Determination of the Potential of the Rotor Field of an Electrostatic Machine

I. P. Guk, V. A. Lukutin.

(Presented by the scientific seminar of the Department of Theoretical Foundations of Electrical Engineering)

In the preceding article a separate consideration of the field of the stator and of the rotor of an electrostatic machine is proposed, and a calculation of the electric field of the stator is given. Below a possible variant is presented for determining the potential function of the rotor field. Before proceeding to consideration of the field, it is necessary briefly to explain the construction of a cylindrical electrostatic machine (ESM), a detailed description of which is given, for example, in [1].

In the cavity formed by two concentric cylinders and filled with a certain gas, a rotor is placed. The rotor is a cylinder made of dielectric, in which rods (conveyors) of conducting material are fixed in the axial direction. There may be \(2m\) such rods, separated from one another by equal distances, with the conveyors in the region \(0 \leq \theta < \pi\) carrying charges \(ql\), while the conveyors placed in the region \(\pi \leq \theta < 2\pi\) have charges \(q'l\), where \(l\) is the length of the rods.

If the end effect at the ends of the rotor is neglected, then the field created by the charged conveyors may be regarded as two-dimensional.

Thus, the problem is reduced to finding the potential function in an annular region with a multilayer dielectric. In the region under consideration, in the general case there are three dielectric media with permittivities \(\varepsilon_1, \varepsilon_2, \varepsilon_3\), whose interfaces have the form of circles with radii \(r_2\) and \(r_3\).

The outermost circles (\(r=r_1\) and \(r=r_4\)) have zero potential. If for this region the normalized Green’s function \(G(r;\theta)\) is determined, then the distribution of the potential from a charge \(q\) located in the region may be found from the formula

\[ \varphi(r;\theta)=qG(r;\theta). \tag{1} \]

We shall regard a charged filament as the limit of a prism uniformly charged over its volume, whose cross-section in the plane of the drawing is bounded by two arcs of circles of radius \(r_0\) and \((r_0+\delta)\), and by two radii \(\theta=\pm\Delta\), where \(\delta\) and \(\Delta\) are very small quantities tending to zero in the limit. The volume density of the charge uniformly distributed within the prism is determined by the relation-offsetof

\[ \rho_0=\frac{q}{2r_0\Delta\sigma}. \tag{2} \]

According to [2], in the region filled with a medium with \(\varepsilon=\varepsilon_i\), the potential must satisfy Poisson’s equation

\[ \nabla^2\varphi^{(i)}=-\frac{\rho^{(i)}}{\varepsilon_i}\equiv f^{(i)}(r) \tag{3} \]

\[ (i=1,\,2,\,3), \]

and at the boundary of two media the following conditions must be satisfied:

\[ \varphi^{(1)}\big|_{r=r_2}=\varphi^{(2)}\big|_{r=r_2}; \qquad \varepsilon_1\frac{\partial\varphi^{(1)}}{\partial r}\bigg|_{r=r_2} = \varepsilon_2\frac{\partial\varphi^{(2)}}{\partial r}\bigg|_{r=r_2}, \tag{4} \]

\[ \varphi^{(2)}\big|_{r=r_3}=\varphi^{(3)}\big|_{r=r_3}; \qquad \varepsilon_2\frac{\partial\varphi^{(2)}}{\partial r}\bigg|_{r=r_3} = \varepsilon_3\frac{\partial\varphi^{(3)}}{\partial r}\bigg|_{r=r_3}. \tag{5} \]

Using the method of particular solutions [2], we write the solution of equation (3) in the cylindrical coordinate system in the form of the series

\[ \varphi^{(i)} = \frac{1}{2}a_0^{(i)} + \sum_{k=1}^{\infty}\left(a_k^{(i)}\cos k\theta+b_k^{(i)}\sin k\theta\right), \tag{6} \]

where the coefficients \(b_k^{(i)}=0\), since the distribution of the potential function is symmetric with respect to the diameter \(\theta=0\), and the coefficients \(a_0^{(i)}\) and \(a_k^{(i)}\) have the following form in the first and third regions \(\bigl(f_k^{(1)}=f_k^{(3)}=0\bigr)\):

\[ \begin{aligned} a_0^{(1)}&=A_0^{(1)}\ln r+B_0^{(1)},\\ a_k^{(1)}&=A_k^{(1)}r^k+B_k^{(1)}r^{-k} \end{aligned} \quad\bigg\}\quad r_1\le r\le r_2, \tag{7} \]

\[ \begin{aligned} a_0^{(3)}&=A_0^{(3)}\ln r+B_0^{(3)},\\ a_k^{(3)}&=A_k^{(3)}r^k+B_k^{(3)}r^{-k} \end{aligned} \quad\bigg\}\quad r_3\le r\le r_4. \tag{8} \]

