# 電勢

## 靜電學裏的電勢

${\displaystyle \phi (\mathbf {r} )\ {\stackrel {def}{=}}\ U_{\mathrm {E} }(\mathbf {r} )/q}$

${\displaystyle U_{\mathrm {E} }(\mathbf {r} )\ {\stackrel {def}{=}}\ W}$

${\displaystyle W=\int _{\mathbb {L} }\mathbf {F} \cdot \mathrm {d} {\boldsymbol {\ell }}=-q\int _{\mathbb {L} }\mathbf {E} \cdot \mathrm {d} {\boldsymbol {\ell }}}$

${\displaystyle U_{\mathrm {E} }(\mathbf {r} )=-q\int _{\infty }^{\mathbf {r} }\mathbf {E} \cdot \mathrm {d} {\boldsymbol {\ell }}}$

${\displaystyle \phi (\mathbf {r} )=-\int _{\infty }^{\mathbf {r} }\mathbf {E} \cdot \mathrm {d} {\boldsymbol {\ell }}}$

${\displaystyle \Delta \phi =\phi (\mathbf {r} _{2})-\phi (\mathbf {r} _{1})=-\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\mathbf {E} \cdot \mathrm {d} {\boldsymbol {\ell }}}$

### 點電荷

${\displaystyle V={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q}{r}}}$

### 疊加原理

${\displaystyle \phi _{t}(\mathbf {r} )=-\int _{\infty }^{\mathbf {r} }\mathbf {E} _{t}\cdot \mathrm {d} {\boldsymbol {\ell }}=-\int _{\infty }^{\mathbf {r} }(\mathbf {E} _{1}+\mathbf {E} _{2})\cdot \mathrm {d} {\boldsymbol {\ell }}=\phi _{1}(\mathbf {r} )+\phi _{2}(\mathbf {r} )}$

### 電勢的微分方程式

${\displaystyle \mathbf {\nabla } \phi (\mathbf {r} )=-\mathbf {\nabla } \int _{\infty }^{\mathbf {r} }\mathbf {E} (\mathbf {r} ')\cdot \mathrm {d} {\boldsymbol {\ell }}^{\,\prime }=-\mathbf {E} (\mathbf {r} )}$

${\displaystyle \mathbf {E} (\mathbf {r} )=-\mathbf {\nabla } \phi (\mathbf {r} )}$

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

${\displaystyle \nabla ^{2}\phi =-\rho /\epsilon _{0}}$

${\displaystyle \nabla ^{2}\phi =0}$

## 拉普拉斯方程式的解答

### 邊界條件

1. 狄利克雷邊界條件：在所有邊界，電勢都已良態給定。具有這種邊界條件的問題稱為狄利克雷問題
2. 紐曼邊界條件：在所有邊界，電勢的法向導數都已良態給定。具有這種邊界條件的問題稱為紐曼問題
3. 混合邊界條件：一部分邊界的電勢都已良態給定，其它邊界的電勢的法向導數也已良態給定。

### 兩個半平面導體案例

${\displaystyle \nabla ^{2}\phi ={\frac {\partial ^{2}\phi }{\partial x^{2}}}+{\frac {\partial ^{2}\phi }{\partial y^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}=0}$

${\displaystyle \nabla ^{2}\phi (y,z)={\frac {\partial ^{2}\phi }{\partial y^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}=0}$

${\displaystyle \phi (y,z)=Y(y)Z(z)}$

${\displaystyle {\frac {1}{Y(y)}}\ {\frac {\mathrm {d} ^{2}Y(y)}{\mathrm {d} y^{2}}}+{\frac {1}{Z(z)}}\ {\frac {\mathrm {d} ^{2}Z(z)}{\mathrm {d} z^{2}}}=0}$

${\displaystyle {\frac {1}{Y(y)}}\ {\frac {\mathrm {d} ^{2}Y(y)}{\mathrm {d} y^{2}}}=C}$
${\displaystyle {\frac {1}{Z(z)}}\ {\frac {\mathrm {d} ^{2}Z(z)}{\mathrm {d} z^{2}}}=-C}$

${\displaystyle Y(y)=A_{1}e^{iky}+A_{2}e^{-iky}}$
${\displaystyle Z(z)=B_{1}e^{kz}+B_{2}e^{-kz}}$

${\displaystyle z}$趨向於無窮大時，${\displaystyle Z(z)}$趨向於零，所以，${\displaystyle B_{1}=0}$。綜合起來，電勢為

${\displaystyle \phi (y,z)=\int _{0}^{\infty }(A_{1}e^{iky}+A_{2}e^{-iky})e^{-kz}\mathrm {d} k}$

