普物二总结——电学部分#

About the tutor

Prof.FangMinghu

[email protected]

• E&M
• Optics
• Modern Physics

Intro#

All of simple physics in 5 equations

Gravity , Electric and Magnetic Force has magical similarity!

Chap. 25 Charge and Coulomb’s law#

Crucial constant nums

$e=1.602\times 10^{-19}C$

$k=\frac{1}{4\pi \varepsilon _0}=9\times 10^{9}N\cdot m^2\cdot C^{-2}$

$1C=6\times 10^{18}e$

$m_e=9.1\times 10^{-31}kg$

25-2#

$1n,1p=3 \mathrm{~quarks}$

$1p(+e)=2\times (\frac{2}{3}e)+(-\frac{1}{3}e)$

$1n(0)=2\times (-\frac{1}{3}e)+\frac{2}{3}e$

25-3#

Electrostatics in matter

Atom force H

Ionic Crystal force NaCl

Covalent Bond force H-H

Metal force Au

NOTE

Coulomb’s laws an exact result for stationary

charges and not an approximation form some higher

law.

• Gravitational vs Electric Force $$\frac{F_e}{F_g}=\cfrac{\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}}{G\frac{m_1m_2}{r^2}}=4.17\times 10^{42}$$

e.g 25-3

for a man ,he push his arms apart with a force of 450N,how many charge can he hold outstretched?

$$F=450N\\ Q=r\sqrt{\frac{F}{k}}=4.47\times 10^{-4}$$

e.g 25-3-2

How many electrons in a person by Feynman

treat people like water

$$\lambda_p=\frac{NA}{M_r(H_2O)}\times e(H_2O)=3.3\times 10^{23}e/g$$

then ,assume he is 80kg

$$n_e=\lambda _p m=2.6\times 10^{28}e$$

1% of a person

$$1\%\times 2.6\times 10^{28}\times 1.6\times 10^{-19}=4.2\times 10^7C$$

then for two people 0.75m apart:

$$F=9\times 10^{9}\times \left(\frac{4.2\times 10^7C}{0.75m}\right)^2=2.8\times 10^{25}\approx W_{earth}=6\times 10^{24}kg\times 9.8m/s^2$$

IMPORTANT

you should distinguish the difference between the vec and scalars

25-4 Conductors and Insulators#

• Insulators : $\leqslant 1 e_c ~\mathrm{per}~ cm^3$,Glass,Plastics,Dry wood
• Conductors : $\approx 10^{23} e_c ~\mathrm{per}~ cm^3$, Aluminum,Copper,Silver…
• Semiconductor : $10^{10}\to 10^{12} ~\mathrm{per}~ cm^3$, Silicon,Germanium
• Superconductor : $R=0,B=0$

(the following picture is from OCR,so some fault occurred )

usage of $B=0$

• NMR(Nuclear Magnetic Resonance)
• Brain research
• **Magnetic Levitation **(Maglev)

25-5 Continuous Charge Distribution#

The charge density

• $\lambda =\frac{\mathrm d q}{\mathrm d x}$
• $\sigma =\frac{\mathrm d q}{\mathrm d A}$
• $\rho =\frac{\mathrm d q}{\mathrm d V}$

to calculate $q_0\to (\mathrm{distribution})\rho$

eg-ring

For a uniform ring of charge :

$$\begin{aligned} &\lambda=\frac{q}{2\pi R} \\ &dF=\frac{1}{4\pi\varepsilon_{0}}\frac{q_{0}dq}{r^{2}}=\frac{1}{4\pi\varepsilon_{0}}\frac{q_{0}\lambda Rd\phi}{(z^{2}+R^{2})} \\ &F_{z}=\int dF_{z}=\int dF\cos\theta \\ &=\int\frac1{4\pi\varepsilon_0}\frac{q_0\lambda Rd\phi}{(z^2+R^2)}\frac z{\sqrt{z^2+R^2}} \\ &=\frac1{4\pi\varepsilon_0}\frac{q_0\lambda Rz}{\left(z^2+R^2\right)^{3/2}}\int_0^{2\pi}d\phi \\ &=\frac1{4\pi\varepsilon_0}\frac{q_0qz}{\left(z^2+R^2\right)^{3/2}} \end{aligned}$$

