mountains, as in any tip, would be formed of the high concentration of charges that would make the field strength there just Romney, up to a constant atmospheric discharges. But we know that even at the highest peaks, tension elektricheskogo field of the atmosphere is only 2.5-3 times higher than normal.
On such reasoning the author gave the experience of which we'll talk further (see figure 1). For it was sadashivan well, which, by means of a pump (in the figure, for simplicity, not shown) via insulated pipe, raises the water to the surface.
 Fig.1. Measurement of the potential difference between the Ground and the water layer
For measuring potentials on the surface of the earth was done the usual, relatively shallow, GND: pin 0.8 m vertically hammered into the ground. The potential difference was measured between this ground surface and the aquifer. When the well depth of 27 meters, the potential difference was approximately 0.35-0.42 Volt, and a negative potential on the surface (surface ground).
Was also measured the current which was very small and amounted to only 5 µa. Since the current depends only on the area of the electrodes that can be done, in principle, any, here will not be considered. In this work we investigate the distribution of electric potential.
The proposed model
According to the above logic, the potential at the surface of the earth must be equal to zero, and all the Coulomb force, at any distance from it, should be compensated. If part of the mantle must contain negative charges and some positive (by this we mean, for example, ionized particles). We also know that the top layer of soil has a negative charge, where we can assume the following distribution of space charge along the radius $$r$$: $\rho (r) = \rho_0 \left(1 - \frac43 \frac{r}{R} \right) \qquad (2)$ where: $$\rho_0$$ — the average volume charge density [4], and $$r$$ changes from zero (the center of the Earth) and $$R$$ — its surface. At a glance, on the surface of the planet, and up to 3/4 of the total radius, the charge is negative, at the point 3/4 of it is missing, and when you move on to the center — positive (Fig. 2). Immediately, we note that the idea you can't do that and need to consider separately the negative and positively charged parts of the world, but the reader can check, the result will be the same as the joint approach to the problem. Why (2) is selected such coefficients will become apparent in the following calculations.
 Fig.2. The charge distribution along the radius of the Earth in relative units Fig.3. The potential distribution along the radius of the Earth in relative units
To calculate the potential distribution along the radius we use the formula (19) from this work. After calculations and reductions, we get the following relationship: $\varphi(r) = {\rho_0 R^2 \over 3 \varepsilon_0} \left(\frac{1}{6} - \frac{\delta^2}{2} + \frac{\delta^3}{3} \right), \quad \delta=\frac{r}{R} \qquad (3)$ As you can see, if $$\delta=1$$ (we are on the Earth's surface), the potential is zero, which is what we needed. From a classical point of view this result can be explained if we remember that in fact we would have to consider two balls: one negatively charged and the second positively. The result would be the same: each ball contributes to the potential, with the result that it is equal to zero at the point $$R$$.
If we move to the center of the Earth, the potential will become: $\varphi(0) = {\rho_0 R^2 \over 18 \varepsilon_0} \qquad (4)$ In the relative units the graph of the potential distribution can be seen in figure (3).
We find the charge of the Earth
Try to estimate the real charge of the Earth. Also, we will need to derive a formula for the potential difference between the surface and the layer at a known depth $$h$$: $\Delta \varphi(h) = |\varphi(R) - \varphi(R-h)| \qquad (5)$ Substituting into it the expression (3) we get this difference: $\Delta \varphi(h) = {\rho_0 h^2 \over 18 \varepsilon_0} (3 - 2 \delta) \qquad (6)$ In our case, $$R \gg h$$, so the formula (6) is further simplified, and its accuracy remains high enough: $\Delta \varphi(h) \approx {\rho_0 h^2 \over 6 \varepsilon_0} \qquad (7)$ Where we will find the average charge density distribution in the Earth's mantle: $\rho_0 \approx \Delta \varphi(h) {6 \varepsilon_0 \over h^2} \qquad (8)$ According to experience, she will be like this: $$\rho_0 = 2.77\cdot 10^{-14}\, [C/m^3]$$, that is, in each cubic meter of the earth, counting from its surface to 3/4 of the depth, there are on average 181 thousand free electrons.
To determine the total charge of the Earth, in every part of the globe, you need to find the integral of the volume charge density and the volume of the planet $$V$$ [4]: $q_E = \int \limits_{(3/4)R}^{R} \rho (r) \Bbb{d} V \qquad (9)$ Substituting in it the charge distribution (2) we get: $q_E = 4\pi \rho_0 \int \limits_{(3/4)R}^{R} \left(1 - \frac43 \frac{r}{R} \right) r^2 \Bbb{d} r = - {9\pi \over 64} \rho_0 R^3 \qquad (10)$ Taking this integral, and substituting the average density from formula (8) into the resulting expression, we get the Earth's charge, which is counted from 3/4 of its radius to its surface and is equal to: $$q_E = 3.1 \cdot 10^6$$ Coulomb's. The minus sign here means that the charge ranging from $$(3/4)R$$ to $$R$$ is negative. From 3/4 of the radius, and to the center, there is exactly the same charge, but with the opposite (positive) sign. This charge distribution is shown in Figure 4. By the way, if integral (10) is taken over the entire radius: $$0..R$$, then it will be equal to zero, which means that our planet should be electrically neutral with respect to other celestial bodies, which is what we needed to get as a result of our calculations.
 Fig.4. A new model for the distribution of electric charge in the Earth Fig.5. The dependence of potential difference (volts) relative to the ground surface and some depth h (meters)
On the chart (Fig. 5) shows the dependence of potential difference on the surface of the earth and some slight depth $$h$$. This graph was built according to the formula (6). Interestingly, at a depth of about 600 meters can be expected industrial potential difference of 220 Volts.
Thus, the total potential energy of the planet Earth, which can be obtained by extracting free electrons from it, is $$7\cdot 10^{15}$$ Joules, which is roughly comparable to the world's electricity generation in one second. Therefore, even taking into account the fast restoring the charge of our planet, it is possible to recommend the use of this energy source only in limited quantities, for example, for supplying electricity to individual households, or as a supplement to the existing electrical network and to other alternative energy sources.
Of course, this hypothesis needs further confirmation in additional and costly experiments. As a minimum, required the study of several wells to a depth of 100, 300 and 1000 meters. But if the hypothesis is confirmed, humanity will get another inexhaustible source of clean energy.
In addition, such studies will help to solve another problem. In connection with the constant threat of accidents caused by sudden explosions of methane in coal mines, despite the fact that electrical equipment of mines has a special protection, the possible presence of portions (seams) of coal and host rocks with the potential To 100-250 at a depth of 500 meters or more from the surface, can be one of the reasons for these sudden explosions of methane.To exclude these factors from proposed measurements of the potentials at the opening of layers and horizons, as well as to existing stations by the Method of Natural Field [5]. Thus, it is proposed to measure a probe of the GIS between the ground surface and formation at a depth of 600m.
The materials used
1. Wikipedia. Earth.
2. Wikipedia. Dielectric permeability.
3. Wikipedia. Coulomb's Law.
4. Wikipedia. The density of the charge.
5. Methods of geophysical studies of wells.