${\displaystyle y>0}$時，${\displaystyle \int _{0}^{\infty }(A_{1}e^{iky}+A_{2}e^{-iky})\mathrm {d} k=+V}$
${\displaystyle y<0}$時，${\displaystyle \int _{0}^{\infty }(A_{1}e^{iky}+A_{2}e^{-iky})\mathrm {d} k=-V}$

${\displaystyle A_{1}(k)={\frac {V}{2\pi }}\left(\int _{0}^{\infty }e^{-iky'}\mathrm {d} y'-\int _{-\infty }^{0}e^{-iky'}\mathrm {d} y'\right)}$
${\displaystyle A_{2}(k)={\frac {V}{2\pi }}\left(\int _{0}^{\infty }e^{iky'}\mathrm {d} y'-\int _{-\infty }^{0}e^{iky'}\mathrm {d} y'\right)}$

{\displaystyle {\begin{aligned}\phi _{1}&={\frac {V}{2\pi }}\int _{0}^{\infty }\mathrm {d} k\left\{\int _{0}^{\infty }e^{ik(y-y')-kz}\mathrm {d} y'-\int _{-\infty }^{0}e^{ik(y-y')-kz}\mathrm {d} y'\right\}\\&=-\ {\frac {V}{2\pi }}\int _{0}^{\infty }{\frac {\mathrm {d} y'}{i(y-y')-z}}+\ {\frac {V}{2\pi }}\int _{-\infty }^{0}{\frac {\mathrm {d} y'}{i(y-y')-z}}\\\end{aligned}}}

{\displaystyle {\begin{aligned}\phi _{2}&={\frac {V}{2\pi }}\int _{0}^{\infty }\mathrm {d} k\left\{\int _{0}^{\infty }e^{-ik(y-y')-kz}\mathrm {d} y'-\int _{-\infty }^{0}e^{-ik(y-y')-kz}\mathrm {d} y'\right\}\\&=-\ {\frac {V}{2\pi }}\int _{0}^{\infty }{\frac {\mathrm {d} y'}{-i(y-y')-z}}+\ {\frac {V}{2\pi }}\int _{-\infty }^{0}{\frac {\mathrm {d} y'}{-i(y-y')-z}}\\\end{aligned}}}

{\displaystyle {\begin{aligned}\phi &={\frac {Vz}{\pi }}\int _{0}^{\infty }{\frac {\mathrm {d} y'}{(y-y')^{2}+z^{2}}}-\ {\frac {Vz}{\pi }}\int _{-\infty }^{0}{\frac {\mathrm {d} y'}{(y-y')^{2}+z^{2}}}\\&={\frac {2V}{\pi }}\ \arctan {\left({\frac {y}{z}}\right)}\\\end{aligned}}}

## 帕松方程式的解答

### 電荷分佈所產生的電勢

${\displaystyle \mathbf {E} (\mathbf {r} )={\frac {q}{4\pi \epsilon _{0}}}\ {\frac {(\mathbf {r} -\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|^{3}}}}$

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

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

${\displaystyle \mathbf {E} (\mathbf {r} )=-\ {\frac {1}{4\pi \epsilon _{0}}}\nabla \int _{\mathbb {V} '}{\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\ \mathrm {d} ^{3}r'}$

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

${\displaystyle \nabla ^{2}\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right)=-4\pi \delta (\mathbf {r} -\mathbf {r} ')}$

${\displaystyle \nabla ^{2}\phi (\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}}}\int _{\mathbb {V} '}\nabla ^{2}\left({\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\right)\ \mathrm {d} ^{3}r'=-\ {\frac {1}{\epsilon _{0}}}\int _{\mathbb {V} '}\rho (\mathbf {r} ')\delta (\mathbf {r} -\mathbf {r} ')\ \mathrm {d} ^{3}r'=-\ {\frac {\rho (\mathbf {r} )}{\epsilon _{0}}}}$

### 邊界條件

${\displaystyle \int _{\mathbb {V} }\left(\phi \nabla ^{2}\psi -\psi \nabla ^{2}\phi \right)\ \mathrm {d} ^{3}r=\oint _{\mathbb {S} }\left(\phi {\partial \psi \over \partial n}-\psi {\partial \phi \over \partial n}\right)\ \mathrm {d} ^{2}r}$

${\displaystyle \int _{\mathbb {V} '}\left[\phi (\mathbf {r} ')\nabla ^{2}\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right)+{\frac {\rho (\mathbf {r} ')}{\epsilon _{0}|\mathbf {r} -\mathbf {r} '|}}\right]\mathrm {d} ^{3}r'=\oint _{\mathbb {S} '}\left[\phi \ {\partial \over \partial n'}\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right)-\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right){\partial \phi \over \partial n'}\right]\mathrm {d} ^{2}r'}$

