For some reason, this topic is poorly covered in science, only a few works are devoted to it, but even they do not contain direct instructions for calculating this parameter. The solitary capacity of our planet \(C_E \), if it is indicated in textbooks, is found by the classical formula for the capacity of a sphere [1]: \[C_E = 4 \pi \, \varepsilon_0 \varepsilon \, R = 7 \cdot 10 ^ {- 4} \, [F] \qquad (1) \] where: \(\varepsilon_0 \) - absolute dielectric constant of vacuum [2], \(\varepsilon \) - relative dielectric constant (RDC), which for some reason is taken as a unit, \(R \) - the radius of the Earth. In this case, the solitary capacity of the planet is about 700 microfarads. Let us remember this figure in order to compare it with the received one - at the end of this work.

It is already clear from this that such an approach does not stand up to criticism. First, electric charges are located not only on the surface of the planet, but also inside it, therefore, the formula for a sphere is not suitable. Second, the composition of the Earth is basically a mixture of oxides, each of which has its own relative permittivity, significantly greater than unity. This means that the average \(\varepsilon \) planet will be more than 1. Based on these inputs, we will sequentially find the missing pieces, and then put them together, and we will receive an answer about a more accurate meaning of the secluded capacity of our planet.

At this stage, we will determine the capacitance without relative permittivity, which we will deal with later. We will only note that we will represent the Earth as a ball. To determine its capacity, we will use two works done by the author several years ago. The first work suggests the distribution of electric charge in the Earth in such a way, so that its total potential for other planets and stars would be neutral. This is possible if the distribution of the charge along the radius of the ball is as follows: \[\rho (r) = \rho_0 \left (1 - \frac43 \frac {r} {R} \right) \qquad (2) \] where: \(\rho_0 \) is the average volumetric charge density, and \(r \) is the current radius, which varies from zero (center of the Earth) to \(R \) - its surface. When this density has positive values, which is possible for \(r = 0 \ldots \frac {3} {4} R \), then this layer contains positive charges (presumably - ions), when it is negative, which is possible for \(r = \frac {3} {4} R \ldots R \), then this means the presence of negative charges (electrons) in this layer. This charge distribution is shown in Figure 1a.

Fig.1. Distribution of charges in the Earth (a); we investigate only a part of the mantle where negative charges are located (b)

The calculation of the solitary capacity of a ball with a charge distributed in it is not considered in the scientific literature, therefore, we have dedicated a separate work to this . From it follows the formula for finding such a capacity: \[C = 4 \pi \int \limits_0 ^ R {\rho (r) r ^ 2 \over \varphi (r)} \Bbb {d} r \qquad (3) \] But in our case, the distribution of the charge in the ball has different signs in its different parts, so further we will look for two capacities: for its positive and negative parts separately.

1. Negatively charged part of the ball
Our condition is shown schematically in Figure 1b. We mentally remove the positive charge in that part of the ball where the radius takes values from zero to ¾R, and leave a negative charge on the part of the ball where the radius takes values from ¾R to R. Also, to simplify the notation, we will introduce the following notation: \[x = {r \over R} \qquad (4) \] Then formula (3) will be written as follows: \[C ^ {-} = 4 \pi R ^ 3 \int \limits_ {3/4} ^ 1 {\rho (x) x ^ 2 \over \varphi (x)} \Bbb {d} x \qquad (5) \] We already know the charge distribution from formula (2), and the potential distribution \(\varphi (r) \), taking into account (4), we will find from of this work: \[\varphi (x) = {R ^ 2 \over \varepsilon_0} \left (\frac {1} {x} \int \limits_ {3/4} ^ xx ^ 2 \rho (x) \, \Bbb {d} x + \int \limits_x ^ 1 x \, \rho (x) \, \Bbb {d} x \right) \qquad (6) \] We note that in this expression, only absolute permeability \(\varepsilon_0 \) is taken into account, but for the full capacity, this value must also be multiplied by the relative permeability \(\varepsilon \), which we will find in the next part of this work. Substitute formula (2) into (6), and calculate the distribution of the ball potential in the selected area: \[\varphi (x) = {R ^ 2 \over \varepsilon_0} \left ({1 \over 18} - {9 \over 256 x} - {x ^ 2 \over 6 } + {x ^ 3 \over 9} \right) \qquad (7) \] This expression must be substituted in (5), calculate this integral, and finally find the capacity of the negatively charged part of the ball: \[C ^ {-} = 11.997 \, \varepsilon_m \varepsilon_0 R \qquad (8) \] Here: \(\varepsilon_m \) is the relative dielectric constant of the Earth's mantle.