for approximation

$$z\to \infty ,F_z\to \frac{1}{4\pi \varepsilon _0}\frac{q_0q}{z^2}$$

eg-disk

For a uniform disk of charge :

$$\begin{aligned} \sigma &=\frac{q}{\pi R^2}\\\mathrm d q&=\sigma \mathrm dA =2\pi \sigma \omega \mathrm d\omega\\\mathrm dF_z&=\frac{1}{4\pi \varepsilon _0}\frac{q_0(2\pi \sigma \omega \mathrm d\omega)}{(z^2+\omega ^2)^{\frac{3}{2}}}\\F_z&=\frac{1}{4\pi \varepsilon _0}q_02\pi \sigma z \int _{0}^R \frac{\omega \mathrm d\omega }{(z^2+\omega ^2)^{\frac{3}{2}}}\\&=\frac{1}{4\pi \varepsilon _0 }\frac{2q_0q}{R^2}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right) \end{aligned}$$

for approximation

$$z\to \infty ,F_z\to \frac{1}{4\pi \varepsilon _0}\frac{q_0q}{z^2}$$

25-6 Charge conservation#

Chap. 26 Electric Fields#

26-1 Field#

• scalar field
• Vector field

Mass-field-mass: not action at a distance

26-2 E field#

26-3 The Electric field of point charge#

we just deal with electrostatics instead of electrodynamics

for a single charge $q$

for the Electric Dipole ($p=2Qa=Ql$)

remember $\vec p=q\cdot\vec l$,the direction is from $-q$ to $+q$

then for y-axis

NaCl e.g.

$2a=0.236nm,p_t=2ea=1.6\times 3.78\times 10^{-29}$

$p_e=3\times 10^{-29}$(measured)

it indicates that the electron is not entirely removed from Na to Cl

for $x>>a$

we can approximate $E$ more precisely

26-4 The Electric field of Continuous Charge Distribution#

once again we turn to distribution densities:$\mathrm d q=\lambda d x=\sigma d A=\rho d V$

then

Summary#

when $r>>a/R$

• Dipole $p=2Qa=Ql$

  • in line : $E=\frac{4Qa}{4\pi \varepsilon_0 r^3 }=\frac{2p}{4\pi \varepsilon _0r^3}$
  • off line : $E=\frac{2Qa}{4\pi \varepsilon _0r^3}=\frac{p}{4\pi \varepsilon _0r^3}$

• Point $E=\frac{q}{4\pi \varepsilon_0 r^2 }$

• Line $E=\frac{2\lambda }{4\pi \epsilon_0 r}$

• Ring $E=\frac{q}{4\pi \varepsilon _0z^2}$

• Disk $E=\frac{q}{4\pi \varepsilon _0z^2}$

26-5 E-Field Lines#

26-6 Point Charge in E-field#

deflecting electrode system of an ink-jet printer

$$\begin{aligned}\text{An ink drop}&:m=1.3\times 10^{-10}kg\quad L=1.6cm\\&q=-1.5\times l0^{-13}C,\quad E=1.4\times 10^6N/C\\&v=18m/s\end{aligned}$$$$y=\frac{1}{2}\frac{qE}{m}\frac{L^2}{v^2}\approx 0.64mm$$

one letter -> about 100 drops

100000 drops/s $\to 1000$ letters/s

bonusthe elementary charge $e=1.602\times 10^{-19}C$

skip

then in an ununiformed field,for example

26-7 Dipole in E-field#

• Torque:$\vec \tau=pE\sin \theta=\vec p\times \vec E$
• $W=\int _{\theta_0}^\theta =pE(\cos \theta -\cos \theta_0)=-\Delta U,U(\theta)=-\vec p\cdot \vec E$

we can divide molecules to

• Dipole-molecule $\mathrm{H_2O}$
• non-dipole molecule $\mathrm{C_2O}$

the max torque on H2O

$$\tau =qE\sin \theta =9.3\times 10^{-26}N\cdot m\\ \\ \theta :\pi\to 0 :W=2pE=1.9\times 10^{-25}J\\ \epsilon _{int}=\frac{3kT}{2}-6.3\times 10^{-21}>>\epsilon _{elect}$$

26-8 Atom Nuclear Model#

• Thomson model plum pudding

Rutherford:

then , the result disobeys the experimental result!