${\displaystyle \nabla ^{2}\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right)=-4\pi \delta (\mathbf {r} -\mathbf {r} ')}$

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}}}\int _{\mathbb {V} '}{\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\ \mathrm {d} ^{3}r'+{\frac {1}{4\pi }}\oint _{\mathbb {S} '}\left[\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right){\partial \phi \over \partial n'}-\phi \ {\partial \over \partial n'}\left({\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\right)\right]\mathrm {d} ^{2}r'}$

### 格林函數

${\displaystyle \nabla ^{2}G(\mathbf {r} ,\mathbf {r} ')=-4\pi \delta (\mathbf {r} -\mathbf {r} ')}$

${\displaystyle \nabla ^{2}H(\mathbf {r} ,\mathbf {r} ')=0}$

• 對於狄利克雷問題，當源位置${\displaystyle \mathbf {r} '}$在邊界表面${\displaystyle {\mathbb {S} '}}$時，規定格林函數${\displaystyle G_{D}(\mathbf {r} ,\mathbf {r} ')=0}$。這樣，從格林第二恆等式，設定${\displaystyle \phi (\mathbf {r} ')}$為在${\displaystyle \mathbf {r} '}$的電勢，${\displaystyle \psi (\mathbf {r} ,\mathbf {r} ')=G_{D}(\mathbf {r} ,\mathbf {r} ')}$，則可得到
${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}}}\int _{\mathbb {V} '}\rho (\mathbf {r} ')G_{D}(\mathbf {r} ,\mathbf {r} ')\ \mathrm {d} ^{3}r'-\ {\frac {1}{4\pi }}\oint _{\mathbb {S} '}\phi (\mathbf {r} ')\ {\partial G_{D}(\mathbf {r} ,\mathbf {r} ') \over \partial n'}\mathrm {d} ^{2}r'}$（2）
• 對於滿足紐曼問題，當源位置${\displaystyle \mathbf {r} '}$在邊界表面${\displaystyle {\mathbb {S} '}}$時，規定格林函數${\displaystyle \oint _{\mathbb {S} '}{\frac {\partial G_{D}(\mathbf {r} ,\mathbf {r} ')}{\partial n'}}\mathrm {d} ^{2}r'=-{\frac {4\pi }{S}}}$

### 無限平面導體案例

${\displaystyle {\begin{matrix}G_{D}(\mathbf {r} ,\mathbf {r} ')={\cfrac {1}{\sqrt {(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}}}\\\qquad \qquad \qquad -\ {\cfrac {1}{\sqrt {(x-x')^{2}+(y-y')^{2}+(z+z')^{2}}}}\\\end{matrix}}}$

${\displaystyle \phi (\mathbf {r} )={\frac {1}{4\pi \epsilon _{0}}}\int _{\mathbb {V} '}\rho (\mathbf {r} ')G_{D}(\mathbf {r} ,\mathbf {r} ')\ \mathrm {d} ^{3}r'}$

{\displaystyle {\begin{aligned}\phi (\mathbf {r} )&={\frac {1}{4\pi \epsilon _{0}}}\int _{\mathbb {V} '}\rho (\mathbf {r} ')\left({\frac {1}{\sqrt {x^{2}+y^{2}+(z-a)^{2}}}}-{\frac {1}{\sqrt {x^{2}+y^{2}+(z+a)^{2}}}}\right)\ \mathrm {d} ^{3}r'\\&={\frac {1}{4\pi \epsilon _{0}}}\left({\frac {q}{\sqrt {x^{2}+y^{2}+(z-a)^{2}}}}-{\frac {q}{\sqrt {x^{2}+y^{2}+(z+a)^{2}}}}\right)\\\end{aligned}}}

#### 導引

${\displaystyle \nabla ^{2}G(\mathbf {r} ,\mathbf {r} ')=-4\pi \delta (\mathbf {r} -\mathbf {r} ')}$

{\displaystyle {\begin{aligned}{\frac {1}{|\mathbf {r} -\mathbf {r} '|}}&\equiv {\frac {1}{2\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\mathrm {d} ^{3}k{\frac {e^{i\mathbf {k} \cdot (\mathbf {r} -\mathbf {r} ')}}{k^{2}}}\\&={\frac {1}{2\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\mathrm {d} k_{x}\ \mathrm {d} k_{y}e^{ik_{x}(x-x')+ik_{y}(y-y')}\int _{-\infty }^{\infty }\ \mathrm {d} k_{z}{\frac {e^{ik_{z}(z-z')}}{k_{x}^{2}+k_{y}^{2}+k_{z}^{2}}}\\\end{aligned}}}

