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--- |
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title: 工程电磁学 |
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titleEN: Engineering Electromagnetics |
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date: 2019-02-26 |
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categories: |
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- notes |
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tags: |
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- maxwell |
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--- |
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{% raw %}<span class=".zh">{% endraw %} |
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麦克斯韦大法好!! |
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{% raw %}</span>{% endraw %} |
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{% raw %}<span class=".en">{% endraw %} |
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Maxwell Dafa is good! ! |
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{% raw %}</span>{% endraw %} |
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<!--more--> |
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{% raw %} |
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<script> |
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session.onload(function(){ |
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if(page.tran.getLang() == 'en'){ |
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tips.warning({ |
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title: 'Caution', |
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position: 'topRight', |
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message: 'This page was translated by Machine!!', |
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buttons: [['<button>Show Original Page</button>', function (instance, toast) { |
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page.tran.setLang('zh'); |
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instance.hide({ transitionOut: 'fadeOut' }, toast, 'button'); |
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}, true]] |
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} |
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</script> |
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{% endraw %} |
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{% raw %}<span class=".zh">{% endraw %} |
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### 先供上麦克斯韦方程 膜拜膜拜( o=^•ェ•)o |
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$$ |
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\begin{eqnarray} |
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\nabla\cdot\vec{E} &=& \frac{\rho}{\varepsilon_0} \\ |
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\nabla\cdot\vec{B} &=& 0 \\ |
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\nabla\times\vec{E} &=& -\frac{\partial B}{\partial t} \\ |
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\nabla\times\vec{B} &=& \mu_0\left(\vec{J}+\varepsilon_0\frac{\partial E}{\partial t} \right) |
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\end{eqnarray} |
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$$ |
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---------------- |
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## 第一话 - 高斯生库伦 |
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- 假设空间中两点电荷$Q_{1}$,$Q_{2}$,相距d,欲求其相互作用的电场力。 |
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- 现以$Q_{1}$为圆心,$d$为半径做球。根据高斯law可知球面上的电通量只与球内电荷量有关,本例中为$\frac{Q_{1}}{\varepsilon_0}$。 |
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- 将上式中电通量除以求表面积可得电场强度$\frac{Q_{1}}{4\pi d^{2}\varepsilon_0}$ |
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- 场强乘以$Q_{2}$即可得库仑力$\frac{Q_{1}Q_{2}}{4\pi d^{2}\varepsilon_0}$ |
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- 令$k=\frac{1}{4\pi \varepsilon_0}$整理得 |
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$$F = \frac{k Q_{1}Q_{2}}{d^{2}}$$ |
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证毕 |
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## 第二话 - 高斯金箍棒 |
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- 假设一均匀带电长度正无穷细杆,电荷密度为$\lambda$,求距其$d$处场强。 |
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- 绕杆画一个过待求点的圆柱,设高为$x$。 |
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- 由于杆长无限,圆柱两底面电场被抵消。 |
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- 圆柱侧面积为$2\pi dx$ |
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- 圆柱内电荷量为$\lambda x$ |
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- 引入高斯,得 |
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$$ |
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2\pi dxE = \frac{\lambda x}{\varepsilon_{0}} |
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$$ |
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整理得 |
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$$ |
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E = \frac{\lambda}{2\pi d\varepsilon_{0}} |
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$$ |
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证毕 |
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## 高斯球球球 |
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- 首先要有一个带电小球,电量$Q$,想求其外部距其圆心$d$处场强。 |
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- 然后可列式 |
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$$ |
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4\pi d^2 E = \frac{Q}{\varepsilon_0} |
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$$ |
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整理得 |
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$$ |
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E = \frac{Q}{4\pi \varepsilon_0 d^2} |
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$$ |
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毕 |
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## 高斯大面 |
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- 首先有一个均匀带电无穷面,带电面密度$\rho$,欲求距其$d$的点场强。 |
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- 以无穷面为中央横截面,做一个底面圆心为待求点的圆柱,半径为$r$。 |
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- 由于电场线皆平行,只有两个底面有电场线穿过。 |
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- 可列式 |
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$$ |
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2\pi r^2 E = \frac{\rho \pi r^2}{\varepsilon_0} |
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$$ |
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整理得 |
