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