# 電容

（重定向自电容

${\displaystyle C={\frac {Q}{V}}}$

${\displaystyle \mathrm {d} W={\frac {q}{C}}\,\mathrm {d} q}$

${\displaystyle W_{\text{charging}}=\int _{0}^{Q}{\frac {q}{C}}\,\mathrm {d} q={\frac {Q^{2}}{2C}}={\frac {1}{2}}CV^{2}=U_{\text{stored}}}$

## 单位

${\displaystyle 1F=10^{6}\mu F=10^{9}nF=10^{12}pF=10^{15}fF.}$  [2]

## 電容器

${\displaystyle \sigma =Q/A}$

${\displaystyle E=\sigma /\varepsilon =Q/\varepsilon A}$

${\displaystyle V=Ed=\sigma d/\varepsilon =Qd/\varepsilon A}$

${\displaystyle C=Q/V=\varepsilon A/d}$

### 電壓依賴性電容器

${\displaystyle \mathrm {d} Q=C(V)\ \mathrm {d} V}$

${\displaystyle E=V/d}$

${\displaystyle P=f(E)=f(V/d)=g(V)}$

${\displaystyle D=P+\epsilon _{0}E=g(V)+\epsilon _{0}V/d}$

${\displaystyle Q=DA=(g(V)+\epsilon _{0}V/d)A}$

${\displaystyle C(V)={\frac {Q}{V}}={\frac {g(V)A}{V}}+{\frac {\epsilon _{0}A}{d}}}$

${\displaystyle C={\frac {Q}{V}}=kA+{\frac {\epsilon _{0}A}{d}}}$

${\displaystyle Q=\int _{0}^{V}C(V')\ \mathrm {d} V'}$

${\displaystyle \mathrm {d} U_{\text{stored}}=Q\mathrm {d} V''=\left[\int _{0}^{V''}\ C(V')\ \mathrm {d} V'\right]\mathrm {d} V''}$

${\displaystyle \int _{a}^{z}f(x)g'(x)\ \mathrm {d} x=\left[f(x)g(x)\right]_{a}^{z}-\int _{a}^{z}f'(x)g(x)\ \mathrm {d} x}$

${\displaystyle \int _{a}^{z}\int _{a}^{x}\ h(y)\ \mathrm {d} y\ \mathrm {d} x=\left[\int _{a}^{x}\ xh(y)\ \mathrm {d} y\right]_{a}^{z}-\int _{a}^{z}xh(x)\ \mathrm {d} x=\int _{a}^{z}zh(y)\ \mathrm {d} y-\int _{a}^{z}yh(y)\ \mathrm {d} y=\int _{a}^{z}\ \left(z-y\right)h(y)\ \mathrm {d} y}$

${\displaystyle U_{\text{stored}}=\int _{0}^{V}\ \left[\int _{0}^{V''}\ C(V')\ \mathrm {d} V'\right]\mathrm {d} V''=\int _{0}^{V}\ \left(V-V'\right)C(V')\ \mathrm {d} V'}$

### 頻率依賴性電容器

${\displaystyle {\boldsymbol {D}}(t)={\frac {\varepsilon _{0}}{\sqrt {2\pi }}}\int _{-\infty }^{t}\mathrm {d} t'\ \varepsilon _{r}(t-t'){\boldsymbol {E}}(t')}$

${\displaystyle {\boldsymbol {D}}(t)={\frac {\varepsilon _{0}}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\mathrm {d} t'\ \varepsilon _{r}(t-t'){\boldsymbol {E}}(t')}$

${\displaystyle {\boldsymbol {D}}(\omega )=\varepsilon _{0}\varepsilon _{r}(\omega ){\boldsymbol {E}}(\omega )}$

${\displaystyle \varepsilon _{r}(\omega )}$ 複函數，其虛值部分與介質的電場能量吸收有關。更詳盡細節，請參閱條目電容率。由於電容與電容率成正比，電容也具有這頻率行為。對於時間做傅立葉變換於高斯定律：

${\displaystyle Q(\omega )=\oint _{\mathbb {S} }\mathbf {D} (\mathbf {r} ,\omega )\cdot \mathrm {d} \mathbf {a} }$

{\displaystyle {\begin{aligned}I(\omega )&=j\omega Q(\omega )=j\omega \oint _{\mathbb {S} }\mathbf {D} (\mathbf {r} ,\omega )\cdot \mathrm {d} \mathbf {a} \\&=\left[G(\omega )+j\omega C(\omega )\right]V(\omega )={\frac {V(\omega )}{Z(\omega )}}\\\end{aligned}}}

