# 电势能

（重定向自電勢能

${\displaystyle U=W}$

1. 其它電荷所產生的電勢。
2. 這點電荷Q的電荷量。

## 儲存於點電荷系統內的電勢能

### 雙點電荷系統

${\displaystyle \mathbf {F} _{c}={\frac {q_{1}q_{2}}{4\pi \epsilon _{0}}}\ {\frac {\hat {\mathbf {r} }}{r^{2}}}}$

${\displaystyle W=-\int _{\mathbb {L} }\mathbf {F} _{c}\cdot \mathrm {d} {\boldsymbol {\ell }}=-\ {\frac {q_{1}q_{2}}{4\pi \epsilon _{0}}}\int _{\mathbb {L} }{\frac {\hat {\mathbf {r} }}{r^{2}}}\cdot \mathrm {d} {\boldsymbol {\ell }}}$

${\displaystyle W=-\ {\frac {q_{1}q_{2}}{4\pi \epsilon _{0}}}\int _{\infty }^{r}{\frac {\mathrm {d} r}{r^{2}}}={\frac {q_{1}q_{2}}{4\pi \epsilon _{0}r}}}$

${\displaystyle W=U(\mathbf {r} )-U(\infty )}$

${\displaystyle U(\mathbf {r} )={\frac {q_{1}q_{2}}{4\pi \epsilon _{0}r}}}$

${\displaystyle U={\frac {1}{4\pi \epsilon _{0}}}\ {\frac {q_{1}q_{2}}{|\mathbf {r} _{2}-\mathbf {r} _{1}|}}={\frac {1}{4\pi \epsilon _{0}}}\ {\frac {q_{1}q_{2}}{r_{12}}}}$

### 三個以上點電荷的系統

${\displaystyle U={\frac {1}{4\pi \epsilon _{0}}}\left({\frac {q_{1}q_{2}}{r_{12}}}+{\frac {q_{1}q_{3}}{r_{13}}}+{\frac {q_{2}q_{3}}{r_{23}}}\right)}$

${\displaystyle W_{i}={\frac {1}{4\pi \epsilon _{0}}}\sum _{j=1}^{i-1}{\frac {q_{i}q_{j}}{r_{ij}}}}$

${\displaystyle U=W=\sum _{i=1}^{n}W_{i}={\frac {1}{4\pi \epsilon _{0}}}\sum _{i=1}^{n}\sum _{j=1}^{i-1}{\frac {q_{i}q_{j}}{r_{ij}}}}$

${\displaystyle U={\frac {1}{8\pi \epsilon _{0}}}\sum _{i=1}^{n}\sum _{j=1,j\neq i}^{n}{\frac {q_{i}q_{j}}{r_{ij}}}}$

${\displaystyle \phi (\mathbf {r} _{i})={\frac {1}{4\pi \epsilon _{0}}}\sum _{j=1,j\neq i}^{n}{\frac {q_{j}}{r_{ij}}}}$

${\displaystyle U={\frac {1}{2}}\sum _{i=1}^{n}q_{i}\phi (\mathbf {r} _{i})}$
• 上述方程式假設電介質是自由空間，其電容率${\displaystyle \epsilon _{0}}$ ，即電常數。假設電介質不是自由空間，而是電容率為 ${\displaystyle \epsilon }$ 的某種電介質，則必需將方程式內的 ${\displaystyle \epsilon _{0}}$ 更換為 ${\displaystyle \epsilon }$

## 儲存於連續電荷分佈的能量

${\displaystyle U={\frac {1}{2}}\int _{\mathbb {V} }\rho (\mathbf {r} )\phi (\mathbf {r} )\ \mathrm {d} ^{3}r}$

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

{\displaystyle {\begin{aligned}U&={\frac {\epsilon _{0}}{2}}\int _{\mathbb {V} }[\mathbf {\nabla } \cdot \mathbf {E} (\mathbf {r} )]\phi (\mathbf {r} )\ \mathrm {d} ^{3}r\\&={\frac {\epsilon _{0}}{2}}\int _{\mathbb {V} }\mathbf {\nabla } \cdot [\mathbf {E} (\mathbf {r} )\phi (\mathbf {r} )]-\mathbf {E} (\mathbf {r} )\cdot \mathbf {\nabla } \phi (\mathbf {r} )\ \mathrm {d} ^{3}r\\\end{aligned}}}

${\displaystyle U={\frac {\epsilon _{0}}{2}}\oint _{\mathbb {S} }[\mathbf {E} (\mathbf {r} )\phi (\mathbf {r} )]\cdot \mathrm {d} ^{2}r-{\frac {\epsilon _{0}}{2}}\int _{\mathbb {V} }\mathbf {E} (\mathbf {r} )\cdot \mathbf {\nabla } \phi (\mathbf {r} )\ \mathrm {d} ^{3}r}$

${\displaystyle U=-{\frac {\epsilon _{0}}{2}}\int _{\mathbb {ALL\ SPACE} }\mathbf {E} (\mathbf {r} )\cdot \mathbf {\nabla } \phi (\mathbf {r} )\mathrm {d} ^{3}r}$

