General Physics II Review

Posted on 2020-08-25 Edited on 2025-04-10 Views: Word count in article: 2.2k Reading time ≈ 8 mins.

Ch. 21 Coulomb’s Law

Charge

Plastic rod rubbed on fur: negatively charged
Glass rod rubbed on silk: positively charged

The total (net) electric charge of an isolated system is conserved.

Charging by induction (without losing its own charge)

$$ e=1.60\times10^{-19}\rm C $$

Quantized Charge
Millikin Oil-Drop Experiment

Coulomb’s Law

$$ F=k\frac{|q_1q_2|}{r^2} $$

$$ k= \frac{1}{4\pi\epsilon_0} = 8.99\times10^9 \rm N\cdot m^2/C^2 $$

$$ \epsilon_0 = 8.85\times10^{-12}\rm C^2/N\cdot m^2 $$

k is the electromagnetic constant
ϵ₀ is the permittivity constant
Principle of Superposition

Ch. 22 Electric Field

Definition

F⃗ = qE⃗.

$$ \vec{E} = \frac{\vec{F}}{q_0} = k\frac{|q|}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{|q|}{r^2} $$

Cases

The key to integrate, is find the relation between dq and charge density (as well as length or area)

Uniformly Charged Line

$$ E_x = \frac{1}{4\pi\epsilon_0}\frac{Q}{2a} \int_{-a}^{+a} \frac{x\mathop{dy}}{(x^2+y^2)^{3/2}} = \frac{Q}{4\pi\epsilon_0} \frac{1}{x\sqrt{x^2+a^2}} $$

If the line is very long, a ≫ x

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

Uniformly Charged Ring

$$ \begin{aligned} E & = \int\mathop{dE}\cos\theta \\ & = \frac{z}{\sqrt{z^2+R^2}}\cdot \frac{\lambda}{4\pi\epsilon_0 (z^2+R^2)} \int_0^{2\pi R}\mathop{ds} \\ & = \frac{qz}{4\pi\epsilon_0 (z^2+R^2)^{3/2}}. \end{aligned} $$

If the charged ring is at large distance, z ≫ R

$$ E = \frac{1}{4\pi\epsilon_0}\frac{1}{z^2}. $$

Uniformly Charged Disk

dq = σdA = σ(2πrdr).

$$ \mathop{dE} = \frac{\sigma z}{4\epsilon_0} \frac{2r\mathop{dr}}{(z^2+r^2)^{3/2}}. $$

$$ E = \int\mathop{dE} = \frac{\sigma}{2\epsilon_0}\left( 1 - \frac{z}{\sqrt{z^2+R^2}} \right). $$

For R → ∞ (infinite sheet)

$$ E = \frac{\sigma}{2\epsilon_0}. $$

Electric Field Line

Electric Dipole *

Electric Field Due to Electric Dipole

$$ \begin{aligned} E & = E_{(+)} - E_{(-)} \\ & = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2_{(+)}} - \frac{1}{4\pi\epsilon_0} \frac{q}{r^2_{(-)}} \\ & = \cdots \\ & = \frac{q}{2\pi\epsilon_0z^3} \frac{d}{\left( 1 - \left(\frac{d}{2z}\right)^2 \right)^2} \end{aligned}. $$

If z ≫ d

$$ E = \frac{1}{2\pi\epsilon_0}\frac{qd}{z^3} $$

qd is the electric dipole moment p⃗

On the axis of the dipole

$$ \vec{E}_\text{dipole} \approx \frac{1}{4\pi\epsilon_0}\frac{\vec{p}}{r^3} $$

On the perpendicular plane

$$ \vec{E}_\text{dipole} \approx -\frac{1}{4\pi\epsilon_0}\frac{\vec{p}}{r^3} $$

Dipole in Electric Field

τ⃗ = Fdsin θ = p⃗ × E⃗.

U = − W = − ∫τdθ = ∫pEsin θdθ = − pEcos θ = − p⃗ ⋅ E⃗

Ch. 23 Gauss’s Law

ϵ₀Φ = ϵ₀∮E⃗ ⋅ d = q_enclosed

$$ \oint_A \vec{E}\cdot\mathop{d\vec{A}} = \int_V \nabla\cdot\vec{E}\mathop{dV} = \frac{q_{\text{enc}}}{\epsilon_0}. $$

Gauss’s Law is equivalent with Coulomb’s Law.
Planar Symmetry
Cylindrical Symmetry

Charged Isolated Conductor

There can be no excess charge at any point within a solid conductor
Electric field inside a conductor needs to be zero
Charge on the inner wall cannot produce and electric field in the shell to affect the charge on the outer wall

Electrostatic Shielding

Ch. 24 Electric Potential

ΔU = U_f − U_i = − W_{by the field}

V = V_f − V_i = − ∫_i^fE⃗ ⋅ d .

i is at ∞, where U is 0, V_i is 0.

