Research website of Vyacheslav Gorchilin
2019-07-02
Calculation secluded capacitance unevenly charged sphere
Sometimes you need to find a solitary capacity of the ball, loaded unevenly, the function of volume distribution of charge is known, and the charge is symmetrically located about its center. In this case the standard formula for determining capacity, expressed as the ratio of charge to potential, is of little use and require a deeper analysis. Here we derive a method for counting such capacity and consider one particular case.
First recall that the electric capacity is [1]: $C = {q \over \varphi} \qquad (1)$ where: $$q$$, electric charge $$\varphi$$ is the potential. This formula is well-suited to bodies with evenly distributed charge or when the charge is on the surface, for example, on the sphere. What if the charge is distributed unevenly? Then this ratio will be different for different parts of the body. How then to calculate the total capacity? Comes to the aid of its main property, which implies that in this case the capacity of the individual parts of this body will add up. Remains to break the body into small enough pieces, and then sum. We go straight to the ball and break it into layers, in each of which the capacity will be like this: $\Delta C_i = {\Delta q_i \over \varphi_i} \qquad (2)$ and the total capacity as follows: $C = \sum \limits_{i=0}^N {\Delta q_i \over \varphi_i} \qquad (3)$ by Reducing the layer height to a very small value, and proceeding to the integration, obtain the General formula: $C = \int \limits_0^R {\Bbb{d} q(r) \over \varphi (r)} \qquad (4)$ where: $$R$$ — Ls the radius of the ball, where $$r$$ is the current radius of the ball. The conditions of the problem we know the picture abetnego charge distribution along the radius, and hence can Express through it and the charge [2]: $q(r) = \int \limits_0^{r} \rho (r) \Bbb{d} V = 4\pi \int \limits_0^{r} \rho (r) r^2 \Bbb{d} r \qquad (5)$ where: $$\rho (r)$$ is a known function of abetnego charge distribution along the radius. Substituting this expression into the previous one, we get: $C = 4\pi \int \limits_0^R {\rho (r) r^2 \over \varphi (r)} \Bbb{d} r \qquad (6)$ This is the General formula for finding the total capacity is unevenly charged sphere if the charge is symmetrically located about its center.
Secluded capacity uniformly charged sphere
First, we find the dependence of the potential on the radius from the formula (19) of this work. Since, in this case, the volume charge density does not depend on the radius and is everywhere constant, then we factor out the brackets: $\varphi(r) = {\rho_0 \over \varepsilon_0} \left( \frac{1}{r} \int \limits_0^r r^2 \Bbb{d} r + \int \limits_r^R R\, \Bbb{d} r \right) \qquad (7)$ $\varphi(r) = {\rho_0 \over \varepsilon_0} \left({R^2 \over 2} - {r^2 \over 6} \right) \qquad (8)$ Now substitute this relationship into formula (6) where and obtain the required capacity: $C = 24 \pi\, \varepsilon_0 R \left(\frac{\sqrt{3}}{2} \ln{\left| {\sqrt{3}+1 \over \sqrt{3}-1}\right| } - 1 \right) \approx 10.595\, \varepsilon_0 R \qquad (9)$ Wonder what the capacity of a uniformly charged sphere is 16% less than similar capacity radius of the sphere. But the surface will be able to accommodate fewer charges than the volume of a sphere. This is a small paradox.
Secluded capacitance is linearly charged sphere
For linearly charged sphere we choose this dependence is $\rho(r)= \rho_0 {r \over R} \qquad (10)$ and then proceed in the same way as in the previous example, we find the dependence of the potential on the radius: $\varphi(r) = {\rho_0 \over \varepsilon_0 R} \left({R^3 \over 3} - {r^3 \over 12} \right) \qquad (11)$ And already here — we define the capacitance of a sphere: $C = 16 \pi\, \varepsilon_0 R\, \ln{4 \over 3} \approx 14.46\, \varepsilon_0 R \qquad (12)$ As you can see, the capacitance is linearly charged sphere is approximately 36% more than the capacity of the ball is uniformly charged. Then you can track the following dependence: the more the nonlinearity of the charge distribution in the ball, the greater its capacity at the same radius.
Secluded capacity back-linearly charged sphere
Another interesting example could be the ball back-a linear dependence of the charge distribution. Ie now at its surface charge, and in the center — it is the maximum: $\rho(r)= \rho_0 {R - r \over R} \qquad (13)$ Potential, in this case, as follows: $\varphi(r) = {\rho_0 \over \varepsilon_0 R} \left( {(R-R)^2 (R+2r) \over 6} + {r^2 (4R-3r) \over 12}\right) \qquad (14)$ just to find the capacitance of a sphere in this case is quite difficult, so we will give an approximate result: $C \approx 39.887\, \varepsilon_0 R \qquad (15)$ the capacity of the ball from the back-a linear dependence is approximately 2.8 times more than straight (see the previous example), and about $$\pi$$ times higher than that of the radius of the sphere.
The materials used
1. Wikipedia. The electric capacity.
2. Wikipedia. The density of the charge.
3. Wikipedia. Dielectric permeability.