电动力学 第13周作业

Chasse_neige

5. 拉普拉斯方程,分离变量法

书2.2, 2.4, 2.18

2.2 在均匀外电场中置入半径为 $R_0$ 的导体球，试用分离变数法求下列两种情况的电势：

(1) 导体球上接有电池，使球与地保持电势差 ${\mathrm{\Phi }}_0$；

假设球心在垂直电场方向上到无穷远处的电势为 ${\varphi }_0$ ，均匀外电场大小为 $E_0$ ，位矢与外电场夹角为 $\theta$

$$\varphi (r,\theta )=\sum _n(A_n{r}^n+B_n\frac{1}{r^{n+1}})P_n(\cos \theta )$$

保留 $n=0$ 以及 $n=1$ 项

$$\varphi (r,\theta )=A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta$$

带入 $\theta =\frac{\pi }{2},r\to \mathrm{\infty },\varphi ={\varphi }_0$，所以 $A_0={\varphi }_0$

当 $r=R_0$ 时，$\varphi ={\mathrm{\Phi }}_0$ ，所以 $A_1{R}_0+\frac{B_1}{R_0^2}=0$

$$A_0+\frac{B_0}{R_0}={\mathrm{\Phi }}_0$$$$B_0=R_0({\mathrm{\Phi }}_0-{\varphi }_0)$$

再利用电场定出剩余系数

$$\begin{array}{c}\overrightarrow{E}=-\mathrm{\nabla }\varphi =-\hat{r}\frac{\partial }{\partial r}(A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta )-\frac{\hat{\theta }}{r}\frac{\partial }{\partial \theta }(A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta )\\ =\frac{B_0}{r^2}\hat{r}-A_1\cos \theta \hat{r}+\frac{2B_1}{r^3}\cos \theta \hat{r}+\frac{1}{r}(A_{1}r+\frac{B_1}{r^2})\sin \theta \hat{\theta }\\ =\frac{R_0({\mathrm{\Phi }}_0-{\varphi }_0)}{r^2}\hat{r}-A_1\cos \theta \hat{r}+A_1\sin \theta \hat{\theta }+\frac{2B_1}{r^3}\cos \theta \hat{r}+\frac{B_1}{r^3}\sin \theta \hat{\theta }\end{array}$$

当 $r\to \mathrm{\infty },\theta =0$ 时，电场大小为 $E_0$ ，所以

$$A_1=-E_0$$$$-E_0{R}_0+\frac{B_1}{R_0^2}=0$$$$B_1=E_0{R}_0^3$$

所以电势分布为

$$\varphi (r,\theta )={\varphi }_0+\frac{R_0({\mathrm{\Phi }}_0-{\varphi }_0)}{r}-E_{0}r\cos \theta +\frac{E_0{R}_0^3}{r^2}\cos \theta (r>R_0)$$

(2) 导体球上带总电荷 $Q$。

假设球心在垂直电场方向上到无穷远处的电势为 ${\varphi }_0$ ，均匀外电场大小为 $E_0$ ，位矢与外电场夹角为 $\theta$

同样保留球坐标下拉普拉斯方程解的前两项

保留 $n=0$ 以及 $n=1$ 项

$$\varphi (r,\theta )=A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta$$

带入 $\theta =\frac{\pi }{2},r\to \mathrm{\infty },\varphi ={\varphi }_0$，所以 $A_0={\varphi }_0$

当 $r=R_0$ 时，$\varphi$ 为常数 ，所以 $A_1{R}_0+\frac{B_1}{R_0^2}=0$

再利用电场定出剩余系数

$$\begin{array}{c}\overrightarrow{E}=-\mathrm{\nabla }\varphi =-\hat{r}\frac{\partial }{\partial r}(A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta )-\frac{\hat{\theta }}{r}\frac{\partial }{\partial \theta }(A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta )\\ =\frac{B_0}{r^2}\hat{r}-A_1\cos \theta \hat{r}+\frac{2B_1}{r^3}\cos \theta \hat{r}+\frac{1}{r}(A_{1}r+\frac{B_1}{r^2})\sin \theta \hat{\theta }\\ =\frac{B_0}{r^2}\hat{r}-A_1\cos \theta \hat{r}+A_1\sin \theta \hat{\theta }+\frac{2B_1}{r^3}\cos \theta \hat{r}+\frac{B_1}{r^3}\sin \theta \hat{\theta }\end{array}$$

