# 电偶极矩本文重定向自 电偶极矩

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

## 简单电偶极子案例

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

${\displaystyle \rho (\mathbf {r} ')=\sum _{i=1}^{N}\,q_{i}\delta (\mathbf {r} '-\mathbf {r} _{i}')}$

${\displaystyle \mathbf {p} (\mathbf {r} )=\sum _{i=1}^{N}\,q_{i}\int _{\mathbb {V} '}\delta (\mathbf {r} '-\mathbf {r} _{i}')\,(\mathbf {r} '-\mathbf {r} )\ d^{3}\mathbf {r} '=\sum _{i=1}^{N}\,q_{i}(\mathbf {r} _{i}'-\mathbf {r} )}$

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

${\displaystyle \mathbf {p} (\mathbf {r} )=\sum _{i=1}^{N}\mathbf {p} _{i}}$

## 电偶极子产生的电势与电场

${\displaystyle \phi (\mathbf {r} )={\frac {q}{4\pi \varepsilon _{0}r_{+}}}-{\frac {q}{4\pi \varepsilon _{0}r_{-}}}}$

{\displaystyle {\begin{aligned}{\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{aligned}}}

${\displaystyle \phi (\mathbf {r} )\approx {\frac {qd\cos {\theta }}{4\pi \varepsilon _{0}r^{2}}}}$

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

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\ {\frac {\mathbf {p} \cdot {\hat {\mathbf {r} }}}{r^{2}}}}$

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

${\displaystyle \mathbf {E} ={\frac {p(2\cos {\theta }\ {\hat {\mathbf {r} }}+\sin {\theta }\ {\hat {\boldsymbol {\theta }}})}{4\pi \varepsilon _{0}r^{3}}}={\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{4\pi \varepsilon _{0}r^{3}}}}$

{\displaystyle {\begin{aligned}\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{aligned}}}

## 电偶极矩密度与电极化强度

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

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int _{\mathbb {V} '}{\frac {{\boldsymbol {\mathfrak {p}}}(\mathbf {r} ')\cdot (\mathbf {r} -\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|^{3}}}\ d^{3}\mathbf {r} '}$

## 介电质内部的自由电荷与束缚电荷

${\displaystyle \rho _{bound}=-\nabla \cdot \mathbf {P} }$

${\displaystyle \rho _{total}=\rho _{free}+\rho _{bound}}$

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

${\displaystyle \nabla \cdot \mathbf {E} =\rho _{total}/\epsilon _{0}}$

${\displaystyle \nabla \cdot \mathbf {P} =-\rho _{bound}}$

${\displaystyle \mathbf {D} \ {\stackrel {\mathrm {def} }{=}}\ \epsilon _{0}\mathbf {E} +\mathbf {P} }$

${\displaystyle \nabla \cdot \mathbf {D} =\rho _{free}}$

### 介电质产生的电势

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int _{\mathbb {V} '}\left[{\frac {\rho _{free}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}+{\frac {{\boldsymbol {\mathfrak {p}}}(\mathbf {r} ')\cdot (\mathbf {r} -\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|^{3}}}+\sum _{i,j=1}^{3}{\frac {{\mathfrak {Q}}_{ij}(\mathbf {r} ')(x_{i}-x_{i}')(x_{j}-x_{j}')}{2|\mathbf {r} -\mathbf {r} '|^{5}}}\dots \right]\ d^{3}\mathbf {r} '}$

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int _{\mathbb {V} '}\left[{\frac {\rho _{free}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}+{\frac {{\boldsymbol {\mathfrak {p}}}(\mathbf {r} ')\cdot (\mathbf {r} -\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|^{3}}}\right]\ d^{3}\mathbf {r} '}$

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

{\displaystyle {\begin{aligned}\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{aligned}}}

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

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int _{\mathbb {V} '}\left[{\frac {\rho _{free}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}-\ {\frac {\nabla '\cdot {\boldsymbol {\mathfrak {p}}}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\right]\ d^{3}\mathbf {r} '}$

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int _{\mathbb {V} '}{\frac {\rho _{total}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\ d^{3}\mathbf {r} '}$

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

${\displaystyle \rho _{bound}=-\nabla \cdot {\boldsymbol {\mathfrak {p}}}}$

${\displaystyle \nabla \cdot \mathbf {P} =-\rho _{bound}}$

### 面束缚电荷密度

{\displaystyle {\begin{aligned}\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{aligned}}}

${\displaystyle \sigma _{bound}={\boldsymbol {\mathfrak {p}}}\cdot {\hat {\mathbf {n} }}}$

