學院物理/連續電荷分布與高斯定律

高斯定律

$\mathrm{\Phi }=\overrightarrow{E}\cdot \overrightarrow{A}=EA\cos \theta =\int \overrightarrow{E}\cdot d\overrightarrow{A}=\frac{kq}{r^2}\cdot 4\pi r^2=\frac{q4\pi r^2}{4\pi ϵr^2}=\frac{q}{ϵ}$

封閉面的總電通量 =

封閉面內的淨電荷
$ϵ$

例題

一

$\overrightarrow{E}\leftarrow$

$\begin{array}{c}\leftarrow x\to \\ \\ al\end{array}$

圖23.2

$\begin{array}{cc}E& =\int dE=k\int \frac{dq}{r^2}=k\lambda {\int }_a^{l+a}\frac{dx}{x^2}=k\lambda {[-\frac{1}{x}]}_a^{l+a}\\ & dq=\lambda dx\\ & =k\frac{Q}{l}(-\frac{1}{l+a}+\frac{1}{a})=\frac{kQ}{a(l+a)}\end{array}$

二

$a$

x

圖23.3

$E_x=\int dE\cos \theta =\int \frac{kdq}{r^2}\frac{x}{r}=\int \frac{kxdq}{r^3}=\frac{kxQ}{{\sqrt{x^2+a^2}}^3}$

三

R

x

r

圖23.4

$\begin{array}{c}{E}_x=\int \frac{kx\sigma rdrd\theta }{{\sqrt{x^2+r^2}}^3}=kx2\pi \sigma {\int }_0^R\frac{rdr}{{\sqrt{x^2+r^2}}^3}=kx\pi \sigma {\int }_0^R\frac{d(x^2+r^2)}{{\sqrt{x^2+r^2}}^3}\\ dq=\sigma dA=\sigma rdrd\theta =2\pi \sigma rdr\\ =kx\pi \sigma {\int }_0^R{(x^2+r^2)}^{-\frac{3}{2}}d(x^2+r^2)=-2kx\pi \sigma {\left[\frac{1}{\sqrt{x^2+r^2}}\right]}_0^R\\ =-2kx\pi \sigma (\frac{1}{\sqrt{x^2+R^2}}-\frac{1}{\sqrt{x^2}})=2kx\pi \sigma (\frac{1}{x}-\frac{1}{\sqrt{x^2+R^2}})\\ =2k\pi \sigma (1-\frac{x}{\sqrt{x^2+R^2}})\end{array}$

四

$\begin{array}{c}\to \\ \to & \overrightarrow{E}\\ \to \end{array}$

圖23.9

${\mathrm{\Phi }}_E=0$

六

$\leftarrow a\to ⟵r⟶$

r

圖23.14

$\begin{array}{c}E(r>a)=\frac{kQ}{r^2}\\ {\mathrm{\Phi }}_E=E\oint dA=\frac{Q}{ϵ}=E4\pi r^2\\ E(r<a)=\frac{kQr}{a^3}\\ E4\pi r^2=\frac{Q_{in}}{ϵ}=Q\frac{r^3}{a^3}/ϵ=\frac{Q{r}^3}{ϵa^3}\\ E=\frac{k{Q}_{in}}{r^2}{Q}_{in}=Q\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi a^3}=Q\frac{r^3}{a^3}\end{array}$

七

 +

$\begin{array}{c}\nwarrow \nearrow \\ ⟵⟶\\ \swarrow \overrightarrow{E}\searrow \\ \\ ⟵l⟶\\ r\end{array}$

圖23.16

$E2\pi rl=\frac{\lambda l}{ϵ}E=\frac{\lambda }{2\pi ϵr}=\frac{2k\lambda }{r}$

八

+++++

+++++

+++++

圖23.17

$2EA=\frac{\sigma A}{ϵ}E=\frac{\sigma }{2ϵ}$