${\displaystyle \varepsilon _{r}(\omega )=\varepsilon _{r}'(\omega )-j\varepsilon _{r}''(\omega )={\frac {1}{j\omega Z(\omega )C_{0}}}={\frac {C(\omega )}{C_{0}}}}$

## 電容矩陣

${\displaystyle V_{1}=P_{11}Q_{1}+P_{12}Q_{2}+P_{13}Q_{3}}$
${\displaystyle V_{2}=P_{21}Q_{1}+P_{22}Q_{2}+P_{23}Q_{3}}$
${\displaystyle V_{3}=P_{31}Q_{1}+P_{32}Q_{2}+P_{33}Q_{3}}$

${\displaystyle Q_{1}=C_{11}V_{1}+C_{12}V_{2}+C_{13}V_{3}}$
${\displaystyle Q_{2}=C_{21}V_{1}+C_{22}V_{2}+C_{23}V_{3}}$
${\displaystyle Q_{3}=C_{31}V_{1}+C_{32}V_{2}+C_{33}V_{3}}$

${\displaystyle V_{i}=\sum _{j=1}^{n}P_{ij}Q_{j},\qquad \qquad i=1,2,\dots ,n}$
${\displaystyle Q_{i}=\sum _{j=1}^{n}C_{ij}V_{j},\qquad \qquad i=1,2,\dots ,n}$

${\displaystyle P_{ij}\ {\stackrel {def}{=}}\ {\frac {\partial V_{i}}{\partial Q_{j}}}}$

${\displaystyle C_{ij}\ {\stackrel {def}{=}}\ {\frac {\partial Q_{i}}{\partial V_{j}}}}$

${\displaystyle P_{ij}=P_{ji}}$
${\displaystyle C_{ij}=C_{ji}}$

${\displaystyle C\ {\stackrel {def}{=}}\ Q/\Delta V}$

${\displaystyle V_{i}=-P_{ii}Q+P_{ij}Q}$
${\displaystyle V_{j}=-P_{ji}Q+P_{jj}Q}$

${\displaystyle C=Q/(V_{j}-V_{i})=1/(P_{ii}+P_{jj}-2P_{ij})}$

## 自電容

${\displaystyle V=Q/4\pi \varepsilon _{0}R}$

${\displaystyle C=Q/V=4\pi \varepsilon _{0}R}$

### 範例

${\displaystyle C=4\pi \varepsilon _{0}R=4\pi \times 8.85\times 10^{-12}\times 0.2\approx 22[pF]}$

${\displaystyle C=4\pi \times 8.85\times 10^{-12}\times 6.378\times 10^{6}\approx 700[\mu F]}$

## 簡單系統的電容

${\displaystyle \varepsilon }$ : 介質的電容率

${\displaystyle \varepsilon }$ : 介質的電容率

${\displaystyle ={\frac {\pi \varepsilon l}{\ \ln \left({\frac {d}{\ 2a\ }}+{\sqrt {{\frac {d^{2}}{\ 4a^{2}\ }}-1\ }}\right)\ }}\approx {\frac {\pi \varepsilon l}{\ln \left({\cfrac {d}{a}}\right)}}\left(1+{\frac {a^{2}}{\ln \left({\cfrac {d}{a}}\right)d^{2}}}\right)\ ,\qquad d\gg a}$

${\displaystyle d}$ : 距離， ${\displaystyle d>a}$
${\displaystyle l}$ : 電線長度

${\displaystyle w_{i}}$ : 導板板寬
${\displaystyle k_{i}=d/(2w_{i}+d)}$
${\displaystyle k^{2}=k_{1}k_{2}}$
${\displaystyle K}$ : 橢圓積分
${\displaystyle l}$ : 長度

${\displaystyle \varepsilon }$ : 介質的電容率

${\displaystyle =2\pi \varepsilon a\left\{1+{\frac {1}{2D}}+{\frac {1}{4D^{2}}}+{\frac {1}{8D^{3}}}+{\frac {1}{8D^{4}}}+{\frac {3}{32D^{5}}}+O\left({\frac {1}{D^{6}}}\right)\right\}}$
${\displaystyle =2\pi \varepsilon a\left\{\ln 2+\gamma -{\frac {1}{2}}\ln \left(2D-2\right)+O\left(2D-2\right)\right\}}$

${\displaystyle a}$ : 半徑
${\displaystyle d}$ : 距離，${\displaystyle d>2a}$
${\displaystyle D=d/2a>1}$
${\displaystyle \gamma }$ 歐拉-馬歇羅尼常數

${\displaystyle d}$ : 距離，${\displaystyle d>a}$
${\displaystyle D=d/a}$

${\displaystyle l}$ : 電線長度
${\displaystyle \Lambda =\ln(l/a)}$

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