Electric potential for a point charge

$$ V = \frac{1}{4\pi\epsilon_0}\frac{q}{r} $$

Potential due to electric dipole

$$ V = V_{(+)} + V_{(-)} = \cdots = \frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2} $$

Field from Potential

$$ E_s = -\frac{\partial V}{\partial s}. $$

E⃗ = − ∇V

Ch. 25 Capacitance

q = CV

Parallel capacitor

$$ C = \frac{q}{V} = \frac{\epsilon_0 EA}{Ed} = \frac{\epsilon_0 A}{d}. $$

Cylindrical capacitor

q = ϵ₀EA = ϵ₀E(2πrL)

$$ V = \int_-^+ E\mathop{ds} = -\frac{q}{2\pi\epsilon_0 L} \int_b^a \frac{\mathop{dr}}{r} = \frac{q}{2\pi\epsilon_0 L} \ln\left(\frac{b}{a}\right) $$

$$ C = \frac{q}{V} = 2\pi\epsilon_0 \frac{L}{\ln(b/a)} $$

Spherical capacitor

$$ \cdots C = 4\pi\epsilon_0 \frac{ab}{b-a}. $$

Series & Parallel
Energy stored

$$ \mathop{dW} = v\mathop{dq} = \frac{q\mathop{dq}}{C} $$

$$ W = \int_0^W \mathop{dW} = \frac{1}{C}\int_0^Q q\mathop{dq} = \frac{Q^2}{2C} \Rightarrow U = \frac{1}{2}CV^2 = \frac{1}{2}QV $$

Energy density

$$ u = \frac{\frac{1}{2}CV^2}{Ad} = \frac{1}{2}\epsilon_0E^2 $$

Dielectric

ϵ = κϵ₀

$$ E = \frac{V}{d} = \frac{q}{Cd} = \frac{q}{d\epsilon_0A/d} = \frac{\sigma}{\epsilon_0} $$

$$ E = \frac{\sigma-\sigma_i}{\epsilon_0} = \frac{E_0}{K} \Rightarrow \sigma_i = \sigma\left(1-\frac{1}{K}\right) \quad\text{(induced surface charge density)} $$

ϵ₀∮κE⃗ ⋅ d = q

Ch. 26 Current and Resistance

Current

$$ i = \frac{\mathop{dq}}{\mathop{dt}} $$

Current is NOT a vector

Current density

$$ i = \frac{q}{t} = \frac{nALe}{L/v_d} = nAev_d. $$

J⃗ = (ne)v⃗_d

Resistance

Resistivity

$$ \rho = \frac{E}{J} $$

Resistance

$$ \rho = \frac{V/L}{I/A} \Rightarrow R = \frac{V}{I} = \frac{\rho L}{A}. $$

Ohm’s Law

For Ohmic contact cases

V = iR

Ch. 27 Circuits

Kirchhoff’s Rules

∑I = 0, ∑V = 0

Mark for RC Circuit

Ch. 28 Magnetic Field

On Moving Charge

F⃗ = qv⃗ × B⃗

Hall Effect

$$ n = \frac{Bi}{Vle} $$

On a Current Carrying Conductor

F⃗_B = il⃗ × B⃗

Magnetic Dipoles

μ = NiA

τ = μ × B

Ch. 29 Magnetic Fields Due to Currents

Biot-Savart Law

$$ \mathop{d\bm{B}} = \frac{\mu_0}{4\pi}\frac{i\mathop{d\bm{l}}\times\bm{e}_r}{r^2} $$

For B-Field of a wire with steady current

$$ B = \frac{\mu_0 i}{2\pi R} $$

B-Field at the center of a circular arc of wire

$$ B = \int\mathop{dB} = \int_0^\phi \frac{\mu_0}{4\pi}\frac{iR\mathop{d\phi}}{R^2} = \frac{\mu_0 i}{4\pi R} \int_0^\phi\mathop{d\phi} = \frac{\mu_0 i\phi}{4\pi R} $$