${\displaystyle \nabla ^{2}H(\mathbf {r} ,\mathbf {r} ')=0}$

{\displaystyle {\begin{aligned}H(\mathbf {r} ,\mathbf {r} ')&={\frac {1}{2\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\ \mathrm {d} k_{x}\ \mathrm {d} k_{y}e^{ik_{x}(x-x')+ik_{y}(y-y')}\int _{-\infty }^{\infty }\ \mathrm {d} k_{z}\left[B(\mathbf {k} ,z')e^{ik_{z}z}+C(\mathbf {k} ,z')e^{-ik_{z}z}\right]\\\end{aligned}}}

${\displaystyle G_{D}(\mathbf {r} ,\mathbf {r} ')={\frac {1}{2\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\ \mathrm {d} k_{x}\ \mathrm {d} k_{y}e^{ik_{x}(x-x')+ik_{y}(y-y')}\int _{-\infty }^{\infty }\ \mathrm {d} k_{z}\left[{\frac {e^{ik_{z}(z-z')}}{k_{x}^{2}+k_{y}^{2}+k_{z}^{2}}}+B(\mathbf {k} ,z')e^{ik_{z}z}+C(\mathbf {k} ,z')e^{-ik_{z}z}\right]}$

${\displaystyle {\frac {e^{-ik_{z}z'}}{k_{x}^{2}+k_{y}^{2}+k_{z}^{2}}}+B(\mathbf {k} ,z')+C\mathbf {k} ,z')=0}$
${\displaystyle B(\mathbf {k} ,z')={\frac {B_{0}e^{-ik_{z}z'}}{k_{x}^{2}+k_{y}^{2}+k_{z}^{2}}}}$
${\displaystyle C(\mathbf {k} ,z')={\frac {C_{0}e^{-ik_{z}z'}}{k_{x}^{2}+k_{y}^{2}+k_{z}^{2}}}}$

${\displaystyle G_{D}(\mathbf {r} ,\mathbf {r} ')={\frac {1}{2\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\ \mathrm {d} k_{x}\ \mathrm {d} k_{y}e^{ik_{x}(x-x')+ik_{y}(y-y')}\int _{-\infty }^{\infty }\ \mathrm {d} k_{z}\left\{{\frac {(1+B_{0})}{k^{2}}}\left[e^{ik_{z}(z-z')}-e^{ik_{z}(z+z')}\right]\right\}}$

{\displaystyle {\begin{aligned}G_{D}(\mathbf {r} ,\mathbf {r} ')&={\frac {1}{2\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\ \mathrm {d} k_{x}\ \mathrm {d} k_{y}e^{ik_{x}(x-x')+ik_{y}(y-y')}\int _{-\infty }^{\infty }\ \mathrm {d} k_{z}\left\{{\frac {1}{k^{2}}}\left[e^{ik_{z}(z-z')}-e^{ik_{z}(z+z')}\right]\right\}\\&={\frac {1}{|\mathbf {r} -\mathbf {r} '|}}-{\frac {1}{|\mathbf {r} -\mathbf {r} ''|}}\\&={\cfrac {1}{\sqrt {(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}}}-\ {\cfrac {1}{\sqrt {(x-x')^{2}+(y-y')^{2}+(z+z')^{2}}}}\\\end{aligned}}}

### 兩個半平面導體案例

${\displaystyle \phi (\mathbf {r} )=-\ {\frac {1}{4\pi }}\oint _{\mathbb {S} '}\phi (\mathbf {r} ')\ {\partial G_{D}(\mathbf {r} ,\mathbf {r} ') \over \partial n'}\mathrm {d} ^{2}r'}$（3）

{\displaystyle {\begin{aligned}{\partial G_{D} \over \partial n'}&=-\ {\partial G_{D} \over \partial z'}\\&=-\ {\cfrac {z-z'}{[(x-x')^{2}+(y-y')^{2}+(z-z')^{2}]^{3/2}}}\ -\ {\cfrac {z+z'}{[(x-x')^{2}+(y-y')^{2}+(z+z')^{2}]^{3/2}}}\\&=-\ {\cfrac {2z}{[(x-x')^{2}+(y-y')^{2}+z^{2}]^{3/2}}}\\\end{aligned}}}

