Research website of Vyacheslav Gorchilin
2019-08-01
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The electrical capacity of the connection. The method of calculation
In this paper, we will introduce you to the method of calculating the capacitance for symmetric bodies, we find the differences between the classical calculation and more realistic — for some types of condensers, we give a definition for communication capacity. Such calculations are needed to solve more complex problems using electrostatic condenser. They are fundamentally important to separate different types of containers.
Here we will consider the capacity with non-standard positions, nevertheless, all the calculations will remain within the classics. From the point of view of classical notions it is, as the ratio of the charge and the potential of the conductor — for a solitary capacity, and the ratio of charge and potential difference for dvuhkletevoj capacity [1]: $C = {q \over \varphi} \qquad (1)$ This definition is suitable for energy occurring in the condenser processes: the energy expended in its charge and received after his discharge. There are no questions. Then we will show the difference in certain properties for certain types of capacitors, and until we come to the container from a different position and will consider it as the ratio of the flux of electric field [2] passing through the conductor plates divided by the potential. Mathematically, this flow must also be multiplied by $$\varepsilon_0$$ is the absolute dielectric constant [3]: $C = {\varepsilon_0 \Phi \over \varphi} \qquad (2)$ it would Seem that it is the same, just written in another form, but we will show the fundamental difference between this formulation of the question. The thread entirely so called: flux of a vector of electric field intensity is the Gauss theorem [2]: $\Phi = \oint \limits _{S} {E\, S} \qquad (3)$ where: $$E$$ is the electric field, $$S$$ is the surface area through which flows the flux $$\Phi$$ (Fig. 1a). For symmetrical bodies with the same symmetrical distribution of the electric field this formula is greatly simplified and will be used further in this form: $\Phi = E\, S \qquad (4)$ Here it should be noted that the lines of electric fields in this case are perpendicular to the surface area they pierced through the darkness. We also know that this flow is multiplied by the dielectric constant gives us the charge, which forms the flow passing through the surface: $q_2 = \varepsilon_0 \Phi = \varepsilon_0 E\, S \qquad (5)$ But we can also find the total flux from elementary source of the charge (Fig. 1a): $q_1 = \varepsilon_0 E\, S_g \qquad (6)$ from Here we can derive two formulas which we will need in the future.
1. We can find the unknown until the field strength is: $E = {q_1 \over \varepsilon_0 S_g} \qquad (7)$ 2. We introduce the coefficient of propagation of the chargethat is displayed on the basis of (5) and (6): $k_q = {q_2 \over q_1} = {S \over S_g} \qquad (8)$ It shows how reduced the charge $$q_2$$, which we measure at the site $$S$$ located at distance $$d$$ from the initial sarada $$q_1$$.
The potential at the site $$S$$ is the classical formula: $\varphi = \int E\, \Bbb{d} x \qquad (9)$ where: $$x$$ — axis as the center of a symmetrical body. We purposely have not put the limits of integration, since for a solitary capacity and duroblade they will be different. We will be set in each case separately. Substituting here the expression (7) we find: $\varphi = {q_1 \over \varepsilon_0} \int {\Bbb{d} x \over S_g} \qquad (10)$ Then the required capacity will be in the following simple formula: $C = {\varepsilon_0 k_q \over \int {\Bbb{d} x \over S_g}} \qquad (11)$ for all the simplifications associated with the application, it reflects well as classic so non-classical notions of capacity. In addition, it is completely absent in the fields and the charges and remains only the geometry of the capacitor. Of course, as in the classics here we introduce the assumption that the surrounding objects are quite remote and do not affect the electric field of the capacitor. Let's check this formula on the classical examples.
 Fig.1. Scheme for finding the capacitance of flat, spherical and co-axial capacitors
Immediately decide that in the next three examples, the ratio of distribution charge will be equal to one, because the whole flows through the first and second electrode are equal: $$k_q=1$$.
Flat capacitor
Take two conductive circle with squares $$S_1$$ and $$S_2$$, and place them coaxially and perpendicular to one another at a distance $$d$$ (Fig. 1b). The axis $$\ell$$, which we will carry out the integration in formula (11), located in the center of the circles and perpendicular to their plane. Calculate the capacity of the system in which square circles are the same: $$S_1=S_2=S$$. Since in this case the flow does not change along the axis, then $$S$$ is taken out of the integral sign and limits of integration, obviously, would be: ($$0, d$$): $C = {\varepsilon_0 \over (d/S)} = {\varepsilon_0 S \over d} \qquad (12)$ i.e. the classical formula of a plane capacitor [1].
The spherical capacitor
Calculating the capacity of such a condenser is also not difficult (Fig. 1c). Suffice it to recall that the area of a sphere is given by: $$S = 4\pi x^2$$ and podstice it in (11): $C = {\varepsilon_0 \over \int {\Bbb{d} x \over 4\pi x^2}} \qquad (13)$ the Limits of integration there will be between the two radii of the spheres: ($$r_1, r_2$$). Then the capacity of a spherical condenser will be like this: $C = {4\pi\varepsilon_0 \over 1/r_1 - 1/r_2} \qquad (14)$ and keeping with the classics. By the way, the number $$\pi$$ in these formulae shows that the studied surface is not only symmetric, but is still a body of rotation.