当 $r\to \mathrm{\infty },\theta =0$ 时，电场大小为 $E_0$ ，所以

$$A_1=-E_0$$$$-E_0{R}_0+\frac{B_1}{R_0^2}=0$$$$B_1=E_0{R}_0^3$$

因为导体球表面带电量为 $Q$ ，所以

$$Q={ϵ}_0\oint d\overrightarrow{S}\cdot \overrightarrow{E}={ϵ}_0\oint dS(\frac{B_0}{R_0^2}-A_1\cos \theta )$$$$Q={ϵ}_{0}4\pi B_0$$$$B_0=\frac{Q}{4\pi {ϵ}_0}$$

所以电势分布为

$$\varphi (r,\theta )={\varphi }_0+\frac{Q}{4\pi {ϵ}_{0}r}-E_{0}r\cos \theta +\frac{E_0{R}_0^3}{r^2}\cos \theta (r>R_0)$$

2.4 均匀介质球(电容率为 ${\epsilon }_1$)的中心置一自由电偶极子 ${\overrightarrow{p}}_t$，球外充满了另一种介质(电容率为 ${\epsilon }_2$)，求空间各点的电势和极化电荷分布。

提示：同上题，$\varphi =\frac{{\overrightarrow{p}}_t\cdot \overrightarrow{r}}{4\pi {ϵ}_1{r}^3}+{\varphi }^{\prime }$，而 ${\varphi }^{\prime }$ 满足拉普拉斯方程。

$${\varphi }^{\prime }(r,\theta )=\sum _n(A_n{r}^n+B_n\frac{1}{r^{n+1}})P_n(\cos \theta )$$

保留前两项

由于 ${\varphi }^{\prime }$ 在原点处不发散，所以在 $r<R_0$ 区域

$${\varphi }^{\prime }(r,\theta )=A_0+\frac{B_0}{r}+(A_{1}r+\frac{B_1}{r^2})\cos \theta$$$$B_0=B_1=0$$

电场

$$\begin{array}{c}\overrightarrow{E}=-\mathrm{\nabla }({\varphi }^{\prime }+\frac{{\overrightarrow{p}}_t\cdot \overrightarrow{r}}{4\pi {ϵ}_1{r}^3})\\ =-\hat{r}\frac{\partial }{\partial r}(A_0+A_{1}r\cos \theta +\frac{p_t\cos \theta }{4\pi {ϵ}_1{r}^2})-\frac{\hat{\theta }}{r}\frac{\partial }{\partial \theta }(A_0+A_{1}r+\cos \theta +\frac{p_t\cos \theta }{4\pi {ϵ}_1{r}^2})\\ =-A_1\cos \theta \hat{r}+A_1\sin \theta \hat{\theta }+\frac{p_t}{2\pi {ϵ}_1{r}^3}\cos \theta \hat{r}+\frac{p_t}{4\pi {ϵ}_1{r}^3}\sin \theta \hat{\theta }\end{array}$$

由于无穷远处电势为0，所以在 $r>R_0$ 区域

$$\varphi (r,\theta )=C_0+\frac{D_0}{r}+(C_{1}r+\frac{D_1}{r^2})\cos \theta$$$$C_0=C_1=0$$

再利用电场确定剩余系数

$$\begin{array}{c}\overrightarrow{E}=-\mathrm{\nabla }\varphi =-\hat{r}\frac{\partial }{\partial r}(\frac{D_0}{r}+\frac{D_1}{r^2}\cos \theta )-\frac{\hat{\theta }}{r}\frac{\partial }{\partial \theta }(\frac{D_0}{r}+\frac{D_1}{r^2}\cos \theta )\\ =\frac{D_0}{r^2}\hat{r}+\frac{2D_1}{r^3}\cos \theta \hat{r}+\frac{D_1}{r^3}\sin \theta \hat{\theta }\end{array}$$

