# 电偶极矩

$\mathbf{p} = q \, \mathbf{d}$

## 簡單電偶極子案例

$\mathbf{p}(\mathbf{r}) = \int_{\mathbb{V}'} \rho(\mathbf{r}')\, (\mathbf{r}'-\mathbf{r}) \ d^3 \mathbf{r}'$

$\rho (\mathbf{r}') = \sum_{i=1}^N \, q_i \delta (\mathbf{r}' - \mathbf{r}_i ')$

$\mathbf{p}(\mathbf{r}) = \sum_{i=1}^N \, q_i \int_{\mathbb{V}'}\delta (\mathbf{r}' - \mathbf{r}_i' )\, (\mathbf{r}'-\mathbf{r}) \ d^3 \mathbf{r}'= \sum_{i=1}^N \, q_i (\mathbf{r}_i'-\mathbf{r})$

$\mathbf{p}(\mathbf{r})=q(\mathbf{r}_+'-\mathbf{r})-q(\mathbf{r}_-' -\mathbf{r})=q (\mathbf{r}_+' - \mathbf{r}_-')=q\mathbf{d}$

$\mathbf{p}(\mathbf{r}) =\sum_{i=1}^{N} \mathbf{p}_i$

## 電偶極子產生的電勢與電場

$\phi(\mathbf{r})= \frac{q}{4 \pi \varepsilon _0 r_+} - \frac{q}{4 \pi \varepsilon _0 r_-}$

\begin{align} \frac {1}{r_{\pm}} & = \left(r^2+\frac{d^2}{4} \mp rd\cos{\theta}\right)^{-1/2} = \frac{1}{r}\left( 1+\frac{d^2}{4r^2} \mp \frac{d\cos{\theta}}{r}\right)^{-1/2} \\ & \approx \frac {1}{r}\left(1\pm\frac{d\cos{\theta}}{2r}\right) \\ \end{align}

$\phi(\mathbf{r}) \approx \frac{qd\cos{\theta}}{4 \pi \varepsilon _0 r^2}$

$\mathbf{p}=q\mathbf{r}_+ - q\mathbf{r}_- =q\mathbf{d}$

$\phi(\mathbf{r})= \frac{1}{4 \pi \varepsilon _0 }\ \frac{\mathbf{p}\cdot \hat{\mathbf{r}}}{r^2}$

$E_r= - \ \frac{\partial \phi(\mathbf{r})}{\partial r}= \frac{p\cos{\theta}}{2 \pi \varepsilon _0 r^3}$
$E_{\theta}= - \ \frac{1}{r}\ \frac{\partial \phi(\mathbf{r})}{\partial \theta}= \frac{p\sin{\theta}}{4 \pi \varepsilon _0 r^3}$
$E_{\varphi}= - \ \frac{1}{r\sin{\theta}} \frac{\partial \phi(\mathbf{r})}{\partial \varphi}=0$

$\mathbf{E}= \frac{p(2\cos{\theta}\ \hat{\mathbf{r}}+\sin{\theta}\ \hat{\boldsymbol{\theta}})}{4 \pi \varepsilon _0 r^3} =\frac{3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}}{4 \pi \varepsilon _0 r^3}$

\begin{align}\mathbf{E} = - \nabla \Phi & =\frac {1} {4\pi\epsilon_0 r^3} \left(3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{p}\right) - \frac{\mathbf{p}}{3\epsilon_0}\delta^3(\mathbf{r}) \\ & = \frac{p}{4 \pi \epsilon_0 r^3} ( 2 \cos \theta \hat{\mathbf{r}} + \sin \theta \hat{\boldsymbol{\theta}} )- \frac{\mathbf{p}}{3\epsilon_0}\delta^3(\mathbf{r}) \end{align}

## 電偶極矩密度與電極化強度

$\mathbf{p}=\int_{\mathbb{V}'}\boldsymbol{\mathfrak{p}}(\mathbf{r}')\ d^3\mathbf{r}'$

$\phi(\mathbf{r})= \frac{1}{4 \pi \varepsilon _0 }\int_{\mathbb{V}'} \frac{\boldsymbol{\mathfrak{p}}(\mathbf{r}')\cdot (\mathbf{r}- \mathbf{r}')}{|\mathbf{r}- \mathbf{r}'|^3}\ d^3\mathbf{r}'$