Chap. 27 Gauss Law#

27-2(3) Flux#

We need to at first define flux and the vector of surface of a closed region.

if no source or sink of fluid

for electrostatic case

Geometry and Surface Integrals#

we can illustrate that Gauss’s Law -> Coulomb’s Law

for a +Q sphere having radius $R$

$E \cdot 4\pi R^2=\frac{Q}{\varepsilon _0}$

27-5(6) Application Gauss’s Law#

Uniform charged sphere

$$q=\frac{4}{3}\pi a^3\rho\Rightarrow E=\frac{\rho a^3}{3\varepsilon _0r^2}(r>a)\\E=\frac{\rho r}{3\varepsilon _0}(r<a)$$

Conductors——Spherical Symmetry#

since E=0 inside a conductor $Q_{inside}=0$, charges are only on the surface

For a conductor,we obtain a cylinder gauss plane:

Line——Cylindrical Symmetry#

Chap27,ACT2

Aline charge $\lambda\left(\text{C/m}\right)$ is placed along the axis of an uncharged conducting cylinder of inner radius $r_i=a$, and outer radius $r_o=b$ as shown.

we need to get $\sigma_o$

TIP$$E_o=\frac{\lambda }{2\pi \varepsilon _0 r}=\frac{\sigma_0b}{\varepsilon _0 r}\Rightarrow \sigma_o=\frac{\lambda}{2\pi b}$$

Sheets——Planar Symmetry#

for two paralleled sheets

always based on gauss plane

Once we connect the two spheres with a wire:,it becomes a whole conductor with $-2Q_1$

eg2 Cylinders

$$\sigma _i =-\frac{\lambda }{2\pi R},\sigma _o = \sigma_t+\frac{\lambda }{2\pi R}$$$$E_r=\begin{cases}\frac{\lambda }{2\pi \varepsilon _0 r}&r<R\\\frac{\lambda }{2\pi \varepsilon _0,r}+\frac{\sigma R}{\varepsilon _0r} &r>R\end{cases}$$

27-7 Experimental Tests of Gauss’ Law and Coulomb’s Law#

WARNING

just skip

Chap. 28 Electric Potential U & V#

28-1 Potential Energy#

The union energy of two points of charge :

The circuit Law of electrostatics field#

The gauss’ law(the same as Math Analysis):

28-2 Electric Potential#

28-3 Calculate Potential from E#

for a point charge:

Dipole

$$r>>a\\V_r=\frac{1}{4\pi \varepsilon _0}\frac{r_2-r_1}{r_1r_2}\\ r_2-r_1\approx 2a\cos \theta ,r_1r_2=r^2\\ V_r=\frac{1}{4\pi \varepsilon_0}\frac{2aq\cos \theta }{r^2}=\frac{\vec p\cdot \hat r}{4\pi \varepsilon_0 r^2}$$

quadrupole

$$\begin{aligned} &V(r)=\sum_{i}V_{i}(r_{i}) \\ &=\frac{1}{4\pi\varepsilon_{0}}(\frac{q}{r-d}+\frac{-2q}{r}+\frac{q}{r+d}) \\ &=\frac{1}{4\pi\varepsilon_{0}}\frac{2qd^{2}}{r(r^{2}-d^{2})} \\ &=\frac{1}{4\pi\varepsilon_{0}}\frac{2qd^{2}}{r^{3}(1-d^{2}/r^{2})}\to \frac{Q}{4\pi \varepsilon _0r^3}(d<<r) \end{aligned}$$

• Point charge $\propto \frac{1}{r}$
• Dipole $\propto \frac{1}{r^2}$
• Quadrupole $\propto \frac{1}{r^3}$

we can trace back to the distribution reflected by n-dipoles.

:::[egs for potential calculation]

for a charged sphere shell

for a ring

for a disk

:::

Sparks#

High electric fields can ionize nonconducting materials(dielectrics)

(Insulator->Conductor)

ball breakdown

we have two ball shell with same potential $V$

Ball 2 is as twice large as Ball 1

as $V$ goes up, the Ball 1 will breakdown first

$$E_s=\frac{Q}{4\pi \varepsilon _0 r^2},V=\frac{Q}{4\pi \varepsilon_0 r}\Rightarrow E=\frac{V}{r}$$

then $r_1<r_2\Rightarrow E_1>E_2$

• $\Delta V>0$ means we go uphill
• $\Delta V<0$ means we go downhill

you can split into two patterns

28-4 Equipotentials#

when in the equipotential surface we can conclude

28-5 Potential of a charged conductors#

IMPORTANT

Claim: The surface of a conductor is an equipotential surfacewhen two sphere conductors are attatched to each others.

