Summary Flashcards
Delta functions?
δ(x − a) = 0 unless x = a
±∞∫ δ(x − a) F(x) dx = F(a)
δ⁽³⁾(r − a) = δ(x − a₁) δ(y − a₂) δ(z − a₃)
Poisson’s equation?
∇²f = −4πg
Laplace’s equation?
∇²f = 0
The general solution to Poission’s equation?
f(r) = ∫ dV’ g(r’)/|r − r’|
Continuity equation?
ρ̇ + ∇ · j = 0
Electron drift velocity
I = −neAv
Force per unit charge?
f = E + v × B
Electromotive force along a line?
ε = ∫E · dl
Electric and magnetic flux through a surface S
Φₑ = ∫ₛE · dA
Φᴮ = ∫ₛB · dA
Gauss’ law
The flux of the electric field through a closed surface S is proportional to the amount of charge enclosed in the volume V that the surface defines.
Φₑ = Q/ε₀
∇ · E = ρ/ε₀
Practice deriving the differential form using the divergence theorem.
No magnetic monopoles
The flux of the magnetic field through any closed surface is zero.
Φᴮ = 0
∇ · B = 0
Practice deriving the differential form using the divergence theorem.
Faraday/Lenz’s laws
The electromotive force induced in a closed loop is equal to the rate of change of magnetic flux linked in the loop.
The electromotive force is generated in such a direction that any current flow will attempt to annul the change in flux.
ε = −d/dt Φᴮ
∇ × E + Ḃ = 0
Practice deriving the differential form using Stoke’s theorem.
Ampere-Maxwell law
The circulation of the magnetic field around a closed loop L is
proportional to the effective current passing through the surface S that the loop defines.
∫ᴸ B · dl = µ₀ ∫jᵉᶠᶠ · dA
∇ × B − µ₀ε₀Ė = µ₀j
where jᵉᶠᶠ = j + jᵈⁱˢᵖ
jᵈⁱˢᵖ = ε₀Ė
Practice deriving the differential form using Stoke’s theorem.
Equations satisfied by the electrostatic field
∇ · E = ρ/ε₀
∇ × E = 0
Electrostatic potential
E = −∇Φ
∇²Φ = −ρ/ε₀
Electrostatic solutions to Poission’s equation
Φ(r) = 1/4πε₀ ∫ dV’ ρ(r’)/|r − r’|
E(r) = 1/4πε₀ ∫ dV’ (r − r’) ρ(r’)/|r − r’|³
Potential difference from E-field
V(A→B) = −ₐᴮ∫ dl · E
Coulomb’s Law
F₁₂ = q₁E₂(r₁) = q₁q₂/4πε₀ (r₁ − r₂)/|r₁ − r₂|³ = −F₂₁
Equations satisfied by the magnetostatic field
∇ × B = µ₀j
∇ · B = 0
Magnetic vector potential with Coulomb gauge (∇ · A = 0)
B = ∇ × A
∇²A = −µ₀j
Magnetostatic solutions to vector Poission’s equation
A(r) = µ₀/4π ∫dV’ j(r’)/|r − r’|
B(r) = µ₀/4π ∫dV’ j(r’) × (r − r’)/|r − r’|³
Biot-Savart law
B(r) = µ₀I/4π ∫dl×(r − r’)/|r − r’|³
I is upper i, l is lower L
Force between two current carrying loops
not expected to memorize formula
F₁₂ = −µ₀I₁I₂/4π ∫∫ dl₁ · dl₂ (r₁ − r₂)/|r₁ − r₂|³
Electric dipole moment about r
p(r) = ∫(r’ − r)ρ(r’)dV’
Electrostatic potential and electric field of an electric dipole