Gauss’ Law for magnetism

∮B⃗ ⋅ d = 0

Ampere’s Law

∮B⃗ ⋅ d = μ₀I_encl

∮_lB ⋅ d𝐥 = μ₀∫_AJ ⋅ d𝐀

B-Field of solenoid

BL = μ₀nLI

B-Field of toroid

$$ B = \frac{\mu_0 NI}{2\pi r} $$

Current sheet

$$ B = \frac{1}{2}\mu_0 J_s $$

Ch. 30 Induction and Inductance

Laws of Induction

Lenz’s Law
Faraday’s Law of Induction

$$ \mathscr{E} = -\frac{\mathop{d\Phi_B}}{\mathop{dt}} $$

$$ -\frac{d}{dt}\int\limits_A\vec{B}\cdot\mathop{d\vec{A}} = \oint\limits_l\vec{E}\cdot\mathop{d\vec{l}} $$

$$ \nabla\times\mathbf{E} = -\frac{d\mathbf{B}}{dt} $$

Eddy Current

Inductors

Self induction

$$ \mathscr{E}_L = -L\frac{di}{dt} $$

$$ L = \frac{N\Phi_B}{i} $$

Inductance of an ideal solenoid

L = μ₀n²lA

Inductance of a toroid

$$ L = \frac{\mu_0N^2b}{2\pi}\ln\frac{r_2}{r_1} $$

RL Circuit

$$ \mathscr{E} -i(t)R - L\frac{di}{dt} = 0 \Rightarrow \frac{di}{dt} + \frac{iR}{L} = \frac{\mathscr{E}}{L} $$

$$ i(t) = \frac{\mathscr{E}}{R}(1 - e^{-t/\tau_L}), \quad \tau_L = L/R $$

Energy stored in magnetic field

$$ i\mathscr{E} = Li\frac{di}{dt} + i^2R \Rightarrow U_B = \int_0^t Li\mathop{di} = \frac{1}{2}Li^2 $$

Energy density of magnetic field

$$ u_B = \frac{Li^2}{2Al} = \frac{U_B}{Al} = \frac{\mu_0n^2Ai^2}{2A} = \frac{\mu_0n^2i^2}{2} = \frac{B^2}{2\mu_0} $$

Mutual Induction

$$ \mathscr{E}_2 = -M\frac{di_1}{dt},\quad \mathscr{E}_1 = -M\frac{di_2}{dt} $$

Ch. 31 EM Oscillation & AC Current *

$$ U_C = \frac{q^2}{2C}, \quad U_B = \frac{Li^2}{2}, \quad $$

$$ -L\frac{di}{dt} - \frac{q}{C} = 0 \Rightarrow \frac{d^2q}{dt^2} + \frac{q}{LC} = 0 $$

$$ \Rightarrow -A\omega^2\cos(\omega t + \phi) + \frac{A}{LC}\cos(\omega t + \phi) = 0, \quad \omega = \frac{1}{\sqrt{LC}} $$

Damped Oscillation in LRC Circuit

Forced Oscillations

capacitive reactance

$$ X_C = \frac{1}{\omega_d C} $$

Voltage leads the current

inductive reactance

X_L = ω_dL

Voltage lags the current

Impedance

$$ Z = \sqrt{R^2 + (X_L - X_C)^2}, \quad I = \frac{\mathscr{E}}{Za} $$

Power in AC Circuits

$$ P_\text{avg} = \mathscr{E}_\text{rms}I_\text{rms}\cos\phi, \quad \cos\phi = \frac{V_R}{\mathscr{E}_m} = \frac{R}{Z} $$

Transformers

$$ V_1I_1 = V_2I_2, \quad \frac{V_2}{V_1} = \frac{N_2}{N_1} $$

Ch. 32 Maxwell’s Equations; Magnetism of Matter *

Integral Form

$$ \left\{ \begin{aligned} & \oint_{\mathbf{A} = \partial V} \mathbf{E}\cdot\mathop{d\mathbf{A}} = \frac{q_\text{enc}}{\epsilon_0} & \quad\text{Gauss' Law} \\ & \oint_{\mathbf{A} = \partial V} \mathbf{B}\cdot\mathop{d\mathbf{A}} = 0 & \quad\text{Gauss' Law for B-Field} \\ & \oint_{\mathbf{l} = \partial \mathbf{A}} \mathbf{E}\cdot\mathop{d\mathbf{l}} = - \frac{d}{dt} \int_\mathbf{A} \mathbf{B}\cdot\mathop{d\mathbf{A}} & \quad\text{Faraday's Law} \\ & \oint_{\mathbf{l} = \partial\mathbf{A}} \mathbf{B}\cdot\mathop{d\mathbf{l}} = \mu_0\left( \int_\mathbf{A} \mathbf{J}\cdot\mathop{d\mathbf{A}} + \epsilon_0\frac{d}{dt} \int_\mathbf{A} \mathbf{E}\cdot\mathop{d\mathbf{A}} \right) & \quad\text{Ampere' Law} \end{aligned} \right. $$