$$\frac{Q_A}{Q_B}=\frac{r_A}{r_B}$$

for example , when a point of charge is placed off-center inside a sphere conductor,the inside surface will be inuniform and the outside surface will be uniform

28-6 Calculate E from Potential#

• Graphically the E-field line is the fastest-descending line of equipotential surfaces

• Math:

  from $\mathrm d W=-q_0 \mathrm d V$

  $$\mathrm d W=\vec F\cdot \mathrm d\vec l=q_0\vec F\cdot \mathrm d \vec l=q_0E\mathrm d l\cos \theta\\E\cos \theta =-\frac{\mathrm d V}{\mathrm dl}$$ $$E_l=-\frac{\mathrm d V}{\mathrm d l}\Rightarrow \vec E=-\nabla V$$

  • Cartesian coordinates:

    $$\nabla V=\frac{\partial V}{\partial x}\bar x+\frac{\partial V}{\partial y}\bar y+\frac{\partial V}{\partial z}\bar z$$

  • Spherical coordinates:

    $$\nabla V=\frac{\partial V}{\partial r}\bar x+\frac{1}{r}\frac{\partial V}{\partial \theta}\bar y+\frac{1}{r\sin \varphi}\frac{\partial V}{\partial \varphi}\bar \varphi$$

    review

    $$\begin{align*} x & = r \sin \theta \cos \phi \\ y & = r \sin \theta \sin \phi \\ z & = r \cos \theta \end{align*}$$$$r=\sqrt{x^2+y^2+z^2}\\\theta=\arccos\frac z{\sqrt{x^2+y^2+z^2}}=\arccos\frac zr=\begin{cases}\arctan\frac{\sqrt{x^2+y^2}}z&\mathrm{if~}z>0\\\\\pi+\arctan\frac{\sqrt{x^2+y^2}}z&\mathrm{if~}z<0\\\\+\frac\pi2&\mathrm{if~}z=0\mathrm{~and~}\sqrt{x^2+y^2}\neq0\\\mathrm{undefined}&\mathrm{if~}x=y=z=0&\end{cases}\\\varphi=\mathrm{sgn}(y)\arccos\frac x{\sqrt{x^2+y^2}}=\begin{cases}\arctan(\frac8x)&\mathrm{if~}x>0,\\\arctan(\frac9x)+\pi&\mathrm{if~}x<0\mathrm{~and~}y\geq0,\\+\frac\pi2&\mathrm{if~}x<0\mathrm{~and~}y>0,\\-\frac\pi2&\mathrm{if~}x=0\mathrm{~and~}y>0,\\\mathrm{undefined}&\mathrm{if~}x=0\mathrm{~and~}y=0.&\end{cases}$$

eg

$$\begin{align*} V=3x^2+2xy-z^2 \\ \vec E=(-6x-2y,-2x,2z)^T\\ \vec E=\frac{2aq}{4\pi \varepsilon _0r^3}(2\cos \theta ,\sin \theta ,0)_{sp}^T \end{align*}$$

then we have two eg

dipole

$$V(r)=\frac{q}{4\pi \varepsilon _0}\frac{r_2-r_1}{r_1r_2}\to \frac{1}{4\pi \varepsilon _0}\frac{2aq\cos \theta}{r^2}(r>>a)$$

then

$$V(r,\theta)=\frac{1}{4\pi \varepsilon _0}\frac{2aq\cos \theta}{r^2}\\E_r=-\frac{\partial V}{\partial r}\\E_\theta=-\frac{\partial V}{\partial \theta}\\\Rightarrow \vec E=\frac{2aq}{4\pi \varepsilon _0 r^3}((2\cos \theta )\hat r+\sin \theta \hat \theta)$$