not expected to memorize electric field
Φ(r) = p · r̂/4πε₀r²
E(r) = 3(p · r̂)r̂−p/4πε₀r³
must be able to derive electric field equation from first principles
Magnetic dipole moment about r
m(r) = I/2 ∫(r’ − r) × dl
Vector potential and magnetic field of a magnetic dipole
not expected to memorize magnetic field formula
A(r) = µ₀m×r̂/4πr²
B(r) = µ₀(3(m · r̂)r̂ − m)/4πr³
the magnitude of the moment (giving rise to the field) from a current loop is the current in the loop multiplied by the area
Lorenz gauge
you will be given this if you need it
∇ · A + 1/c² Φ̇ = 0
Evolution equations (inhomogeneous wave equations)
(1/c² ∂²/∂t² − ∇²)Φ = ρ/ε₀
(1/c² ∂²/∂t² − ∇²)A = µ₀j
Ohm’s Law
j = σE
Relaxation time
tᵣ = ε₀/σ
Relative permittivity / dielectric constant
εᵣ = C/C₀
where C = Q/V is the capacitance and C₀ is the capacitance in a vacuum
Potential energy and torque of electric dipoles in an external field
Uₑₓₜ = −p · Eₑₓₜ
τₑₓₜ = p × Eₑₓₜ
Polarization
P = np = χₑε₀E
second equality is only true in a linear isotropic dielectric, only these materials are studied this course
Electric susceptibility
not expected to remember formula
χₑ = n(α + p²ⁱⁿᵗʳⁱⁿˢⁱᶜ/3ε₀kT)
α ≡ 4πR₀³ is thhe atomic polarization
First term is the atomic polarization due to induced dipoles (electron moves within atom to create electron-nucleus dipole)
Second term is due to the alignment of intrinsic dipole moments.
Bound charge and surface charge when polarized
ρᵇᵒᵘⁿᵈ = −∇ · P
σ = P · n̂
Electrostatics in a dielectric
∇ · D = ρᶠʳᵉᵉ
where D = ε₀E + P and ∇ × E = 0
D = (1 + χₑ)ε₀E = εᵣε₀E
∫ₐ D · dA = Qᶠʳᵉᵉ
∫ᴸ E · dl = 0
U = ½ ∫D · E dV
D is the electric displacement vector.
Junction conditions in a dielectric
D⊥ is continuous and E|| is continuous
Relative permeability
µᵣ = L/L₀
where L = −V/ İ is the inductance and L₀ is the inductance in a vacuum.
Magnetic dipoles in an external field
Uₑₓₜ = −m · Bₑₓₜ
τₑₓₜ = m × Bₑₓₜ
Magnetization
M = nm = χᴮ/µ₀ B
m = IAn̂ is the average magnetic dipole moment, and A is the area of the dipole loop
second equality only true in a linear isotropic magnetic material ie. one which is diamagnetic (χB < 0) or paramagnetic (χB > 0).
The response is not linear in ferromagnetic materials such as iron, nickel and cobalt.
Magnetic susceptibility
not be expected to remember nor derive formula
χᴮ = nµ₀(m²ⁱⁿᵗʳⁱⁿˢⁱᶜ/3kT − Ze²r₀²/6mₑ)
First term is due to paramagnetism, the alignment of intrinsic dipole moments.
Second term is due to diamagnetism, the induced dipoles due
to an external field
Must know the (different) directions of the induced and intrinsic
effects.