Maxwell’s Law of Induction

$$ \oint_\mathbf{l} \mathbf{B}\cdot\mathop{d\mathbf{l}} = \mu_0\epsilon_0\frac{d\Phi_E}{dt} = \mu_0\epsilon_0\frac{d}{dt} \int_\mathbf{A} \mathbf{E}\cdot\mathop{d\mathbf{A}} $$

displacement current

$$ i_d = \epsilon_0\frac{d\Phi_E}{dt} $$

Differential Form

$$ \left\{ \begin{aligned} & \nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0} \\ & \nabla\cdot\mathbf{B} = 0 \\ & \nabla\cdot\mathbf{E} = - \frac{\partial\mathbf{B}}{\partial t} \\ & \nabla\cdot\mathbf{B} = \mu_0\left( \mathbf{J} + \epsilon_0\frac{\partial\mathbf{E}}{\partial t} \right) \\ \end{aligned} \right. $$

Ch. 33 Electromagnetic Waves *

$$ \left\{ \begin{aligned} & \nabla^2\mathbf{B} - \mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2} = 0 \\ & \nabla^2\mathbf{E} - \mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2} = 0 \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & \mathbf{E}(\mathbf{r},t) = \mathbf{E}_0 \cos(\omega t - \mathbf{k}\cdot\mathbf{r} + \phi_0) \\ & \mathbf{B}(\mathbf{r},t) = \mathbf{B}_0 \cos(\omega t - \mathbf{k}\cdot\mathbf{r} + \phi_0) \end{aligned} \right. $$

$$ k = |\mathbf{k}| = \frac{\omega}{c} = \frac{2\pi}{\lambda} $$

$$ \left|\frac{\mathbf{E}_0}{\mathbf{B}_0}\right| = c = \frac{1}{\sqrt{\mu_0\epsilon_0}} $$

Energy Transfer in EM Waves

$$ u_\text{total} = \epsilon_0E^2 = \epsilon_0EBc = \frac{EB}{\mu_0} $$

Poynting Vector

$$ \vec{S} = \frac{1}{\mu_0}\vec{E}\times\vec{B} $$

Intensity

$$ I = S_\text{avg} = \frac{E_0B_0}{2\mu_0} = \frac{E_0^2}{2\mu_0c} = \frac{E^2_\text{rms}}{\mu_0c} $$

$$ \frac{\text{average power}}{4\pi r^2} \propto \frac{1}{r^2} $$

Polarization

When a unpolarized light passes through a polarizer, half of the intensity is loss

$$ I - \frac{1}{2}I_0 $$

I₂ = I₁cos²ϕ

Reflection & Refraction

Snell’s Law

$$ \frac{\sin\theta_1}{\sin\theta_2} = \frac{n_\text{water}}{n_\text{air}},\quad n = c/v = \frac{\lambda_\text{vacuum}}{\lambda_\text{medium}} $$

Fermat’s Principle in Optics
Total Internal Reflection

$$ \sin\theta_\text{critical} = \frac{n_2}{n_1} $$

Light in a Raindrop *
Polarization bt reflection

$$ \theta_B = \tan^{-1}\frac{n_2}{n_1} $$

Ch. 35 Interference *

Young’s Interference

$$ d\sin\theta = m\lambda\Rightarrow y_m = R\tan\theta_m\approx R\sin\theta_m = R\frac{m\lambda}{d} $$

Intensity of interference pattern

$$ \phi = \frac{2\pi d}{\lambda}\sin\theta, \quad I = 4I_0\cos^2\frac{1}{2}\phi $$

Thin-Film Interference

Half wavelength shift

Ch. 36 Diffraction

Single Slit Diffraction

(of dark fringes)

$$ \sin\theta = \frac{m\lambda}{a} $$

$$ I = I_0\left[\frac{\sin(\beta/2)}{\beta/2}\right]^2, \quad \beta = \frac{2\pi}{\lambda}a\sin\theta $$

Single-Slit diffraction envelope in double slit interference

Diffraction on Multiple Slits

Principal Maxima

sin θ_n = nλ/d

Interference Minima

(Nd is the total width of N-slits)

sin θ_s = sλ/(Nd)

Subsidiary Maxima
Single-Slit Diffraction Envelope

Diffraction Gratings

Dispersion

The angular separation of two lines whose wavelengths differ by a certain amount

$$ D = \frac{\Delta\theta}{\Delta\lambda} = \frac{m}{d\cos\theta} $$

Resolving Power

$$ \Delta\theta_\text{hw} = \frac{\lambda}{Nd\cos\theta} \Rightarrow R = \frac{\lambda}{\Delta\lambda} = Nm $$

Rayleigh’s criterion of resolvability

Ch. 37 Special Relativity

Lorentz’s Factor

$$ \gamma = \frac{1}{\sqrt{1 - u^2/c^2}} $$

Time dilation

$$ t = \frac{1}{\sqrt{1 - (v/c)^2}}t' = \gamma t' $$

Length contraction

$$ L = L'\sqrt{1 - (v/c)^2} = L'/\gamma $$

Lorentz Transformation

…

$$ u = \frac{u' + v}{1 + vu'/c^2} $$