it’s easy to see that $\arg \max_{\theta } ||\vec E||=\frac{\pi}{2}$

eg. disk

$$\begin{aligned} &dq=2\pi\omega\cdot d\omega\cdot\sigma \\ &dV=\frac{dq}{4\pi\varepsilon_{0}\sqrt{z^{2}+\omega^{2}}}=\frac{2\pi\omega\cdot d\omega\cdot\sigma}{4\pi\varepsilon_{0}\sqrt{z^{2}+\omega^{2}}} \\ &V=\int_{0}^{R}\frac{2\pi\omega\cdot d\omega\cdot\sigma}{4\pi\varepsilon_{0}\sqrt{z^{2}+\omega^{2}}}=\frac{\sigma}{2\varepsilon_{0}}\big(\sqrt{z^{2}+R^{2}}-z\big) \\ &E_{z}=-\frac{\partial V}{\partial z}=-\frac{\sigma}{2\varepsilon_{0}}(\frac{2z}{2\sqrt{R^{2}+z^{2}}}-1) \\ &=\frac{\sigma}{2\varepsilon_{0}}(1-\frac{1}{\sqrt{1+(R/z)^{2}}}) \end{aligned}$$

a e.g for getting $v$ with $\Delta U$

$$\frac{1}{2}mv^2=K=-\Delta U=-q\Delta V\Rightarrow v=\sqrt{\frac{2q\Delta V}{m}}$$

we can measure $\alpha$ partical’s large velocity by measuring eletrostatic information like difference of potential

Appendix method of images#

we can see

we can see that the induced charge distribution generated with a point of charge and a conductor sheet is the same as which is made by two charges. It can be proved with symmetry theorem.

Chap. 30 Capacitance and Dielectrics#

30-1 Capacitors#

Classic Capacitors#

Application

• flashbulbdraw energy from battery ($s$) then release it through bulb($ms$)
• Laser pulse
• thermonuclear fusion $10^{14}W,10^{-9}s,10^8K$

Definition of Capacitance : two spatially seperated conductors($+q/-q$) $C=\frac{q}{\Delta V}$

IMPORTANT

one single conductor is capacitor!

egPlate capacitor

$$q=\sigma A\\\Delta V=-\int _{A}^B \vec E\cdot \mathrm{d}\vec l$$

egPlate capacitor

$$q=\sigma A\\\Delta V=-\int _{A}^B \vec E\cdot \mathrm{d}\vec l$$

since

$$EA\varepsilon _0= \sigma A\Rightarrow E=\frac{\sigma}{\varepsilon _0}\Rightarrow \Delta V=\frac{q}{A\varepsilon_0}d\Rightarrow C=\frac{\varepsilon _0 A}{d}$$

Application

• condenser:$C \propto \frac{1}{d}\xrightarrow{\mathrm{fixed~}\Delta V} Q\propto\frac{1}{d},I=\frac{\mathrm{d} Q}{\mathrm d t}$vibration -> different $I$

egCapacitor

$+Q,-Q$ on surface,$\Delta V$

$$2\pi r \cdot L\cdot E\cdot \varepsilon_0=Q\Rightarrow E=\frac{Q}{2\pi \varepsilon _0 Lr}$$$$\Delta V=\int_{b}^a \vec E\cdot \mathrm d\vec l =\frac{Q}{2\pi \varepsilon L\ln \frac{b}{a}},C=\frac{2\pi \varepsilon _0 L}{\ln\frac{b}{a}}$$

exsignals tansmit/coaxial cable

$a=r_i=0.15,b=r_o=2.1$

$$\frac{C}{L}=\frac{2\pi\varepsilon_{0}}{\ln(b/a)}=\frac{2\pi\times8.85\times10^{-12}}{\ln(2.1/0.15)}=21\times10^{-12}F / m=21 pF / m$$

egcapacitor

$$\vec E=\frac{q\hat r}{4\pi \varepsilon _0 r^2}\Rightarrow \Delta V=\frac{q}{4\pi \varepsilon_0 }\left(\frac{1}{a}-\frac{1}{b}\right)\Rightarrow C=\frac{4\pi \varepsilon _0ab}{b-a}$$

for earth $R=6.37\times 10^6 m\Rightarrow C=7.1\times 10^{-4}\mathrm F=710\mu \mathrm F$

Summary#

• Parallel Plate:$C=\frac{\varepsilon _0 A}{d}$
• Cylindrical Capacitor:$C=\frac{2\pi \varepsilon _0 L}{\ln\frac{b}{a}}$
• Spherical $C=4\pi \varepsilon_0\frac{ab}{b-a}$

Parallel & Series#

for capacitors in parallel

for capacitors in series