在 $r=R_0$ 处电势连续

$$\begin{array}{c}{A}_0=\frac{D_0}{R_0}\\ A_1{R}_0+\frac{p_t}{4\pi {ϵ}_1{R}_0^2}=\frac{D_1}{R_0^2}\end{array}$$

在$r=R_0$ 处电场强度切向连续

$$A_1+\frac{p_t}{4\pi {ϵ}_1{R}_0^3}=\frac{D_1}{R_0^3}$$

在$r=R_0$ 处电位移矢量法向连续

$$\begin{array}{c}{ϵ}_1{B}_0={ϵ}_2{D}_0=0\\ {ϵ}_1(\frac{p_t}{2\pi {ϵ}_1{R}_0^3}-A_1)={ϵ}_2\frac{2D_1}{R_0^3}\end{array}$$

可以解出

$$\begin{array}{c}{A}_0=D_0=0\\ A_1=\frac{({ϵ}_1-{ϵ}_2)p_t}{2\pi {ϵ}_1({ϵ}_1+2{ϵ}_2)R_0^3}\\ D_1=\frac{3p_t}{4\pi ({ϵ}_1+2{ϵ}_2)}\end{array}$$

所以电势分布为

$$\varphi =\{\begin{array}{ll}\frac{3({\overrightarrow{p}}_t\cdot \overrightarrow{r})}{4\pi ({ϵ}_1+2{ϵ}_2)r^3}& (r>R_0)\\ \frac{{\overrightarrow{p}}_t\cdot \overrightarrow{r}}{4\pi {ϵ}_1{r}^3}+\frac{2({ϵ}_1-{ϵ}_2)({\overrightarrow{p}}_t\cdot \overrightarrow{r})}{4\pi {ϵ}_1({ϵ}_1+2{ϵ}_2)R_0^3}& (r<R_0)\end{array}$$

极化电荷分布

球心处有极化偶极子 $\overrightarrow{p}=(\frac{{\epsilon }_0}{{\epsilon }_1}-1){\overrightarrow{p}}_t$

球面上有极化面电荷 ${\sigma }_p=\frac{3({\epsilon }_1-{\epsilon }_2){\epsilon }_0{\overrightarrow{p}}_t}{2\pi {\epsilon }_1({\epsilon }_1+2{\epsilon }_2)R_0^3}\cos \theta$

2.18 一半径为 $R_0$ 的球面，在球坐标 $0<\theta <\frac{\pi }{2}$ 的半球面上电势为 ${\varphi }_0$，在 $\frac{\pi }{2}<\theta <\pi$ 的半球面上电势为 $-{\varphi }_0$，求空间各点电势。

$$\varphi (r,\theta )=\sum _n(A_n{r}^n+B_n\frac{1}{r^{n+1}})P_n(\cos \theta )$$

分为 $r<R_0$ 以及 $r>R_0$ 两块空间求解

对于 $r<R_0$

由于电势在球心处不发散，所以 $B_n=0$

$$\varphi (r,\theta )=\sum _n{A}_n{r}^n{P}_n(\cos \theta )$$

利用勒让德多项式的正交性

$$\begin{gathered} {\int }_0^{\pi }\varphi (R_0,\theta )P_n(\cos \theta )\sin \theta d\theta \\ ={\int }_{-1}^0-{\varphi }_0{P}_n(x)dx+{\int }_0^1{\varphi }_0{P}_n(x)dx=(-1{)}^{n+1}{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1+{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1 \\ =\{\begin{array}{ll}0& (n\text{是偶数})\\ 2{\varphi }_0\frac{P_{n-1}(0)-P_{n+1}(0)}{2n+1}& (n\text{是奇数})\end{array} \end{gathered}$$

所以

$${\int }_{-1}^1{A}_n{R}_0^n{P}_n^2(x)dx=\{\begin{array}{ll}0& (n\text{是偶数})\\ 2{\varphi }_0\frac{P_{n-1}(0)-P_{n+1}(0)}{2n+1}& (n\text{是奇数})\end{array}$$

