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The potential of a charged sphere. The method of calculation
This work allows to simplify the method of finding the distribution of electric potential inside a charged sphere. The volumetric electric charge density can change along the radius according to the known law, but should be symmetrical about its center. This task is usually solved with the help of a long transformation, and only for uniform charge density. But sometimes it takes a more realistic approach, for example, unevenly distributed charge inside the Earth that the classical approach will greatly complicate the calculations. Here we find a simple solution and derive a General formula for such a task.
The problem can be solved in two ways: using the electric field and with the help of the Laplace operator. The second approach allows to get the required double integral, but poorly explains the physical essence of the phenomenon.
Заряженный шар
Fig.1. A charged balloon
So always go by the first way in which the desired capacity is searched by the classic formula: \[\varphi(r) = - \int \limits^r E(r)\, \Bbb{d} r + const \qquad (1)\] where: \(E(r)\) is the change of the electric field inside the sphere, along a radius \(r\) and \(const\) is a constant.
In the initial conditions of our task there is only a volume charge density inside the ball (\(\rho\)) and, in order to find the field strength, you first need to do some exercises: to Express it through the charge, and then charge through its density. It is necessary to remind you that field strength and charge are linked through the theorem of Gauss [1], the meaning of which is very simple: if you measure the field strength at any distance from the system of charges, knowing the area covered by this measurement, it is possible to find the sum of these charges. Under the terms of the task, all our charges are located symmetrically relative to the center of the sphere, so this law greatly simplified: \[E(r)\ S = { q(r) \over \varepsilon_0 } \qquad (2)\] Here: \(S\) — the area covered by the dimension of the sphere is equal to \(4\pi r^2\), \(\varepsilon_0\) is the absolute dielectric permittivity of vacuum [2]. Express hence the tension; we need it later: \[E(r) = { q(r) \over 4\pi \varepsilon_0 r^2 } \qquad (3)\] Then the formula (1) is changed like this: \[\varphi(r) = - {1 \over 4\pi \varepsilon_0} \int \limits^r {q(r) \over r^2} \Bbb{d} r + const \qquad (4)\] It remains to Express the charge through its bulk density [3]: \[q(r) = \int \limits_{V} \rho (r)\, \Bbb{d} V \qquad (5)\] If we substitute in this formula the classical volume of a sphere: \(V = \frac43 \pi r^3\), it will be suitable for our solution: \[q(r) = 4 \pi \int \limits_0^r r^2 \rho (r)\, \Bbb{d} r \qquad (6)\] Substitute it in (4) and get the General solution of the problem: \[\varphi(r) = - {1 \over \varepsilon_0} \int \limits^r {\Bbb{d} r \over r^2} \int \limits_0^r r^2 \rho (r)\, \Bbb{d} r + const \qquad (7)\] This kind of is not very conducive to revealing the physical meaning of this phenomenon as well, is a fairly complex double integral, which further confuses the situation. Let us try to simplify and even bring in some new patterns.
Simplify the decision
For this we turn to formula (3) and predifferentiated its radius: \[\dot E_r = \frac{1}{4\pi \varepsilon_0} \left({\dot q_r \over r^2} - \frac{2}{r^3} q(r) \right) \qquad (8)\] it is clear that the derivative of the charge radius is: \[\dot q_r = 4\pi r^2 \rho (r) \qquad (9)\] To (8) substitute the value of the charge from (6) and (9): \[\dot E_r = \frac{1}{\varepsilon_0} \left(\rho (r) - \frac{2}{r^3} \int \limits_0^r r^2 \rho (r)\, \Bbb{d} r \right) \qquad (10)\] the Formula (1) convertible to this type \[\varphi(r) = - E(r) r + \int \limits^r r\, \dot E_r\, \Bbb{d} r + const \qquad (11)\] and substitute it obtained earlier (10): \[\varphi(r) = - E(r) r + \frac{1}{\varepsilon_0} \int \limits^r \left(r\, \rho (r) - \frac{2}{r^2} \int \limits_0^r r^2 \rho (r)\, \Bbb{d} r \right) \Bbb{d} r + const \qquad (12)\] Equating (7) and (12) we derive the double integral as a sum of single: \[\int \limits^r {\Bbb{d} r \over r^2} \int \limits_0^r r^2 \rho (r)\, \Bbb{d} r = \int \limits^r r\, \rho (r) \Bbb{d} r - \varepsilon_0 E(r) r \qquad (13)\] the intensity of the field will take from formulas (3) and (6) and finally simplify the double integral: \[\int \limits^r {\Bbb{d} r \over r^2} \int \limits_0^r r^2 \rho (r) \Bbb{d} r = \int \limits^r r\, \rho (r) \Bbb{d} r - \frac{1}{r} \int \limits_0^r r^2 \rho (r)\, \Bbb{d} r \qquad (14)\] Now we substitute all this in (7) and get a simpler solution: \[\varphi(r) = {1 \over \varepsilon_0} \left( \frac{1}{r} \int \limits_0^r r^2 \rho (r) \Bbb{d} r \int \limits^r r\, \rho (r) \Bbb{d} r \right) + const \qquad (15)\] so, instead of the double integral we got two singles, which in itself may be of interest to theoretical physics.