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$$ |
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E = \frac{\rho}{2\varepsilon_0} |
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$$ |
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完事 |
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----------- |
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To Be Continued... |
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{% raw %}</span>{% endraw %} |
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{% raw %}<span class=".en">{% endraw %} |
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### First offer to Maxwell's equation, worship ( o=^•ェ•)o |
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$$ |
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\begin{eqnarray} |
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\nabla\cdot\vec{E} &=& \frac{\rho}{\varepsilon_0} \\ |
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\nabla\cdot\vec{B} &=& 0 \\ |
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\nabla\times\vec{E} &=& -\frac{\partial B}{\partial t} \\ |
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\nabla\times\vec{B} &=& \mu_0\left(\vec{J}+\varepsilon_0\frac{\partial E}{\partial t} \right) |
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\end{eqnarray} |
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$$ |
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---------------- |
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## Chapter One-Gossian Cullen |
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- Suppose two electric charges $Q_{1}$ and $Q_{2}$ in space are separated by d, and the electric field force they want to interact with. |
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- Now take $Q_{1}$ as the center of the circle and $d$ as the radius to make the ball. According to the Gaussian law, the electric flux on the sphere is only related to the amount of charge in the sphere, which is $\frac{Q_{1}}{\varepsilon_0}$ in this example. |
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- Divide the electric flux in the above formula by the surface area to get the electric field intensity $\frac{Q_{1}}{4\pi d^{2}\varepsilon_0}$ |
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- Field strength is multiplied by $Q_{2}$ to get Coulomb force $\frac{Q_{1}Q_{2}}{4\pi d^{2}\varepsilon_0}$ |
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- Let $k=\frac{1}{4\pi \varepsilon_0}$ to get |
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$$F = \frac{k Q_{1}Q_{2}}{d^{2}}$$ |
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Completed |
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## Chapter 2-Gauss Golden Cudgel |
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- Assuming that a uniformly charged length is positively infinitely thin and the charge density is $\lambda$, find the field strength $d$ away from it. |
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- Draw a cylinder around the rod and set the height to be $x$. |
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- Due to the infinite length of the rod, the electric fields on the two bottom surfaces of the cylinder are cancelled. |
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- The cylindrical side area is $2\pi dx$ |
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- The amount of charge in the cylinder is $\lambda x$ |
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- Introduce Gaussian, get |
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$$ |
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2\pi dxE = \frac{\lambda x}{\varepsilon_{0}} |
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$$ |
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Organized |
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$$ |
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E = \frac{\lambda}{2\pi d\varepsilon_{0}} |
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$$ |
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Completed |
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## Gauss Ball Ball |
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- First, there must be a charged ball with electricity $Q$, and I want to find the field strength at the distance of $d$ from the center of the ball. |
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- Then columnable |
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$$ |
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4\pi d^2 E = \frac{Q}{\varepsilon_0} |
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$$ |
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Organized |
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$$ |
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E = \frac{Q}{4\pi \varepsilon_0 d^2} |
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$$ |
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complete |
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## Gaussian Noodles |
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- First, there is a uniformly charged infinite surface, the charged surface density is $\rho$, and the field strength of the point $d$ is desired. |
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- Using the infinite plane as the central cross section, make a cylinder with the center of the bottom surface as the point to be found, and the radius is $r$. |
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- Since the electric field lines are all parallel, only two bottom surfaces have electric field lines passing through. |
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- Columnable |
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$$ |
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2\pi r^2 E = \frac{\rho \pi r^2}{\varepsilon_0} |
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$$ |
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Organized |
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$$ |
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E = \frac{\rho}{2\varepsilon_0} |
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$$ |
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Finished |
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----------- |
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To Be Continued... |
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{% raw %}</span>{% endraw %}
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