## 介電質內部的自由電荷與束縛電荷

$\rho_{bound} = -\nabla\cdot\mathbf{P}$

$\rho_{total} = \rho_{free} + \rho_{bound}$

$\sigma_{bound}= \mathbf{P}\cdot\hat{\mathbf{ n}}_\mathrm{out}$

$\nabla\cdot\mathbf{E} = \rho_{total}/\epsilon_0$

$\nabla\cdot\mathbf{P} = - \rho_{bound}$

$\mathbf{D}\ \stackrel{\mathrm{def}}{=}\ \epsilon_{0} \mathbf{E} + \mathbf{P}$

$\nabla\cdot\mathbf{D}=\rho_{free}$

### 介電質產生的電勢

$\phi(\mathbf{r}) = \frac {1}{4 \pi \varepsilon_0}\int_{\mathbb{V}'}\left[ \frac{\rho_{free} (\mathbf{r}' )}{|\mathbf{r}- \mathbf{r}'|} +\frac{\boldsymbol{\mathfrak{p}}(\mathbf{r}' )\cdot(\mathbf{r}- \mathbf{r}')}{|\mathbf{r}- \mathbf{r}'|^3} +\sum_{i,j=1}^3\frac{ \mathfrak{Q}_{ij}(\mathbf{r}')(x_i - x_i') (x_j - x_j')}{2|\mathbf{r}- \mathbf{r}'|^5}\dots \right]\ d^3 \mathbf{ r}'$

$\phi(\mathbf{r}) = \frac {1}{4 \pi \varepsilon_0} \int_{\mathbb{V}'}\left[ \frac{\rho_{free}(\mathbf{r}' )}{|\mathbf{r}- \mathbf{r}'|} +\frac{\boldsymbol{\mathfrak{p}}(\mathbf{r}' )\cdot(\mathbf{r}- \mathbf{r}')}{|\mathbf{r}- \mathbf{r}'|^3} \right]\ d^3 \mathbf{ r}'$

$\nabla'\left(\frac {1}{|\mathbf{r} - \mathbf{r}'|}\right)= \frac {\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3}$

\begin{align} \int_{\mathbb{V}'}\frac{\boldsymbol{\mathfrak{p}}(\mathbf{r}' )\cdot(\mathbf{r}- \mathbf{r}')}{|\mathbf{r}- \mathbf{r}'|^3}\ d^3 \mathbf{ r}' & =\int_{\mathbb{V}'}\boldsymbol{\mathfrak{p}}(\mathbf{r}')\cdot\nabla'\left(\frac {1}{|\mathbf{r} - \mathbf{r}'|}\right) \ d^3 \mathbf{ r}' \\ & =\int_{\mathbb{V}'}\nabla'\cdot\left(\frac {\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\right)\ d^3 \mathbf{ r}' - \int_{\mathbb{V}'}\frac {\nabla'\cdot\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\ d^3 \mathbf{ r}' \\ \end{align}

$\int_{\mathbb{V}'}\nabla'\cdot\left(\frac {\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\right)\ d^3 \mathbf{ r}' =\oint_{\mathbb{S}'}\left(\frac {\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\right)\cdot\ d\mathbf{a}'$

$\phi(\mathbf{r}) = \frac {1}{4 \pi \varepsilon_0} \int_{\mathbb{V}'}\left[ \frac{\rho_{free}(\mathbf{r}' )}{|\mathbf{r}- \mathbf{r}'|} - \ \frac{\nabla'\cdot\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\right] \ d^3 \mathbf{ r}'$

$\phi(\mathbf{r}) = \frac {1}{4 \pi \varepsilon_0} \int_{\mathbb{V}'} \frac{\rho_{total}(\mathbf{r}' )}{|\mathbf{r}- \mathbf{r}'|} \ d^3 \mathbf{ r}'$

$\rho_{total}=\rho_{free}+\nabla\cdot\boldsymbol{\mathfrak{p}}(\mathbf{r})$

$\rho_{bound}= - \nabla\cdot\boldsymbol{\mathfrak{p}}$

$\nabla\cdot\mathbf{P}= - \rho_{bound}$

### 面束縛電荷密度

\begin{align} \phi(\mathbf{r}) & = \frac {1}{4 \pi \varepsilon_0} \int_{\mathbb{V}'}\frac{\boldsymbol{\mathfrak{p}}(\mathbf{r}' )\cdot(\mathbf{r}- \mathbf{r}')}{|\mathbf{r}- \mathbf{r}'|^3}\ d^3 \mathbf{ r}' \\ & =\frac {1}{4 \pi \varepsilon_0}\oint_{\mathbb{S}'}\left(\frac {\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\right)\cdot\ d\mathbf{a}' - \frac {1}{4 \pi \varepsilon_0}\int_{\mathbb{V}'}\frac {\nabla'\cdot\boldsymbol{\mathfrak{p}}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\ d^3 \mathbf{ r}' \\ \end{align}