dia- para- and ferro- magnetism
χB < 0 corresponds to diamagnetism – equivalent to induced polarization. to do with change in atoms’ angular momentum. oposite direction to B
χB > 0 corresponds to paramagnetism – equivalent to alignment polarization. comes from electrons orbiting in atoms. same direction as B
M = M(B), aka nonlinear magnetization, is ferromagnetism
Bound current and surface current
jᵇᵒᵘⁿᵈ = ∇ × M
jˢᵘʳᶠᵃᶜᵉ = M × n̂
Magnetostatics in a magnet
∇ × H = jᶠʳᵉᵉ
where H = 1/µ₀ B − M and ∇ · B = 0
B = µ₀H/1−χᴮ = µᵣµ₀H
∫ᴸ H · dl = Iᶠʳᵉᵉ
∫ₐ B · dA = 0
U = ½∫H · B dV
Junction conditions between magnetic materials
B⊥ is continuous and H|| is continuous
Ideal ferromagnets
In an ideal ferromagnet M = constant which implies
that jᵇᵒᵘⁿᵈ = 0 and jˢᵘʳᶠᵃᶜᵉ = M × n̂
Magnetostatics in the absence of a free current
∇ × H = 0
∇ · H = ρₘ = −∇ · M
H = −∇ψ → ∇²ψ = −ρₘ = ∇ · M
ψ = −1/4π ∇ · ∫dV’ M(r’)/|r − r’|
Non-linear magnetism
B = f(H) and µᵣ = 1/µ₀ ∂B/∂H
Work done per unit volume:
w = −∫H · dB = area of hysteresis curve
Maxwell’s equations in free space - ie no charge or current
∇ · E = 0
∇ · B = 0
Ė = c²∇ × B
Ḃ = −∇ × E
Wave equations satisfied by E and B
1/c² Ë = ∇²E
1/c² B̈ = ∇²B
Plane wave solutions
E = E₀ℜ[exp( i(k·r−ωt))]
B = B₀ℜ[exp( i(k·r−ωt))] = k×E₀/ω ℜ[exp( i(k·r−ωt))]
|B₀|=|E₀|/c and k̂ = Ê₀ × B̂₀
Poynting Vector
Microscopic definition:
* N = E×B/µ₀, which is the energy flux.
* For the plane wave solution |N|ₘₐₓ = |E₀|²/Z₀ and the irradiance, or intensity, is given by ⟨|N|⟩ = |E₀| where Z₀ = √µ₀/ε₀’ = 377Ω is the impedance of free space.
Macroscopic definition (materials): N = E × H. Only equal to the above if µᵣ = 1.
The difference reflects whether the energy spent driving currents relates to j or jᶠʳᵉᵉ
Radiation Pressure
Pᵣ = ⟨|N|⟩/c
Fᵣ = A⟨|N|⟩/c
if all the radiation is absorbed
Fᵣ = 2A⟨|N⟩|/c
if all the radiation is reflected
EM waves in the presence of current
1/c²Ë = ∇²E − µ₀ ∂j/∂t
1/c²B̈ = ∇²B + µ₀∇ × j
Dispersion relation of a conductor
k² = ω²/c² + iµ₀σω
ω²/c² ≪ µ₀σω, so this simplifies to k² = iµ₀σω
Skin depth
In a conductor, the displacement current is negligible
0 = ∇²E − µ₀ ∂j/∂t
E = E₀exp(−z/δ)exp( i(√µ₀σω/2’ z−ωt))
where δ = √2/µ0σω’ is the skin depth.
Current in a plasma
∂j/∂t = e²nₑ/mₑ E
Dispersion relation of plasma
ck = ±√ω² − ωₚ²’
Plasma frequency
Waves do not
propagate for frequencies below the plasma frequency which is
fₚ = 1/2π ωₚ = 1/2π √e²nₑ/ε₀mₑ’
Reflection of EM waves at a perfect conductor
- The phase of incident and reflected waves must be continuous at the boundary.
- This leads to the conclusion that the angle of incidence and the angle of reflection are equal.
- E = 0 in the conductor and E|| is continuous at the surface.
- The polarization of the reflected wave can be calculated from that of the incident wave.
- The sum of the two, Eₜₒₜ = E⁽ᴿ⁾ + E⁽ᴵ⁾, corresponds to standing waves.
