电流

\[ \begin{gathered} I=\frac{\mathrm dq}{\mathrm dt}=\frac{n(\Delta S\cdot v\mathrm dt)e}{\mathrm dt}=ne(\Delta S)v\\ j=\frac{I}{\Delta S}=nev\\ \end{gathered} \]

\[ \begin{aligned} j\mathrm d S=\mathrm dI&=-\frac{\mathrm d U}{\mathrm dR}\\ &=-\frac{\mathrm dU}{\rho\frac{\mathrm d l}{\mathrm d S}}\\ &=(-\frac{\mathrm dU}{\mathrm dl})\cdot\frac{1}{\rho}\cdot\mathrm dS\\ &=\frac{E}{\rho}\mathrm dS \end{aligned} \]

\[ j = \frac{E}{\rho}=\sigma E \quad (\sigma=\frac{1}{\rho}) \]

\[ \begin{aligned} Q&=I^2R\Delta t\\ \mathrm{d}Q&=(j\mathrm{d}S)^2(\rho\frac{\mathrm{d}l}{\mathrm{d}S})\mathrm{d}t\\ &=j^2\rho(\mathrm{d}S\mathrm{d}l)\mathrm{d}t\\ &=(\sigma^2 E^2)\frac{1}{\sigma}\mathrm{d}V\mathrm{d}t\\ &=\sigma E^2\mathrm{d}V\mathrm{d}t \end{aligned} \]

\[ w = \frac{\mathrm{d}Q}{\mathrm{d}V\mathrm{d}t}=\sigma E^2 \]

\[ \begin{aligned} j&=ne\bar v=\sum_{i=1}^nev_i=\sum_{i=1}^nv_{0i}+\sum_{i=1}^n\frac{e^2E}{m}t_i\\ &=\sum_{i=1}^n\frac{e^2E}{m}t_i\text{ (初始速度任意)} \end{aligned} \]

\[ j = \frac{ne^2\tau}{m}E=\sigma E\Rightarrow \sigma=\frac{ne^2\tau}{m} \]

\[ \tau = \frac{\lambda}{\bar v}=\lambda\sqrt{\frac{\pi m}{8kT}}\propto\frac{1}{\sqrt T} \]

\[ \begin{aligned}\mathrm{d} \epsilon-iR-\frac{q}{C}&=0\\ R\mathrm{d}q&=\left(\epsilon-\frac{q}{C}\right)\mathrm{d}t\\ \frac{\mathrm{d}q}{C\epsilon-q}&=\frac{\mathrm dt}{CR}\\ \ln|C\epsilon-q|&=-\frac{t}{CR}+C\\ C\epsilon-q&=Ae^{-\frac{t}{CR}}\\ \text{当 }t=0&,q=0\Rightarrow A=C\epsilon\\ q&=C\epsilon(1-e^{-\frac{t}{CR}})\\ \text{令 }\tau=RC\text{，则 }q(t)&=C\epsilon(1-e^{-\frac{t}{\tau}})\\ i(t)&=\frac{\epsilon}{R}e^{-\frac{t}{\tau}}\\ \Delta V_C(t)&=\frac{q(t)}{C}=\epsilon(1-e^{-\frac{t}{\tau}})\ \end{aligned} \]

\[ \begin{aligned} \frac{q}{C}&=R\frac{\mathrm dq}{\mathrm dt}\\ \ln q&=\frac{t}{\tau}+C\\ q&=Ae^{\frac{t}{\tau}}\\ \text{当 }t=0&,q=C\epsilon\Rightarrow A=C\epsilon\\ q(t)&=C\epsilon e^{\frac{t}{\tau}}\\ i(t)&=\frac{\epsilon}{R}e^{-\frac{t}{\tau}}\\ \Delta V_C(t)&=\frac{q(t)}{C}=\epsilon(1-e^{-\frac{t}{\tau}})\ \end{aligned} \]