Electrostatic Interaction of Bilayer Macroparticles

Abstract

The effect of a dielectric film on the surface of conducting dust particles on their electrostatic interaction is investigated. Special attention is paid to the case when the radius of one of particles is much larger than the radius of the other particle and to a nonuniform distribution of the surface charge with variants of equilibrium free charge distribution on each of the macroparticles over the entire surface and over the left and/or right hemispheres. The technique for calculating of slowly converging series is worked out using the hypergeometric Gauss functions and by introducing new functions for which recurrent relations and numerical calculation technique were determined.

Appendices

PROCEDURE OF CALCULATION OF THE HYPERGEOMETRIC GAUSS FUNCTION

Hypergeometric Gauss function ₂F₁ is defined by relation [64, 65]

$$_{2}{{{\text{F}}}_{1}}(a,b;c;z) = \sum\limits_{n = 0}^\infty {\frac{{{{{(a)}}_{n}}{{{(b)}}_{n}}}}{{{{{(c)}}_{n}}}}\frac{{{{z}^{n}}}}{{n!}},} $$

(72)

where (a)_n is the Pochhammer symbol,

$${{(a)}_{n}} = \frac{{\Gamma (a + n)}}{{\Gamma (a)}} = a(a + 1)...(a + n - 1),$$

(73)

Γ is the gamma function. It should be noted that n! = (1)_n, and

$$C_{{n + k}}^{k} = \frac{{(n + k)!}}{{n!k!}} = \frac{{{{{(k + 1)}}_{n}}}}{{{{{(1)}}_{n}}}}.$$

For calculating the hypergeometric Gauss function, we used the recurrent relation [64, 65]:

$$\begin{gathered} _{2}{{{\text{F}}}_{1}}(a - 1,b;c,z) \\ \, + {{(2a - c - az + bz)}_{2}}{{{\text{F}}}_{1}}(a,b;c;z) \\ \, + a{{(z - 1)}_{2}}{{{\text{F}}}_{1}}(a + 1,b;c;z) = 0 \\ \end{gathered} $$

(74)

and the following expressions [64, 65]:

$$\begin{gathered} _{2}{{{\text{F}}}_{1}}(a,b;c;z) = {{\,}_{2}}{{{\text{F}}}_{1}}(b,a;c;z), \\ _{2}{{{\text{F}}}_{1}}(0,b;c;z) = {{\,}_{2}}{{{\text{F}}}_{1}}(b,a;c;0) = 1, \\ _{2}{{{\text{F}}}_{1}}(a,b;b;z) = {{(1 - z)}^{{ - a}}}, \\ \end{gathered} $$

(75)

$$\begin{gathered} _{2}{{{\text{F}}}_{1}}(1,1;c;z) = \frac{{(c - 1)z}}{{{{{(1 - z)}}^{2}}}} \\ \times \left[ {\sum\limits_{k = 2}^{c - 1} {\frac{1}{{c - k}}{{{\left( {\frac{{z - 1}}{z}} \right)}}^{k}} - {{{\left( {\frac{{z - 1}}{z}} \right)}}^{c}}\ln (1 - z)} } \right], \\ c = 2,3,4,..., \\ \end{gathered} $$

(76)

$$\begin{gathered} _{2}{{{\text{F}}}_{1}}(1,b;c;z) = \frac{{(c - 1)!}}{{(c - b - 1)!}}\frac{1}{z} \\ \times \,\,\left\{ {\sum\limits_{k = 1}^{c - b - 1} {\frac{{(c - b - k - 1)!}}{{(c - k - 1)!}}} {{{\left( {\frac{{z - 1}}{z}} \right)}}^{{k - 1}}}} \right. \\ \left. { - \,\,\frac{z}{{(b - 1)!}}{{{\left( {\frac{{z - 1}}{z}} \right)}}^{{c - b - 1}}}\left[ {\sum\limits_{k = 1}^{b - 1} {\frac{{{{z}^{{ - k}}}}}{{z - k}}} + {{z}^{{ - b}}}\ln (1 - z)} \right]} \right\}, \\ c > b, \\ \end{gathered} $$