即 $n$ 是奇数时

$$\begin{array}{c}{A}_n=\frac{{\varphi }_0}{R_0^n}(P_{n-1}(0)-P_{n+1}(0))\\ =(-1{)}^{\frac{n-1}{2}}\frac{{\varphi }_0}{R_0^n}\frac{(n-2)!!}{(n+1)!!}(2n+1)\end{array}$$

所以

$$\varphi (r,\theta )=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(4n+3)\frac{(2n-1)!!}{(2n+2)!!}{\varphi }_0{\left(\frac{r}{R_0}\right)}^{2n+1}{P}_{2n+1}(\cos \theta )$$

对于 $r>R_0$

由于电势在无穷远处不发散，所以 $A_n=0$

$$\varphi (r,\theta )=\sum _n\frac{B_n}{r^{n+1}}{P}_n(\cos \theta )$$

利用勒让德多项式的正交性

$$\begin{gathered} {\int }_0^{\pi }\varphi (R_0,\theta )P_n(\cos \theta )\sin \theta d\theta \\ ={\int }_{-1}^0-{\varphi }_0{P}_n(x)dx+{\int }_0^1{\varphi }_0{P}_n(x)dx=(-1{)}^{n+1}{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1+{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1 \\ =\{\begin{array}{ll}0& (n\text{是偶数})\\ 2{\varphi }_0\frac{P_{n-1}(0)-P_{n+1}(0)}{2n+1}& (n\text{是奇数})\end{array} \end{gathered}$$

所以

$${\int }_{-1}^1\frac{B_n}{R_0^{n+1}}{P}_n^2(x)dx=\{\begin{array}{ll}0& (n\text{是偶数})\\ 2{\varphi }_0\frac{P_{n-1}(0)-P_{n+1}(0)}{2n+1}& (n\text{是奇数})\end{array}$$

即 $n$ 是偶数时

$$\begin{array}{c}{B}_n={\varphi }_0{R}_0^{n+1}(P_{n-1}(0)-P_{n+1}(0))\\ =(-1{)}^{\frac{n-1}{2}}{\varphi }_0{R}_0^{n+1}\frac{(n-2)!!}{(n+1)!!}(2n+1)\end{array}$$

所以

$$\varphi (r,\theta )=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(4n+3)\frac{(2n-1)!!}{(2n+2)!!}{\varphi }_0{\left(\frac{R_0}{r}\right)}^{2n+2}{P}_{2n+1}(\cos \theta )$$

所以电势分布为

$$\varphi (r,\theta )=\{\begin{array}{l}\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(4n+3)\frac{(2n-1)!!}{(2n+2)!!}{\varphi }_0{\left(\frac{r}{R_0}\right)}^{2n+1}{P}_{2n+1}(\cos \theta )(r<R_0)\\ \sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(4n+3)\frac{(2n-1)!!}{(2n+2)!!}{\varphi }_0{\left(\frac{R_0}{r}\right)}^{2n+2}{P}_{2n+1}(\cos \theta )(r>R_0)\end{array}$$

提示：

$${\int }_0^1{P}_n(x)dx={\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1$$$$P_n(1)=1$$$$P_n(0)=\{\begin{array}{ll}0& (n=\mathrm{奇}\mathrm{数})\\ (-1{)}^{\frac{n}{2}}\cdot \frac{1\cdot 3\cdot 5\cdots (n-1)}{2\cdot 4\cdot 6\cdots n}& (n=\mathrm{偶}\mathrm{数})\end{array}$$

6. 格林函数

书2.19

2.19 上题能用格林函数方法求解吗？结果如何？

可以使用格林函数求解

写出对于球形边界满足第一类边界条件的格林函数

$$G(\overrightarrow{r},\overrightarrow{r^{\prime }})=\frac{1}{4\pi {ϵ}_0}(\frac{1}{|\overrightarrow{r}-\overrightarrow{r^{\prime }}|}-\frac{\frac{R_0}{r^{\prime }}}{|\overrightarrow{r}-\frac{R_0^2}{r^{\prime 2}}\overrightarrow{r^{\prime }}|})$$