For the second region \((r_2\le r\le r_3)\)

\[ \left. \begin{aligned} a_0^{(2)} &= A_0^{(2)}\ln r+B_0^{(2)} + \int_{r_2}^{r} f_0^{(2)}(\zeta)\,\zeta\ln\frac{r}{\zeta}\,d\zeta, \\[6pt] a_k^{(2)} &= A_k^{(2)}r^k+B_k^{(2)}r^{-k} + \int_{r_2}^{r}\frac{1}{2k}f_k^{(2)}(\zeta)\,\zeta \left[ \left(\frac{r}{\zeta}\right)^k - \left(\frac{\zeta}{r}\right)^k \right]d\zeta \end{aligned} \right\} \tag{9} \]

where the integrals in the limit \((\delta\to0,\ \Delta\to0)\) of any function \(F(\zeta)\), finite and continuous in the interval \(r_2\le\zeta\le r_3\), are determined as follows [2]:

\[ \lim_{\substack{\delta\to0\\ \Delta\to0}} \int_{r_2}^{r} f_k^{(2)}(\zeta)F(\zeta)\,d\zeta = \begin{cases} 0, & \text{for } r_2\le r<r_0,\\[4pt] -\dfrac{q}{\pi\varepsilon_2 r_0}\,F(r_0), & \text{for } r_0\le r\le r_3, \end{cases} \tag{10} \]

where

\[ f_k^{(2)}=\frac{1}{\pi}\int_{-\pi}^{+\pi} f^{(2)}\cos k\theta\,d\theta. \]

To determine the constants \(A_0^{(i)}, B_0^{(i)}, A_k^{(i)}\), and \(B_k^{(i)}\), we write the sy-

system of equations in accordance with (4), (5) and the condition that the potential is equal to zero at \(r=r_1,\ r=r_4\) and, bearing in mind the uniqueness of the expansion of a function in a Fourier series, we equate the coefficients of the corresponding \(\cos k\theta\) and \(\sin k\theta\):

\[ \left. \begin{aligned} &A_0^{(1)}\ln r_1+B_0^{(1)}=0;\\ &A_0^{(1)}\ln r_2+B_0^{(1)}=A_0^{(2)}\ln r_2+B_0^{(2)};\\ &A_0^{(1)}\varepsilon_1=A_0^{(2)}\varepsilon_2;\\[2mm] &A_0^{(2)}\ln r_3+B_0^{(2)}-\frac{q}{\pi\varepsilon_2}\ln\frac{r_3}{r_0} =A_0^{(3)}\ln r_3+B_0^{(3)};\\[2mm] &A_0^{(2)}\frac{\varepsilon_2}{r_3}-\frac{q}{\pi} =A_0^{(3)}\frac{\varepsilon_3}{r_3};\\[2mm] &A_0^{(3)}\ln r_4+B_0^{(3)}=0; \end{aligned} \right\} \tag{11} \]

\[ \left. \begin{aligned} &A_k^{(1)}r_1^k+B_k^{(1)}r_1^{-k}=0;\\ &A_k^{(1)}r_2^k+B_k^{(1)}r_2^{-k}=A_k^{(2)}r_2^k+B_k^{(2)}r_2^{-k};\\ &\bigl(A_k^{(1)}r_2^{k-1}k-B_k^{(1)}r_2^{-k-1}k\bigr)\varepsilon_1 =\bigl(A_k^{(2)}r_2^{k-1}k-B_k^{(2)}r_2^{-k-1}k\bigr)\varepsilon_2;\\ &A_k^{(2)}r_3^k+B_k^{(2)}r_3^{-k} -\frac{q}{\pi\varepsilon_2 k}\operatorname{sh}\!\left(k\ln\frac{r_3}{r_0}\right) =A_k^{(3)}r_3^k+B_k^{(3)}r_3^{-k};\\ &\left[ A_k^{(2)}r_3^{k-1}k-B_k^{(2)}r_3^{-k-1}k -\frac{q}{\pi\varepsilon_2 r_3}\operatorname{ch}\!\left(k\ln\frac{r_3}{r_0}\right) \right]\varepsilon_2\\ &\hspace{40mm}= \bigl(A_k^{(3)}r_3^{k-1}k- B_k^{(3)}r_3^{-k-1}k\bigr)\varepsilon_3;\\ &A_k^{(3)}r_4^k+B_k^{(3)}r_4^{-k}=0. \end{aligned} \right\} \tag{12} \]