Coaxial capacitor
For the geometric calculation of the capacitor present here the methodology, it is important to choose the axis of symmetry is $$x$$ along which will be integration. In the first two cases it was obvious, and in the case of the coaxial capacitor on it you need more detail (Fig. 1d). This axis must be directed along the electric field lines and at the same time to be in the middle of them. For cylinders, consists of a coaxial capacitor, the axis may pass through the center of the cylinder and is directed perpendicular to their axes. Then the area of the capacitor plates (cylinder) is: $$S = 2\pi x \ell$$, where $$\ell$$ is the length of the cylinder. Substitute this area in equation (11): $C = {\varepsilon_0 \over \int {\Bbb{d} x \over 2\pi x \ell}} \qquad (15)$ the Limits of integration there will be between the two radii of these cylinders: ($$r_1, r_2$$). Finally we find the required capacity $C = {2\pi \varepsilon_0 \ell \over \ln (r_2 / r_1)} \qquad (16)$ which also coincides with the classical [1].
Capacity connection
If you find the value of capacitance between two conductive plates placed at a certain distance, it will be found that, for example, flat, spherical and co-axial capacitors are calculated without taking into account their solitary tanks and two spheres — including those. Thus, the capacitance between two conductive spheres (balls) in the classical manner is calculated as follows [1,4,5]: $C = 2\pi \varepsilon_0 r \left(1 + \frac{1}{2D} + \frac{1}{4D^2} + ... \right) \qquad (17)$ where: $$D=d/(2r)$$. Hence it follows directly that, for large values of $$d$$ (Fig. 2c), the capacitance ceases to change, is fixed on the value of $$C = 2\pi \varepsilon_0 r$$, and no longer depends on the distance between the spheres. This is absolutely correct from the point of view of power relations. Indeed, if charged spheres of opposite charges, if between them stretching wire connected in series with it resistance, and to discharge the thus the sphere, then resistance will stand out as the energy that was spent on charging them. Moreover, the distance between the spheres does not play any role. All right.
But look what an interesting picture is obtained if we want such areas to arrange radio or energy relationship (figure 2a). Times at some point, the capacitance ceases to change from a distance, by well-known scheme (Fig. 2b) we could transmit information and energy to unlimited distance. Moreover, even between planets and zvetami! However, as we know from practice, with the distance power of the receiving signal decreases, which is logical, otherwise one transmitter we could collect an infinite amount of energy all the way :) Where is the error?
 Fig.2. The relationship between the two spheres (a,b) and the scheme for calculating communication capacity between the two spheres (c,d)
This error, rather, the misunderstanding will vanish if you select in the calculation of the capacitors is a separate category, in which calculation parameters will be carried out without taking into account their private capacity. Capacity, calculated in this way, we will refer to the capacity connection, or a coupling capacitor.
The vast majority of cases, the capacitor is exactly what is calculated. For example, the described method of the capacitor we can calculate solely by its capacity of communication. Also, the classical capacity of flat, spherical and co-axial capacitors are considered without regard to solitary containers of their plates.
The capacity of the connection between the two spheres
Now let's calculate the capacitance between two spheres with radius $$r$$ and the distance between them is $$d$$. Axis integration will hold between the centers of the spheres (Fig. 2c). From the theory of electrostatics it is known that the charge on the sphere can be mentally placed in its center and then, on this basis, to make the calculations. So we're going to do with the first (left in the figure) of the sphere (Fig. 2d). From this point, will be radially guided electric power line $$\bar E$$, part of which will pass through the second sphere. Then, biting her thread will cover spherical sector (displayed in green), the area of which is from the formulas of trigonometry: $S = 2\pi \sqrt{d^2 - r^2} (d - \sqrt{d^2 - r^2}) \qquad (18)$ Recall that the electrical power lines must be perpendicular to the area they pierced through the darkness. Also recall, as is the area of the sphere encompassing the whole thread: $S_g = 4\pi d^2 \qquad (19)$ Then the coefficient of propagation of the charge here would be: $k_q = \frac12 \sqrt{1 - \delta^2} (1 - \sqrt{1 - \delta^2}),\, \delta = \frac{r}{d} \qquad (20)$ Substituting $$d$$ on $$x$$ area and substituting (19) into formula (11), we obtain the required capacity: $C = {\varepsilon_0 k_q \over \int {\Bbb{d} x \over 4\pi x^2}} \qquad (21)$ Border integration here are: ($$r, d-r$$). Solving the integral and substituting these bounds, we obtain the capacity of the connection between the two fields: $C = {4\pi r \varepsilon_0 {1 - \delta \over 1 - 2\delta} k_q }, \quad \delta \lt 1/2 \qquad (22)$ If we take $$\delta = 1/2$$, i.e. when the spheres are in contact, the capacitance will tend to infinity, which is equivalent to the circuit between its conclusions. This is consistent with our initial assumptions (Fig. 2b). If the distance between the spheres is large, i.e., if $$d \gg r$$, this formula is simplified, the volume is: $C \approx {\pi r^3 \varepsilon_0 \over d^2} \qquad (23)$ Now everything fell into place: with increasing distance, the capacitance is reduced when all the interval.
The materials used
1. Wikipedia. The electric capacity.
2. Wikipedia. Theorem Of Gauss.
3. Wikipedia. The absolute dielectric permittivity of vacuum.
4. Problems in electrostatics. Lesson 13.
5. Rawlins, A. D. Note on the Capacitance of Two Closely Separated Spheres // IMA Journal of Applied Mathematics. 1985. — Vol. 34, No. 1. — P. 119-120