# 學院物理/連續電荷分布與高斯定律

## 高斯定律

${\displaystyle \Phi ={\vec {E}}\cdot {\vec {A}}=EA\cos \theta =\int {\vec {E}}\cdot d{\vec {A}}={\frac {kq}{r^{2}}}\cdot 4\pi r^{2}={\frac {\quad q\;\;4\pi r^{2}}{4\pi \epsilon \,r^{2}\quad \;\;}}={\frac {q}{\epsilon }}}$

${\displaystyle \epsilon }$

## 例題

### 一

${\displaystyle {\vec {E}}\gets }$
{\displaystyle {\begin{aligned}\gets x\to \\\\a\qquad l\end{aligned}}}

{\displaystyle {\begin{aligned}E&=\int dE=k\int {\frac {dq}{r^{2}}}=k\lambda \int _{a}^{l+a}{\frac {dx}{x^{2}}}=k\lambda \left[-{\frac {1}{x}}\right]_{a}^{l+a}\\&\qquad \qquad \qquad dq=\lambda dx\\&=k{\frac {Q}{l}}\left(-{\frac {1}{l+a}}+{\frac {1}{a}}\right)={\frac {kQ}{a\left(l+a\right)}}\end{aligned}}}

### 二

${\displaystyle a}$
x

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

### 三

R
x
r

{\displaystyle {\begin{aligned}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\left(x^{2}+r^{2}\right)}{{\sqrt {x^{2}+r^{2}}}^{3}}}\\dq=\sigma dA=\sigma rdrd\theta =2\pi \sigma rdr\qquad \qquad \qquad \qquad \qquad \qquad \quad \\=kx\pi \sigma \int _{0}^{R}\left(x^{2}+r^{2}\right)^{-{\frac {3}{2}}}d\left(x^{2}+r^{2}\right)=-2kx\pi \sigma \left[{\frac {1}{\sqrt {x^{2}+r^{2}}}}\right]_{0}^{R}\\=-2kx\pi \sigma \left({\frac {1}{\sqrt {x^{2}+R^{2}}}}-{\frac {1}{\sqrt {x^{2}}}}\right)=2kx\pi \sigma \left({\frac {1}{x}}-{\frac {1}{\sqrt {x^{2}+R^{2}}}}\right)\\=2k\pi \sigma \left(1-{\frac {x}{\sqrt {x^{2}+R^{2}}}}\right)\end{aligned}}}

### 四

{\displaystyle {\begin{aligned}\to \\\to &{\vec {E}}\\\to \end{aligned}}}

${\displaystyle \Phi _{E}=0}$

### 六

${\displaystyle \gets a\to \longleftarrow r\longrightarrow }$
r

{\displaystyle {\begin{aligned}E\left(r>a\right)={\frac {kQ}{r^{2}}}\\\Phi _{E}=E\oint dA={\frac {Q}{\epsilon }}=E4\pi r^{2}\\E\left(r

### 七

+
${\displaystyle {\begin{matrix}\nwarrow \quad \;\nearrow \\\longleftarrow \quad \longrightarrow \\\swarrow {\vec {E}}\searrow \\\\\longleftarrow l\longrightarrow \\\qquad \quad \;\,r\end{matrix}}}$

${\displaystyle E2\pi rl={\frac {\lambda l}{\epsilon }}\qquad E={\frac {\lambda }{2\pi \epsilon r}}={\frac {2k\lambda }{r}}}$

### 八

+++++

+++++

+++++

${\displaystyle 2EA={\frac {\sigma A}{\epsilon }}\qquad E={\frac {\sigma }{2\epsilon }}}$