${\displaystyle \mathbb {V} '}$的邊界閉曲面在無窮遠位置的電勢為0，所以，只需要計算xy-平面給出的貢獻，就可以得到在${\displaystyle \mathbb {V} '}$內部任意位置的電勢。將上述方程式代入方程式（3）：[4]

{\displaystyle {\begin{aligned}\phi (\mathbf {r} )&={\frac {2z}{4\pi }}\left\{\int _{0+}^{\infty }\int _{-\infty }^{\infty }{\cfrac {V\mathrm {d} x'\mathrm {d} y'}{[(x-x')^{2}+(y-y')^{2}+z^{2}]^{3/2}}}+\int _{-\infty }^{0-}\int _{-\infty }^{\infty }{\cfrac {-V\mathrm {d} x'\mathrm {d} y'}{[(x-x')^{2}+(y-y')^{2}+z^{2}]^{3/2}}}\right\}\\&=\ {\frac {zV}{\pi }}\left\{\int _{0+}^{\infty }{\frac {\mathrm {d} y'}{(y-y')^{2}+z^{2}}}-\int _{-\infty }^{0-}{\frac {\mathrm {d} y'}{(y-y')^{2}+z^{2}}}\right\}\\&={\frac {2V}{\pi }}\ \arctan {\left({\frac {y}{z}}\right)}\\\end{aligned}}}

## 推廣至電動力學

${\displaystyle \mathbf {B} \ {\stackrel {def}{=}}\ \mathbf {\nabla } \times \mathbf {A} }$

${\displaystyle \nabla \cdot \mathbf {F} (\mathbf {r} )=D(\mathbf {r} )}$
${\displaystyle \nabla \times \mathbf {F} (\mathbf {r} )=\mathbf {C} (\mathbf {r} )}$ ;

${\displaystyle \mathbf {F} (\mathbf {r} )=-\nabla \left({\frac {1}{4\pi }}\int _{\mathbb {V} '}{\frac {D(\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\mathrm {d} ^{3}r'\right)+\nabla \times \left({\frac {1}{4\pi }}\int _{\mathbb {V} '}{\frac {\mathbf {C} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\mathrm {d} ^{3}r'\right)}$ ;

${\displaystyle \mathbf {\nabla } \cdot \mathbf {A} =0}$

${\displaystyle \mathbf {A} (\mathbf {r} )=\nabla \times \left({\frac {1}{4\pi }}\int _{\mathbb {V} '}{\frac {\mathbf {B} (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\mathrm {d} ^{3}r'\right)={\frac {1}{4\pi }}\int _{\mathbb {V} '}\mathbf {B} (\mathbf {r} ')\times {\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}\mathrm {d} ^{3}r'}$

${\displaystyle \mathbf {A} (\mathbf {r} ,\,t)={\frac {1}{4\pi }}\int _{\mathbb {V} '}\mathbf {B} (\mathbf {r} ',\,t)\times {\frac {\mathbf {r} -\mathbf {r} '}{|\mathbf {r} -\mathbf {r} '|^{3}}}\mathrm {d} ^{3}r'}$

${\displaystyle \nabla \times \mathbf {G} =\nabla \times \mathbf {E} +\nabla \times \partial \mathbf {A} /\partial t=0}$

${\displaystyle \mathbf {E} =-\mathbf {\nabla } \phi -{\frac {\partial \mathbf {A} }{\partial t}}}$

${\displaystyle \int _{a}^{b}\mathbf {E} \cdot \mathrm {d} {\boldsymbol {\ell }}\neq \phi (b)-\phi (a)}$

## 參考文獻

1. Halliday, David; Robert Resnick, Jearl Walker, Fundamental of Physics 7th, USA: John Wiley and Sons, Inc.: pp. 630ff, 2005, ISBN 0-471-23231-9
2. ^ 電勢. 中華語文知識庫. [2016-03-03]. （原始內容存檔於2016年3月6日） （中文（中國大陸）‎）.
3. ^ Jackson 1999，第70-72頁
4. Beyer, William, CRC Standard Mathematical Table 28th, CRC Press, 1987, ISBN 0-8493-0628-0 pp. 241, formula #43, pp. 252, formula#165
5. Jackson 1999，第35-40頁
6. ^ Jackson 1999，第127-129頁
7. ^ Griffiths, David J. Introduction to Electrodynamics 3rd. Prentice Hall. 1998: pp. 555–557. ISBN 0-13-805326-X.
• Jackson, John David, Classical Electrodynamic 3rd., USA: John Wiley & Sons, Inc., 1999, ISBN 978-0-471-30932-1
• Wangsness, Roald K. Electromagnetic Fields 2nd., Revised, illustrated. Wiley. 1986. ISBN 9780471811862.