Polarization
Linear, Circular, Pure right circular, Pure left circular
Linear polarization: E = (E₀ₓx̂ + E₀ᵧŷ) ℜ[exp( i(k·r−ωt+θ))]
Circular polarization: E = ℜ[(Eₗl̂ + Eᵣr̂)exp( i(k·r−ωt))]
where l̂ = 1/√2’ (x̂ + iŷ) and r̂ = 1/√2’ (x̂ − iŷ)
Pure right circular polarization: E = 1/√2’ Eᵣ(x̂cos(k·r − ωt) + ŷsin(k·r − ωt))
Pure left circular polarization : E = 1/√2’ Eᵣ(x̂cos(k·r − ωt) − ŷsin(k·r − ωt))
Stokes parameters
I = |Eₓ|² + |Eᵧ|²
Q = |Eₓ|² − |Eᵧ|²
U = |Eₐ|² − |Eᵇ|² = EᵧEₓ + EₓEᵧ = 2ℜ(EᵧEₓ)
V = |Eₗ|² − |Eᵣ|² = i(EᵧEₓ − EᵧEₓ) = 2{Im} EᵧEₓ)
Not expected to remember these but should know that
* I² = Q² + U² + V²
* The fractional linear polarization is √Q² + U²’/I
* The fractional circular polarization is V/I
EM waves in a dielectric and magnetic media
In time dependent cases the bound current is modified to jᵇᵒᵘⁿᵈ = ∇×M+Ṗ and Maxwell’s equations can be written as
∇ · D = ρᶠʳᵉᵉ
∇ · B = 0
Ḃ = −∇ × E
Ḋ = ∇ × H − jᶠʳᵉᵉ
where D = ε₀E + P and B = µ₀(H + M)
Refractive index
When there is no free charge or current and D = εᵣε₀E and B = µᵣµ₀H (ie linear isotropic materials)
Ė = v²∇ × B → 1/v² Ë = ∇²E
where v = c/n and n = √εᵣµᵣ’ is the refractive index.
Snell’s law
At the boundary between two materials with refractive indices nᵢ and nₜ (I refers to incident and T to transmitted)
nᵢ sinθᵢ = nₜ sinθₜ
Conductivity
σ dS = 1/R dL
What is relaxation time?
tᴿ is timescale for movement of conduction electrons to smooth out non-uniformities in the charge density.
Typical values:
8 × 10⁻¹⁹ s for a metal with σ ≈ 10⁷ Ω⁻¹ m⁻¹
8 × 10³ s for an insulator with σ ≈ 10⁻¹⁵ Ω⁻¹ m⁻¹
ᴿ should be subscript
Describe the method of images
Questions invlolving perfect conductors can be solved using an “image charge”. Electric potential is zero in a conductor, which is the same as if there was a charge-anticharge pair and the midpoint was where the conducting plane should be.
Poissons equation can be solved for this fictional scenario, and due to the uniqueness theorem this must be the solution for the real plane.
Force due to an “image charge”
F = qEⁱᵐᵃᵍᵉ(r = a)
where Eⁱᵐᵃᵍᵉ = −q/4πε₀ r+a/|r+a|³
Capacitance
C = Q/∆V = ε₀A/d
Energy of a parallel plate capacitor
U = ½ε₀∫|E|²dV = ½ε₀(∆V/d)²Ad = ½C(∆V)²
In terms of electric displacement vector
U = ½CV² = ½Aεᵣε₀/d (Eᶻd)² = ½AdDᶻEᶻ = ½∫D · E dV
What are dielectrics
Electrical insulators with very few free electrons.
When an electric field is applied to them, intrinsic dipoles become aligned and atoms/molecules with no intrinsic dipole moment become polarized, creating dipoles
Interface between dielectrics
E|| continuous, so E₁sinθ₁ = E₂sinθ₂
D⟂ continuous, so D₁cosθ₁ = D₂cosθ₂
D₁/E₁ cotθ₁ = D₂/E₂ cotθ₂
εᵣ⁽¹⁾cotθ₁ = εᵣ⁽²⁾cotθ₂
Inductance and relative permeability
Inductance
L ≡ −ε/İ = µ₀N²πr²l
(unit is Henry [H])
Derived using amperes law in solenoid, calculating magnetic flux, and considering faradays law
Relative permeability
µᵣ ≡ L/Lᵛᵃᶜ
where Lᵛᵃᶜ is the inductance in vacuum
İ is capital i with dot and l is lower L
Relative permeability of a ferromagnet
µᵣ = 1/µ₀ ∂B/∂H
Magnetic intensity vector
H ≡ 1/µ0(B − M)
Ampere’s law becomes
∇ × H = jfree
∮H · dL = Ifree