Solving the system of equations (11) with respect to \(A_0^{(i)}\) and \(B_0^{(i)}\) gives:

\[ A_0^{(1)}=\frac{q}{\pi\varepsilon_1}\Lambda_0, \tag{13} \]

\[ B_0^{(1)}=-\ln r_1 A_0^{(1)}, \tag{14} \]

\[ A_0^{(2)}=\frac{q}{\pi\varepsilon_2}\Lambda_0, \tag{15} \]

\[ B_0^{(2)}=\left(\frac{\varepsilon_2}{\varepsilon_1}\ln\frac{r_2}{r_1}-\ln r_2\right)A_0^{(2)}, \tag{16} \]

\[ A_0^{(3)}=\frac{q}{\pi\varepsilon_2}(\Lambda_0-1), \tag{17} \]

\[ B_0^{(3)}=-\ln r_4 A_0^{(3)}, \tag{18} \]

where

\[ \Lambda_0= \frac{\displaystyle \frac{\varepsilon_2}{\varepsilon_3}\ln\frac{r_4}{r_3}+\ln\frac{r_3}{r_0}} {\displaystyle \frac{\varepsilon_2}{\varepsilon_3}\ln\frac{r_4}{r_3}+\ln\frac{r_3}{r_2}+\frac{\varepsilon_2}{\varepsilon_1}\ln\frac{r_2}{r_1}}. \]

And solving the system of equations (12) and introducing the notation

\[ \varepsilon_{12}=(\varepsilon_2+\varepsilon_1)-(\varepsilon_2-\varepsilon_1)\left(\frac{r_1}{r_2}\right)^{2k}, \]

\[ \varepsilon_{21}=(\varepsilon_2+\varepsilon_1)-(\varepsilon_2-\varepsilon_1)\left(\frac{r_2}{r_1}\right)^{2k}, \]

\[ \varepsilon_{34}=(\varepsilon_2+\varepsilon_3)-(\varepsilon_2-\varepsilon_3)\left(\frac{r_3}{r_4}\right)^{2k}, \]

\[ \varepsilon_{43}=(\varepsilon_2+\varepsilon_3)-(\varepsilon_2-\varepsilon_3)\left(\frac{r_4}{r_3}\right)^{2k}, \]

we obtain that

\[ A_k^{(1)}=\frac{q}{\pi k r_0^k}\, \frac{r_4^{2k}\varepsilon_{34}-r_0^{2k}\varepsilon_{43}} {r_4^{2k}\varepsilon_{12}\varepsilon_{34}-r_1^{2k}\varepsilon_{21}\varepsilon_{43}}, \tag{19} \]

\[ B_k^{(1)}=-r_1^{2k}A_k^{(1)}, \tag{20} \]

\[ A_k^{(2)}=\frac{\varepsilon_{12}}{2\varepsilon_2}A_k^{(1)}, \tag{21} \]

\[ B_k^{(2)}=-\frac{\varepsilon_{21}}{2\varepsilon_2}r_1^{2k}A_k^{(1)}, \tag{22} \]

\[ A_k^{(3)}=\frac{q}{\pi k r_0^k}\, \frac{r_1^{2k}\varepsilon_{21}-r_0^{2k}\varepsilon_{12}} {r_4^{2k}\varepsilon_{12}\varepsilon_{34}-r_1^{2k}\varepsilon_{21}\varepsilon_{43}}, \tag{23} \]

\[ B_k^{(3)}=-r_4^{2k}A_k^{(3)}. \tag{24} \]