2. Positively charged part of the ball
Here we take only that part of the ball, which contains positive charges (Fig. 2), while the radius can take values from zero to ¾R. Next, we will proceed in the same way as in the previous case. First, we write down the general formula for the capacity: \[C ^ {+} = 4 \pi R ^ 3 \int \limits_0 ^ {3/4} {\rho (x) x ^ 2 \over \varphi (x)} \Bbb {d} x \qquad (9) \] Let us find the potential distribution in the selected part of the ball: \[\varphi (x) = {R ^ 2 \over \varepsilon_0} \left (\frac {1} {x} \int \limits_0 ^ {x} x ^ 2 \rho (x) \, \Bbb {d} x + \int \limits_x ^ {3/4} x \, \rho (x) \, \Bbb {d} x \right) \qquad (10) \] Taking this integral we get: \[\varphi (x) = {R ^ 2 \over \varepsilon_0} \left ({(4x - 3) ^ 2 (8x + 3) \over 288} - {(x - 1) x ^ 2 \over 3} \right) \qquad (11) \] Putting this result in (5), we finally get the capacity of the positively charged part of the ball: \[C ^ {+} = 6.521 \, \varepsilon \varepsilon_0 R \qquad (12) \] In fact, this capacity should be made up of two components. Indeed, from ¾ R to about 0.547R, the Earth's mantle continues with its permeability \(\varepsilon_m \), and from 0.547R - to the center of the Earth is its core [3-4], which has its own permeability \(\varepsilon_c \). This distribution is shown in the following figure.

Fig.2. Positively charged part of the Earth, consisting of a core 0-0.547R, and a part of the mantle 0.547R-¾R

Our reader, using formulas (9-11), can independently recalculate these two capacities, and we will immediately write down the answer: \[C_m ^ {+} = 2.35 \, \varepsilon_m \varepsilon_0 R \qquad (13) \] \[C_c ^ {+} = 4.171 \, \varepsilon_c \varepsilon_0 R \qquad (14) \] In sum, they give the capacity from expression (12).

Formulas (8) and (12-14) determine the capacities of the negatively and positively charged parts of our planet. But for the final solution, it is necessary to find the NDP of the mantle and core of the planet.

Generally speaking, the relative permittivity is frequency dependent. But here we make the assumption that the wavelength of the field is much larger than the characteristic dimensions of the system and the polarization processes keep pace with its change.

It is known that for a statistical mixture of a multicomponent medium, the generalized formula from [5] can be applied: \[\varepsilon ^ s = \sum v_i \, \varepsilon_i ^ s \qquad (15) \] where: \(\varepsilon \) is the average ODP for a mixture consisting of \(i \) components, \(v_i \) - weight part of the i-th component, \(\varepsilon_i \) - RDC of the i-th component. Moreover, the sum of all \(v_i\) is equal to one. Landau and Lifshitz proposed a coefficient \(s\) equal to 1/3, the application of which gives good accuracy for this formula.

Let us find the RDC of the Earth's mantle using data from this table and expressions (15): \[\varepsilon_m \approx 6.5 \qquad (16) \] which is in good agreement with the data on ceramics of a similar composition [6].

The core composition mainly consists of iron, with small inclusions of nickel and sulfur [7]. Applying the same principle, and assuming that the RDC of iron and nickel is about 2.6, and sulfur is about 3, we can find the average RDC of the Earth's core: \[\varepsilon_c \approx 2.7 \qquad (17) \]

Putting together all the formulas obtained and the data of this work, we can find the secluded capacity of our planet. It will consist of two components: the capacity of the negatively charged part of the Earth \[C ^ {-} = 11.997 \, \varepsilon_m \varepsilon_0 R \approx 4.4 \cdot 10 ^ {- 3} \, (F) \qquad (18) \] and the capacity of the positively charged part of the Earth \[C ^ {+} = (2.35 \, \varepsilon_m + 4.171 \, \varepsilon_c) \varepsilon_0 R \approx 1.5 \cdot 10^{-3} \, (F) \qquad (19) \] So, for the negatively charged part of our planet, its secluded capacity is approximately 4400 microfarads, and for the positively charged part - 1500 microfarads. Whether these containers can be stacked together is a big question, and most likely, they must be used separately for calculations.

As we can see, the results obtained quantitatively and qualitatively differ from those announced at the beginning of this work. Of course, they can be updated as new data from geophysical earth science becomes available. But the same technique can be used to calculate the secluded capacity of any other planet, if its parameters are known: radius, depth of layers and their chemical composition.