${\displaystyle \mathbf {E} =-\nabla \phi }$

${\displaystyle U={\frac {\epsilon _{0}}{2}}\int _{\mathbb {ALL\ SPACE} }[E(\mathbf {r} )]^{2}\mathrm {d} ^{3}r}$

${\displaystyle u(\mathbf {r} )={\frac {\epsilon _{0}}{2}}[E(\mathbf {r} )]^{2}}$

## 自身能與交互作用能

${\displaystyle U={\frac {1}{8\pi \epsilon _{0}}}\sum _{i=1}^{n}\sum _{j=1,j\neq i}^{n}{\frac {q_{i}q_{j}}{r_{ij}}}}$
${\displaystyle U={\frac {\epsilon _{0}}{2}}\int _{\mathbb {ALL\ SPACE} }[E(\mathbf {r} )]^{2}\mathrm {d} ^{3}r}$

${\displaystyle \mathbf {E} =\mathbf {E} _{1}+\mathbf {E} _{2}={\frac {q_{1}}{4\pi \epsilon _{0}}}\ {\frac {\mathbf {r} -\mathbf {r} _{1}}{|\mathbf {r} -\mathbf {r} _{1}|^{3}}}+{\frac {q_{2}}{4\pi \epsilon _{0}}}\ {\frac {\mathbf {r} -\mathbf {r} _{2}}{|\mathbf {r} -\mathbf {r} _{2}|^{3}}}}$

${\displaystyle u={\frac {\epsilon _{0}}{2}}E^{2}={\frac {\epsilon _{0}}{2}}(E_{1}\,^{2}+E_{2}\,^{2}+2\mathbf {E} _{1}\cdot \mathbf {E} _{2})}$

${\displaystyle U_{int}=\int _{\mathbb {V} }u_{int}\ \mathrm {d} ^{3}r=\epsilon _{0}\int _{\mathbb {V} }\mathbf {E} _{1}\cdot \mathbf {E} _{2}\ \mathrm {d} ^{3}r={\frac {q_{1}q_{2}}{16\pi ^{2}\epsilon _{0}}}\int _{\mathbb {V} }{\frac {\mathbf {r} -\mathbf {r} _{1}}{|\mathbf {r} -\mathbf {r} _{1}|^{3}}}\ \cdot \ {\frac {\mathbf {r} -\mathbf {r} _{2}}{|\mathbf {r} -\mathbf {r} _{2}|^{3}}}\ \mathrm {d} ^{3}r}$

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

{\displaystyle {\begin{aligned}U_{int}&={\frac {q_{1}q_{2}}{16\pi ^{2}\epsilon _{0}}}\int _{\mathbb {V} }\nabla \left({\frac {1}{|\mathbf {r} -\mathbf {r} _{1}|}}\right)\ \cdot \ \nabla \left({\frac {1}{|\mathbf {r} -\mathbf {r} _{2}|}}\right)\mathrm {d} ^{3}r\\&={\frac {q_{1}q_{2}}{16\pi ^{2}\epsilon _{0}}}\int _{\mathbb {V} }\nabla \ \cdot \ \left[{\frac {1}{|\mathbf {r} -\mathbf {r} _{1}|}}\nabla \left({\frac {1}{|\mathbf {r} -\mathbf {r} _{2}|}}\right)\right]-\ \left({\frac {1}{|\mathbf {r} -\mathbf {r} _{1}|}}\right)\nabla ^{2}\left({\frac {1}{|\mathbf {r} -\mathbf {r} _{2}|}}\right)\mathrm {d} ^{3}r\\\end{aligned}}}

${\displaystyle \int _{\mathbb {V} }\nabla \ \cdot \ \left[{\frac {1}{|\mathbf {r} -\mathbf {r} _{1}|}}\nabla \left({\frac {1}{|\mathbf {r} -\mathbf {r} _{2}|}}\right)\right]\mathrm {d} ^{3}r=\oint _{\mathbb {S} }\left[{\frac {1}{|\mathbf {r} -\mathbf {r} _{1}|}}\nabla \left({\frac {1}{|\mathbf {r} -\mathbf {r} _{2}|}}\right)\right]\cdot \mathrm {d} ^{2}r}$

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

${\displaystyle U_{int}={\frac {q_{1}q_{2}}{4\pi \epsilon _{0}}}\int _{\mathbb {ALL\ SPACE} }{\frac {\delta (\mathbf {r} -\mathbf {r} _{2})}{|\mathbf {r} -\mathbf {r} _{1}|}}\ \mathrm {d} ^{3}r={\frac {1}{4\pi \epsilon _{0}}}\ {\frac {q_{1}q_{2}}{|\mathbf {r} _{1}-\mathbf {r} _{2}|}}}$

## 參考文獻

1. ^ Halliday, David; Resnick, Robert; Walker, Jearl. Electric Potential. Fundamentals of Physics 5th. John Wiley & Sons. 1997. ISBN 0-471-10559-7.
2. Jackson, John David, Classical Electrodynamic 3rd., USA: John Wiley & Sons, Inc.: pp. 40–43, 1999, ISBN 978-0-471-30932-1