The solution to this problem there is another point that sometimes causes difficulty for understanding real processes. Mathematically it is a constant \(const\) and is still an unknown quantity. In fact, there is nothing difficult, because this constant takes into account the potential outside the ball, which is always equal to the following expression [4]: \[\varphi(r) = {1 \over 4\pi \varepsilon_0 } {q(R) \over r}, \quad r \gt R \qquad (16)\] where: \(R\) is the radius of the ball (Fig. 1). Substituting here the expression (6), we get: \[\varphi(r) = {1 \over \varepsilon_0 r} \int \limits_0^R R^2 \rho (r)\, \Bbb{d} r, \quad r \gt R \qquad (17)\] From the condition of continuity of the potential, we can immediately conclude that at the surface, where \(R = r \), the formula (15) and (17) must be equal to: \[\varphi(R) = {1 \over \varepsilon_0 r} \int \limits_0^R R^2 \rho (r)\, \Bbb{d} r = {1 \over \varepsilon_0} \left( \frac{1}{r} \int \limits_0^R R^2 \rho (r) \Bbb{d} r \int \limits^R R\, \rho (r) \Bbb{d} r \right) + const \qquad (17)\] Hence our constant displays is very simple: \[const = {1 \over \varepsilon_0} \int \limits^R R\, \rho (r) \Bbb{d} r \qquad (18)\]
The final decision
We get if we substitute in the formula (15) previously found a constant: \[\varphi(r) = {1 \over \varepsilon_0} \left( \frac{1}{r} \int \limits_0^r r^2 \rho (r) \Bbb{d} r + \int \limits_r^R R\, \rho (r) \Bbb{d} r \right) \qquad (19)\] As you can see, after substitution, it was enough to change the boundaries of integration are worth paying special attention: in the first integral they change from \(0\) to \(r\) and the second from \(r\) and \(R\). This brings us to revealing the physical meaning of this formula, and proves that the potential inside the sphere is formed by two independent components. The first integral determines the contribution of the inner layer of the ball is \(0..r\), the second contribution of the outer layer \(r..R\). This is the physical interpretation of formula (19).
Example: uniformly charged ball
A special case for this formula is to find the potential distribution within a uniformly charged sphere. If so, then the charge density on the radius is independent and is taken out of the integral sign, and the classic example now is solved very simply: \[\varphi(r) = {\rho \over \varepsilon_0} \left( \frac{1}{r} \int \limits_0^r r^2 \Bbb{d} r + \int \limits_r^R \Bbb{d} r \right) = {\rho \over \varepsilon_0} \left( \frac{r^2}{3} + \frac{R^2 - r^2}{2} \right) \qquad (20)\] By the way, the potential outside the ball is always in the formula (16), regardless of the distribution of charge density inside it.
The materials used
  1. Wikipedia. Theorem Of Gauss.
  2. Wikipedia. The absolute dielectric permittivity of vacuum.
  3. Wikipedia. The density of the charge.
  4. Sudopedia. Field and potential of a charged sphere.