電容

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

1法拉等於1庫侖伏特，即電容為1法拉的電容器，在正常操作範圍內，每增加1伏特的電勢差可以多儲存1庫侖的電荷。

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

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

電容器

$\sigma= Q/A$

$E=\sigma/\varepsilon= Q/\varepsilon A$

$V=Ed=\sigma d/\varepsilon= Q d/\varepsilon A$

$C=Q/V= \varepsilon A/d$

電壓依賴性電容器

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

$E= V/d$

$P=f(E)=f(V/d)=g(V)$

$D=P+\epsilon_0 E=g(V)+\epsilon_0 V/d$

$Q=DA=(g(V)+\epsilon_0 V/d)A$

$C(V)=\frac{Q}{V}=\frac{g(V)A}{V}+\frac{\epsilon_0A}{d}$

$C=\frac{Q}{V}=kA+\frac{\epsilon_0A}{d}$

$Q =\int_0^V C(V') \ \mathrm{d}V'$

$\mathrm{d}U_\text{stored} =Q \mathrm{d}V'' =\left[ \int_0^{V''}\ C(V') \ \mathrm{d}V' \right] \mathrm{d}V''$

$\int_a^z f(x) g'(x)\ \mathrm{d}x = \left[ f(x) g(x) \right]_a^z - \int_a^z f'(x) g(x)\ \mathrm{d}x$

$\int_a^z \int_a^x \ h(y) \ \mathrm{d}y\ \mathrm{d}x=\left[\int_a^x \ xh(y)\ \mathrm{d}y \right]_a^z - \int_a^z xh(x) \ \mathrm{d}x = \int_a^z zh(y) \ \mathrm{d}y - \int_a^z yh(y) \ \mathrm{d}y= \int_a^z \ \left(z-y\right) h(y)\ \mathrm{d}y$

$U_\text{stored}=\int_0^{V}\ \left[ \int_0^{V''}\ C(V') \ \mathrm{d}V' \right] \mathrm{d}V'' =\int_0^V \ \left(V- V'\right) C(V')\ \mathrm{d}V'$

頻率依賴性電容器

$\boldsymbol D (t) = \frac{\varepsilon_0}{\sqrt{2\pi}} \int_{-\infty}^t \mathrm{d}t' \ \varepsilon_r (t-t') \boldsymbol E (t')$

$\boldsymbol D (t) = \frac{\varepsilon_0}{\sqrt{2\pi}} \int_{-\infty}^\infty \mathrm{d}t' \ \varepsilon_r (t-t') \boldsymbol E (t')$

$\boldsymbol D(\omega) = \varepsilon_0 \varepsilon_r(\omega)\boldsymbol E (\omega)$

$\varepsilon_r(\omega)$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：

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

\begin{align}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{align}

$\varepsilon_r(\omega) = \varepsilon_{r}'(\omega) - j \varepsilon_{r}''(\omega) = \frac{1}{j\omega Z(\omega) C_0} = \frac{C(\omega)}{C_0}$

電容矩陣

$V_1 = P_{11}Q_1 + P_{12}Q_2 + P_{13}Q_3$
$V_2 = P_{21}Q_1 + P_{22}Q_2 + P_{23}Q_3$
$V_3 = P_{31}Q_1 + P_{32}Q_2 + P_{33}Q_3$

$Q_1 = C_{11}V_1 + C_{12}V_2 + C_{13}V_3$
$Q_2 = C_{21}V_1 + C_{22}V_2 + C_{23}V_3$
$Q_3 = C_{31}V_1 + C_{32}V_2 + C_{33}V_3$

$V_i=\sum_{j=1}^n P_{ij}Q_j,\qquad\qquad i=1,2,\dots,n$
$Q_i=\sum_{j=1}^n C_{ij}V_j,\qquad\qquad i=1,2,\dots,n$

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

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

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

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

$V_i = -P_{ii}Q + P_{ij}Q$
$V_j = -P_{ji}Q + P_{jj}Q$

$C=Q/(V_j - V_i)=1/(P_{ii}+P_{jj} - 2P_{ij})$

自電容

$V=Q/4\pi\varepsilon_0 R$

$C=Q/V=4\pi\varepsilon_0R$

範例

$C=4\pi\varepsilon_0R=4\pi\times 8.85\times 10^{-12}\times 0.2\approx 22[pF]$

$C=4\pi\times 8.85\times 10^{-12}\times 6.378\times10^{6}\approx 700[\mu F]$

簡單系統的電容

$=2\pi \varepsilon r\left\{ 1+\frac{1}{2D}+\frac{1}{4D^{2}}+\frac{1}{8D^{3}}+\frac{1}{8D^{4}}+\frac{3}{32D^{5}}+O\left( \frac{1}{D^{6}}\right) \right\}$
$=2\pi \varepsilon r\left\{ \ln 2+\gamma -\frac{1}{2}\ln \left( 2D-2\right) +O\left( 2D-2\right) \right\}$
$d>2r$
$D = d/2r > 1$
$\gamma$歐拉-馬歇羅尼常數

$D = d/r$

參考文獻

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