(77)

$$\begin{gathered} _{2}{{{\text{F}}}_{1}}(1,b;c;z) = (c - 1){{(1 - z)}^{{c - b - 1}}} \\ \times \,\,\sum\limits_{k = 0}^{c - b - 1} {\frac{{{{{(c - b)}}_{k}}}}{{k!(c + k - 1)!}}} {{z}^{k}},\quad b \geqslant c. \\ \end{gathered} $$

(78)

PROCEDURE FOR CALCULATING FUNCTION $T_{n}^{a}$

For function $T_{n}^{a}$ (34), we can easily find recurrent relations

$$\begin{gathered} T_{n}^{a}({{y}^{2}}) = \frac{{{{y}^{{ - n + 1}}}}}{{n - 1}}\frac{{\partial {{y}^{n}}T_{{n - 1}}^{a}({{y}^{2}})}}{{\partial y}} \\ = \frac{n}{{n - 1}}T_{{n - 1}}^{a}({{y}^{2}}) + \frac{y}{{n - 1}}y\frac{\partial }{{\partial y}}T_{{n - 1}}^{a}({{y}^{2}}), \\ \end{gathered} $$

(79)

$$T_{n}^{a} = \frac{2}{{n - 1}}T_{{n - 1}}^{{a - 1}} + \frac{{b - 2a}}{{n - 1}}T_{{n - 1}}^{a}.$$

(80)

Further, we introduce function

$${{R}_{m}} = {{\left( {\frac{1}{{1 - 4{{y}^{2}}}}} \right)}^{{m + 1/2}}}$$

(81)

and operator Ω_n, which transforms $T_{n}^{a}$ into $T_{{n + 1}}^{a}$ (see expression (79)):

$$\begin{gathered} T_{{n + 1}}^{a}({{y}^{2}}) = {{\Omega }_{n}}\left[ {\frac{1}{n}T_{n}^{a}({{y}^{2}})} \right] \\ = \frac{{n + 1}}{n}T_{n}^{a}({{y}^{2}}) + \frac{y}{n}\frac{\partial }{{\partial y}}T_{n}^{a}({{y}^{2}}). \\ \end{gathered} $$

(82)

The action of this operator on R_m is defined by expression

$${{\Omega }_{n}}({{R}_{m}}) = \left( {1 - \frac{{2m}}{n}} \right){{R}_{m}} + \frac{{2m + 1}}{n}{{R}_{{m + 1}}}.$$

(83)

The procedure of calculating functions $T_{{n + 1}}^{2}$, $T_{{n + 1}}^{3}$ (n = 1, 2, …, n_B + 1) required for determining Q_n (35) is as follows.

1. We calculate functions R_m = (1 – 4y²)^{–m – 1/2} for m = 1, 2, …, n_B at y² = –x²/4.

2. We calculate

$$T_{2}^{1}({{y}^{2}}) = \frac{2}{{{{{(1 - 4{{y}^{2}})}}^{{3/2}}}}} = 2{{R}_{1}}.$$

3. We calculate $T_{n}^{1}$ for n = 3, 4, …, n_B + 2 using operator (83).

4. We calculate

$$\begin{gathered} T_{2}^{2}({{y}^{2}}) = \sum\limits_{k = 0}^\infty {\frac{{(2k + 2)!}}{{k!(k + 2)!}}{{y}^{{2k}}}} \\ = \frac{4}{{{{{(1 - 4{{y}^{2}})}}^{{1/2}}}{{{[1 + {{{(1 - 4{{y}^{2}})}}^{{1/2}}}]}}^{2}}}}. \\ \end{gathered} $$

5. Using recurrent relation (79), we calculate $T_{n}^{2}$ for n = 3, 4, …, n_B + 2.

6. We calculate

$$T_{2}^{3}({{y}^{2}}) = \sum\limits_{k = 0}^\infty {\frac{{(2k + 3)!}}{{k!(k + 3)!}}} {{y}^{{2k}}} = \frac{8}{3}\frac{1}{{{{{[1 + {{{(1 - 4{{y}^{2}})}}^{{1/2}}}]}}^{3}}}}.$$

7. Using recurrent relation (79), we calculate $T_{n}^{3}$ for n = 3, 4, …, n_B + 2.

Ultimately, all functions $T_{n}^{a}$ required for calculating coefficients b_n (22) and force F_σ (56) have been calculated.