所以电势分布为

$$\varphi (\overrightarrow{r})=-{ϵ}_0{\oint }_S\varphi (R_0,\alpha )d{S}^{\prime }\frac{\partial G(\overrightarrow{r},\overrightarrow{r^{\prime }})}{\partial n^{\prime }}$$$$\begin{array}{c}\frac{\partial G(\overrightarrow{r},\overrightarrow{r^{\prime }})}{\partial n^{\prime }}=\widehat{r^{\prime }}\cdot {\mathrm{\nabla }}^{\prime }\frac{1}{4\pi {ϵ}_0}(\frac{1}{|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }|}-\frac{\frac{R_0}{r^{\prime }}}{|\overrightarrow{r}-\frac{R_0^2}{r^{\prime 2}}{\overrightarrow{r}}^{\prime }|})\\ =\widehat{r^{\prime }}\cdot \frac{1}{4\pi {ϵ}_0}(\frac{\overrightarrow{r}-\overrightarrow{r^{\prime }}}{|\overrightarrow{r}-\overrightarrow{r^{\prime }}{|}^3}+\frac{\frac{R_0\overrightarrow{r^{\prime }}}{r^{\prime 3}}}{|\overrightarrow{r}-\frac{R_0^2}{r^{\prime 2}}{\overrightarrow{r}}^{\prime }|}+\frac{R_0}{r^{\prime }}\frac{\overrightarrow{r}-\frac{R_0^2}{r^{\prime 2}}{\overrightarrow{r}}^{\prime }}{|\overrightarrow{r}-\frac{R_0^2}{r^{\prime 2}}{\overrightarrow{r}}^{\prime }{|}^3}\cdot (\frac{2R_0^2}{r^{\prime 4}}\overrightarrow{r^{\prime }}\overrightarrow{r^{\prime }}-\frac{R_0^2}{r^{\prime 2}}\overleftrightarrow{I}))\end{array}$$

在 $r^{\prime }=R_0$ 处

$${\frac{\partial G(\overrightarrow{r},\overrightarrow{r^{\prime }})}{\partial n^{\prime }}|}_{r^{\prime }=R_0}=-\frac{1}{4\pi {ϵ}_0{R}_0}\frac{|r^2-R_0^2|}{|\overrightarrow{r}-\overrightarrow{r^{\prime }}{|}^3}$$

所以

$$\varphi (\overrightarrow{r})={\oint }_S\varphi (R_0,\alpha )\frac{1}{4\pi R_0}\frac{|r^2-R_0^2|}{|\overrightarrow{r}-\overrightarrow{r^{\prime }}{|}^3}d{S}^{\prime }$$

展开被积函数为勒让德多项式

$$\frac{1}{\sqrt{1-2xt+t^2}}=\sum _n{t}^n{P}_n(x)$$

两侧对 $t$ 偏导，所以

$$\frac{x-t}{(1-2xt+t^2{)}^{\frac{3}{2}}}=\sum _{n}n{t}^{n-1}{P}_n(x)$$$$\begin{array}{c}\frac{1-t^2}{(1-2xt+t^2{)}^{\frac{3}{2}}}=\frac{1-2xt+t^2+2xt-2t^2}{(1-2xt+t^2{)}^{\frac{3}{2}}}=\frac{1}{\sqrt{1-2xt+t^2}}+\frac{2t(x-t)}{(1-2xt+t^2{)}^{\frac{3}{2}}}\\ =\sum _n(1+2n)t^n{P}_n(x)\end{array}$$

带入 $x=\cos \gamma$ 以及 $t=\frac{r_{<}}{r_{>}}$ （$\gamma$ 为 $\overrightarrow{r}$ 与 $\overrightarrow{r^{\prime }}$ 的夹角，$\alpha$ 为积分点和 $z$ 轴的夹角，$\theta$ 为场点和$z$ 轴的夹角）