$\sigma_{bound}=\boldsymbol{\mathfrak{p}}\cdot\hat{\mathbf{n}}$

$\phi(\mathbf{r})=\frac{1}{4 \pi \varepsilon_0} \oint_{\mathbb{S}'}\frac{\sigma_{bound}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\ da' +\frac {1}{4 \pi \varepsilon_0}\int_{\mathbb{V}'}\frac{\rho_{bound}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\ d^3 \mathbf{ r}'$

### 範例：處於均勻外電場的介電質球

$\phi(r,\theta)=\sum_{l=0}^{\infty} (A_l\ r^l +B_l\ r^{-(l+1)}) P_l(\cos{\theta})$

$\phi_{in}(r,\theta)=\sum_{l=0}^{\infty} A_l\ r^l P_l(\cos{\theta})$

$\phi_{out}(r,\theta)= - E_{\infty} r\cos{\theta}+ \sum_{l=0}^{\infty} B_l r^{- (l+1)} P_l(\cos{\theta})$

$\phi_{in}(R,\theta)=\phi_{out}(R,\theta)$
$\epsilon_r\left.\frac{\partial\phi_{in}(r,\theta)}{\partial r}\right|_{r=R} =\left.\frac{\partial\phi_{out}(r,\theta)}{\partial r}\right|_{r=R}$

$A_1 R= - E_{\infty} R+B_1R^{-2}$
$A_l R^l=B_l R^{-(l+1)}, \qquad\qquad l\ne 1$

$\epsilon_rA_1 = - E_{\infty} - 2B_1R^{-3}$
$\epsilon_r l A_l R^{(l-1)}=- (l+1)B_l R^{-(l+2)}, \qquad\qquad l\ne 1$

$A_1 = - \ \frac{3E_{\infty}}{\epsilon_r+2}$
$B_1 =\frac{(\epsilon_r - 1)R^3 E_{\infty}}{\epsilon_r+2}$

$A_l =B_l =0, \qquad\qquad l\ne 1$

$\phi_{out}(r,\theta)= - E_{\infty} r\cos{\theta}+ \frac{(\epsilon_r - 1)R^3 E_{\infty} \cos{\theta}}{(\epsilon_r+2)r^2}$

$\phi_{in}(r,\theta)= - \frac{3}{\epsilon_r +2} E_{\infty} r \cos{\theta}$

$\mathbf{E}_{in}= - \nabla \phi_{in}(r,\theta)=\frac{3}{\epsilon_r +2} \mathbf{E}_{\infty}=\left(1 - \ \frac{\epsilon_r - 1}{\epsilon_r +2}\right) \mathbf{E}_{\infty}$

$\mathbf{E}_{p}= \mathbf{E}_{in}- \mathbf{E}_{\infty}= - \ \left(\frac{\epsilon_r - 1}{\epsilon_r +2}\right) \mathbf{E}_{\infty}= - \frac{\boldsymbol{\mathfrak{p}}}{3\epsilon_0}$

$\boldsymbol{\mathfrak{p}}=\epsilon_0(\epsilon_r - 1)\mathbf{E}_{in}$

$\rho_{bound}=-\nabla\cdot\boldsymbol{\mathfrak{p}}=0$

$\sigma_{bound} = {3}\varepsilon_0\frac {\epsilon_r -1}{\epsilon_r +2} E_{\infty} \cos{\theta} =\boldsymbol{\mathfrak{p}}\cdot \hat{\mathbf{r}}$

## 註釋

1. ^ 粒子物理學裏，有三種重要的離散對稱性：電荷共軛對稱性是粒子與其反粒子的對稱性，又稱「正反共軛對稱性」。宇稱對稱性是關於粒子位置 $\mathbf{r}$$- \mathbf{r}$ 的對稱性，時間反演對稱性是時間 $t$$- t$ 的對稱性。
2. ^ 時間反演變換將 $t$ 改變為 $- t$ 。一個載流迴圈的磁偶極矩 $\boldsymbol{\mu}$ 是其所載電流 $I$ 乘於迴圈面積 $\mathbf{a}$ ，以方程式表示為 $\boldsymbol{\mu}=I\mathbf{a}=\frac{\mathrm{d} q}{\mathrm{d} t}\mathbf{a}$ 。注意到電流是電荷量對於時間的導數，所以，時間反演會逆反磁偶極矩的方向。電偶磁矩的兩個參數，電荷量和位移向量都跟時間反演無關，所以，時間反演不會改變電偶極矩的方向。
3. ^ 空間反演（宇稱）變換是粒子位置坐標對於參考系原點的反射。電偶極矩是極向量polar vector），而磁偶極矩是軸向量axial vector），所以，空間反演（宇稱）會逆反電偶極矩的方向，不會改變磁偶極矩的方向。

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