# Chapter 1 Introduction to Electrostatics
# 1.1 Coulomb’s Law (Coulomb’s Inverse-Square Law)
F=4πϵ01r2qq1er
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F∝q,q1
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F∝r21
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F∥er(central force)
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qq1>0 repulsive
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qq1<0 attractive
Also, principle of linear superposition
# 1.2 Electric Field
x→0limE(x)≡q→0limqF
⇒E(x)=4πϵ0q1r2er=4πϵ0q1∣x−x1∣3x−x1
principle of linear superposition
⇒E(x)=4πϵ01i=1∑nqi∣x−xi∣3x−xi
continuous charge distribution
⇒E(x)=4πϵ01∫ρ(x′)∣x−x′∣3x−x′d3x′
discrete set of point charges
ρ(x′)=i=1∑nqiδ(x′−xi)
E(x)=4πϵ01∫i∑qiδ(x′−xi)∣x−x′∣3x−x′d3x′
Dirac Delta Function
Paul Dirac 1927 “The Physical Interpretation of the Quantum Mechanics”
∗∣ψ>=∫f(x)∣x>dx
f(a)=<a∣ψ>=∫f(x)δ(x−a)<a∣x>dx
Kronecker Delta
i=m∑nfiδia={fa0if m≤a≤n, a∈Zotherwise
↓continuous analogy
Dirac Delta Function δ(x−a)
∫a1a2f(x)δ(x−a)dx={f(a)0if a1<a<a2∈Rotherwise
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δ(x−a)=0 , for x=a
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∫a1a2δ(x−a)dx=1 , if a∈(a1,a2)
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∫a1a2δ′(x−a)dx=f(x)δ(x−a)∣a1a2−∫a1a2f′(x)δ(x−a)dx=−f′(a)
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δ(f(x))=i∑∣dxdf(xi)∣1δ(x−xi)
( f(x) has simple zeros at x=xi )
# 1.3 Gauss’s Law (Gauss’s Flux Theorem)
Electric Flux 電通量 ϕ
dϕ=E⋅nda=4πϵ0qr2cosθda=4πϵ0qdΩ
ϕ=∮SE⋅nda={ϵ0q0q inside Sq outside S
Integral Form of Gauss’s Law
superpositon∮SE⋅nda=i∑ϵ0qi
continuous limit∮SE⋅nda=ϵ01∫Vρ(x)
高斯散度定理 (Gauss’s theorem)
∮SE⋅nda=ϵ01∫Vρ(x)d3x=∫Vfor arbitrary volume∇⋅Ed3x
(Using the divergence theorem)
∮SA⋅nda=∫V∇⋅Ad3x
⇔Mathematically Equivalent∇⋅E=ϵ0ρDifferential Form of Gauss’s Law
Note
- Gauss’s law also applies to moving charges(while Coulomb’s law doesn’t)
- Newton’s gravitational force field : similar form(inverse square, central force, superposition)
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r21→∇⋅E does not depend on the shape of the surface, only charge density
Accuracy of the Inverse Square Law
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r2+ϵ1
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Φ(r)=4πϵ01rqe−μr where μ=ℏmr/c
Concentric Shell Experiment
Inverse square law => Shell Theorems :
- No E inside an uniform spherical shell of charges
- Uniform spherical shell of charges exerts E as if its charges were concentrated at the center point
Q : Gauss’s Law ⇒ Coulomb’s Law ?
No ! Need special assumptions : electrostatic -> E-field symmetric with point charge
Mathematically : Need ∇⋅E , ∇×E + Boundary condition to find E
# Delta Function in 3D
- Cartesian Coordinates
x=(x1,x2,x3)=(x,y,z)
δ(x−x0)≡δ(x−x0)δ(y−y0)δ(z−z0)
d3x=dxdydz
∫free spaceδ(x−x0)d3x=∫−∞∞δ(x−x0)dx∫−∞∞δ(y−y0)dy∫−∞∞δ(z−z0)dz
- Cylindrical Coordinates
x=(ρ,ϕ,z)
δ(x−x0)≡ρ1δ(ρ−ρ0)δ(ϕ−ϕ0)δ(z−z0)
d3x=ρdρdϕdz
∫free spaceδ(x−x0)d3x=∫0∞δ(ρ−ρ0)dρ∫02πδ(ϕ−ϕ0)dϕ∫−∞∞δ(z−z0)dz
- Spherical Coordinates
x=(r,θ,ϕ)
δ(x−x0)≡⎩⎨⎧r2sinθ1δ(r−r0)δ(θ−θ0)δ(ϕ−ϕ0)r21δ(r−r0)δ(cosθ−cosθ0)δ(ϕ−ϕ0)
d3x={r2sinθdrdθdϕr2drd(cosθ)dϕ
∫free spaceδ(x−x0)d3x=⎩⎨⎧∫0∞δ(r−r0)dr∫0πδ(θ−θ0)dθ∫02πδ(ϕ−ϕ0)dϕ∫0∞δ(r−r0)dr∫−11δ(cosθ−cosθ0)dcosθ∫02πδ(ϕ−ϕ0)dϕ
# Volume Charge Density with Delta Functions
- point charge at (x0,y0,z0)
ρ(x′)=qδ(x′−x0)δ(y′−y0)δ(z′−z0)
E(x)=4πϵ01∫∣x−x′∣3ρ(x′)(x−x′)d3x′=4πϵ0q∣x−x0∣3(x−x0)∝r21
- line charge (infinite wire along the z-axis)
ρ(x′)=λδ(x′)δ(y′)
E(x)=∫−∞∞4πϵ0λ∣cex−z′ez∣3cex−z′ezdz′=4πϵ0λ∫−∞∞(c2+z′2)23excdz′=4πϵ0λex[(c2+z′2)21z′/c]z′=−∞∞=4πϵ01c2λex∝r1
- surface charge (infinite plate in the x-y plane)
ρ(x′)=σδ(z)
E(x)=4πϵ0σ∫∣(−x′)ex+(−y′)ey+cez∣3(−x′)ex+(−y′)ey+cezdx′dy′=4πϵ0σcez∫(r′2+c2)23r′drdθ=4πϵ0σ∫02πdθ′[r′2+c2−1]r′=0∞=⎩⎨⎧−2ϵ0σezc>02ϵ0σezc<0∝r01non-divergent
ref: 古典電力學一 王喬萱教授 上課講義