所以在球内（取 $\alpha$ 和 $\varphi$ 为积分用的球坐标，其中$\varphi$ 的起始位置为 $\overrightarrow{r^{\prime }}$ 的投影位置）

$$\begin{array}{c}\varphi (r,\theta )=\frac{1}{4\pi R_0^2}{\oint }_S\varphi (R_0,\alpha )\sum _n(1+2n)(\frac{r}{R_0}{)}^n{P}_n(\cos \gamma )d{S}^{\prime }\\ =\sum _n\frac{1+2n}{4\pi }{\int }_0^{\pi }{\int }_0^{2\pi }(\frac{r}{R_0}{)}^n\varphi (R_0,\alpha )P_n(\sin \theta \sin \alpha \cos \varphi +\cos \theta \cos \alpha )\sin \alpha d\alpha d\varphi \\ =\sum _n\frac{1+2n}{4\pi }{\int }_0^{\pi }{\int }_0^{2\pi }(\frac{r}{R_0}{)}^n\varphi (R_0,\alpha )P_n(\cos \theta )P_n(\cos \alpha )\sin \alpha d\alpha d\varphi \\ =\sum _n\frac{1+2n}{2}{\int }_0^{\pi }(\frac{r}{R_0}{)}^n\varphi (R_0,\alpha )P_n(\cos \theta )P_n(\cos \alpha )\sin \alpha d\alpha \\ =\sum _n\frac{1+2n}{2}(\frac{r}{R_0}{)}^n{P}_n(\cos \theta ){\int }_{-1}^1\varphi (R_0,\alpha )P_n(\cos \alpha )\sin \alpha d(\cos \alpha )\\ =\sum _n\frac{1+2n}{2}(\frac{r}{R_0}{)}^n{P}_n(\cos \theta )({\int }_{-1}^0-{\varphi }_0{P}_n(x)dx+{\int }_0^1{\varphi }_0{P}_n(x)dx)\\ =\sum _n\frac{1+2n}{2}(\frac{r}{R_0}{)}^n{P}_n(\cos \theta )((-1{)}^{n+1}{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1+{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1)\\ =\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(4n+3)\frac{(2n-1)!!}{(2n+2)!!}{\varphi }_0{\left(\frac{r}{R_0}\right)}^{2n+1}{P}_{2n+1}(\cos \theta )\end{array}$$

同理，在球外

$$\begin{array}{c}\varphi (r,\theta )=\frac{1}{4\pi R_0^2}{\oint }_S\varphi (R_0,\alpha )\sum _n(1+2n)(\frac{R_0}{r}{)}^{n+1}{P}_n(\cos \gamma )d{S}^{\prime }\\ =\sum _n\frac{1+2n}{4\pi }{\int }_0^{\pi }{\int }_0^{2\pi }(\frac{R_0}{r}{)}^{n+1}\varphi (R_0,\alpha )P_n(\sin \theta \sin \alpha \cos \varphi +\cos \theta \cos \alpha )\sin \alpha d\alpha d\varphi \\ =\sum _n\frac{1+2n}{4\pi }{\int }_0^{\pi }{\int }_0^{2\pi }(\frac{R_0}{r}{)}^{n+1}\varphi (R_0,\alpha )P_n(\cos \theta )P_n(\cos \alpha )\sin \alpha d\alpha d\varphi \\ =\sum _n\frac{1+2n}{2}{\int }_0^{\pi }(\frac{R_0}{r}{)}^{n+1}\varphi (R_0,\alpha )P_n(\cos \theta )P_n(\cos \alpha )\sin \alpha d\alpha \\ =\sum _n\frac{1+2n}{2}(\frac{R_0}{r}{)}^{n+1}{P}_n(\cos \theta ){\int }_{-1}^1\varphi (R_0,\alpha )P_n(\cos \alpha )\sin \alpha d(\cos \alpha )\\ =\sum _n\frac{1+2n}{2}(\frac{R_0}{r}{)}^{n+1}{P}_n(\cos \theta )({\int }_{-1}^0-{\varphi }_0{P}_n(x)dx+{\int }_0^1{\varphi }_0{P}_n(x)dx)\\ =\sum _n\frac{1+2n}{2}(\frac{R_0}{r}{)}^{n+1}{P}_n(\cos \theta )((-1{)}^{n+1}{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1+{\varphi }_0{\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}|}_0^1)\\ =\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(4n+3)\frac{(2n-1)!!}{(2n+2)!!}{\varphi }_0{\left(\frac{R_0}{r}\right)}^{2n+2}{P}_{2n+1}(\cos \theta )\end{array}$$