Now, substituting the values of the constants into equation (6), we obtain the expression for the normalized Green’s function:

\[ G(r;\theta)= \begin{cases} \displaystyle A_0^{(1)}\ln\frac{r}{r_1} +\sum_{k=1}^{\infty} A_k^{(1)} \frac{r^{2k}-r_1^{2k}}{r^k}\cos k\theta, \\[1.2em] \displaystyle \hfill (r_1\leqslant r\leqslant r_2), \\[1.2em] \displaystyle A_0^{(2)} \left( \frac{\varepsilon_2}{\varepsilon_1}\ln\frac{r_2}{r_1} -\ln\frac{r}{r_2} \right) -\nu\,\frac{1}{\pi\varepsilon_2}\ln\frac{r}{r_0} \\[0.8em] \displaystyle \quad +\sum_{k=1}^{\infty} \left( A_k^{(2)}r^k+B_k^{(2)}r^{-k} -\frac{\nu\,\operatorname{sh}\!\left[k\ln\frac{r}{r_0}\right]} {2\pi\varepsilon_2 k r_0^k} \right)\cos k\theta, \\[1.2em] \displaystyle \hfill (r_2\leqslant r\leqslant r_3), \\[1.2em] \displaystyle A_0^{(3)}\ln\frac{r}{r_4} +\sum_{k=1}^{\infty} A_k^{(3)} \frac{r^{2k}-r_4^{2k}}{r^k}\cos k\theta, \\[1.2em] \displaystyle \hfill (r_3\leqslant r\leqslant r_4), \end{cases} \tag{25} \]

where

\[ \nu= \begin{cases} 0, & \text{for } r_2\leqslant r<r_0,\\ 1, & \text{for } r_0\leqslant r\leqslant r_3. \end{cases} \]

It is not difficult to verify that expression (25) has the properties which the Green’s function must possess. Inside the ring it has a single singularity at \(r=r_0\) and \(\theta=0\), and is equal to zero on the inner and outer circumferences.

Let us now apply expression (25) to the calculation of the electric field of an electrostatic machine with conductor conveyors, the construction of which we explained above. The cross section of a generator of this type, for a conductive excitation circuit in the short-circuit regime, will correspond to Fig. 1, in which one must take not one, but \(2m\) charged conductors; moreover, the coordinates of the \(n\)-th conveyor will be

\[ r=r_0 \quad \text{and} \quad \theta=\frac{n-1}{m}\pi. \]

We now extend relation (25) to a sequence of \(2m\) line charges, remembering that the conveyors from the first to the \(m\)-th carry

Fig. 1.

charge \(q\), and the remaining ones \(q^{1}\). And, using the method of superposition, we obtain the formula for determining the potential

\[ \varphi(r;\theta) = q\sum_{n=1}^{m} G_n(r;\theta) + q^{1}\sum_{n=m+1}^{2m} G_n(r;\theta), \tag{26} \]

where \(G_n(r;\theta)\) is the normalized Green’s function for the \(n\)-th transporter, defined by expression (25), in which \(\theta\) must be replaced by \(\left(\theta-\frac{n-1}{m}\pi\right)\). Using the boundary conditions in the short-circuit mode

\[ \varphi(r_0;\pi)=0 \quad \text{and} \quad \varphi(r_0;0)=U_{\mathrm{v}}, \]

(where \(U_{\mathrm{v}}\) is the voltage of the excitation source), we obtain that the potential at any point of the field is

\[ \varphi(r;\theta) = \left[ \frac{\displaystyle \sum_{n=1}^{2m} G_n(r;\theta)} {\displaystyle \sum_{n=1}^{2m} G_n(r_0;0)} + \frac{\displaystyle \sum_{n=1}^{m} G_n(r;\theta) - \displaystyle \sum_{n=m+1}^{2m} G_n(r;\theta)} {G_1(r_0;0)} \right] \frac{U_{\mathrm{v}}}{2}. \]

REFERENCES

1. V. I. Levitov, A. G. Lapin. Electrostatic generators with a rigid rotor, I, M., 1963.
2. G. A. Grinberg. Selected Questions of the Mathematical Theory of Electrical and Magnetic Phenomena, §§ 15, 18, M., 1948.