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\oint _{\mathbb {S} '}{\frac {\sigma _{bound}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\ da'+{\frac {1}{4\pi \varepsilon _{0}}}\int _{\mathbb {V} '}{\frac {\rho _{bound}(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\ d^{3}\mathbf {r} '}$

### 范例：处于均匀外电场的介电质球

${\displaystyle \phi (r,\theta )=\sum _{l=0}^{\infty }(A_{l}\ r^{l}+B_{l}\ r^{-(l+1)})P_{l}(\cos {\theta })}$

${\displaystyle \phi _{in}(r,\theta )=\sum _{l=0}^{\infty }A_{l}\ r^{l}P_{l}(\cos {\theta })}$

${\displaystyle \phi _{out}(r,\theta )=-E_{\infty }r\cos {\theta }+\sum _{l=0}^{\infty }B_{l}r^{-(l+1)}P_{l}(\cos {\theta })}$

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

${\displaystyle A_{1}R=-E_{\infty }R+B_{1}R^{-2}}$
${\displaystyle A_{l}R^{l}=B_{l}R^{-(l+1)},\qquad \qquad l\neq 1}$

${\displaystyle \epsilon _{r}A_{1}=-E_{\infty }-2B_{1}R^{-3}}$
${\displaystyle \epsilon _{r}lA_{l}R^{(l-1)}=-(l+1)B_{l}R^{-(l+2)},\qquad \qquad l\neq 1}$

${\displaystyle A_{1}=-\ {\frac {3E_{\infty }}{\epsilon _{r}+2}}}$
${\displaystyle B_{1}={\frac {(\epsilon _{r}-1)R^{3}E_{\infty }}{\epsilon _{r}+2}}}$

${\displaystyle A_{l}=B_{l}=0,\qquad \qquad l\neq 1}$

${\displaystyle \phi _{out}(r,\theta )=-E_{\infty }r\cos {\theta }+{\frac {(\epsilon _{r}-1)R^{3}E_{\infty }\cos {\theta }}{(\epsilon _{r}+2)r^{2}}}}$

${\displaystyle \phi _{in}(r,\theta )=-{\frac {3}{\epsilon _{r}+2}}E_{\infty }r\cos {\theta }}$

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

${\displaystyle \mathbf {E} _{p}=\mathbf {E} _{in}-\mathbf {E} _{\infty }=-\ \left({\frac {\epsilon _{r}-1}{\epsilon _{r}+2}}\right)\mathbf {E} _{\infty }=-{\frac {\boldsymbol {\mathfrak {p}}}{3\epsilon _{0}}}}$

${\displaystyle {\boldsymbol {\mathfrak {p}}}=\epsilon _{0}(\epsilon _{r}-1)\mathbf {E} _{in}}$

${\displaystyle \rho _{bound}=-\nabla \cdot {\boldsymbol {\mathfrak {p}}}=0}$

${\displaystyle \sigma _{bound}={3}\varepsilon _{0}{\frac {\epsilon _{r}-1}{\epsilon _{r}+2}}E_{\infty }\cos {\theta }={\boldsymbol {\mathfrak {p}}}\cdot {\hat {\mathbf {r} }}}$

## 注释

1. ^ 粒子物理学里，有三种重要的离散对称性：电荷共轭对称性是粒子与其反粒子的对称性，又称“正***轭对称性”。宇称对称性是关于粒子位置 ${\displaystyle \mathbf {r} }$${\displaystyle -\mathbf {r} }$ 的对称性，时间反演对称性是时间 ${\displaystyle t}$${\displaystyle -t}$ 的对称性。
2. ^ 时间反演变换将 ${\displaystyle t}$ 改变为 ${\displaystyle -t}$ 。一个载流循环的磁偶极矩 ${\displaystyle {\boldsymbol {\mu }}}$ 是其所载电流 ${\displaystyle I}$ 乘于循环面积 ${\displaystyle \mathbf {a} }$ ，以方程表示为 ${\displaystyle {\boldsymbol {\mu }}=I\mathbf {a} ={\frac {\mathrm {d} q}{\mathrm {d} t}}\mathbf {a} }$ 。注意到电流是电荷量对于时间的导数，所以，时间反演会逆反磁偶极矩的方向。电偶磁矩的两个参数，电荷量和位移矢量都跟时间反演无关，所以，时间反演不会改变电偶极矩的方向。
3. ^ 空间反演（宇称）变换是粒子位置坐标对于参考系原点的反射。电偶极矩是极矢量（polar vector），而磁偶极矩是轴矢量axial vector），所以，空间反演（宇称）会逆反电偶极矩的方向，不会改变磁偶极矩的方向。