# Chapter 1 Introduction to Electrostatics

# 1.1 Coulomb’s Law (Coulomb’s Inverse-Square Law)

F=14πϵ0qq1r2er \begin{gathered} \vec{F}=\frac 1 {4\pi{\epsilon}_0} \frac {qq_1} {r^2} \vec{e_r} \end{gathered}
  • Fq,q1 \vec{F} \propto q,q_1
  • F1r2 \vec{F} \propto \frac 1 {r^2}
  • Fer \vec{F} \parallel \vec{e_r} (central force)
  • qq1>0 qq_1 > 0 repulsive
  • qq1<0 qq_1 < 0 attractive

Also, principle of linear superposition

# 1.2 Electric Field

limx0E(x)limq0Fq \begin{gathered} \lim_{x \to 0} \vec{E}(\vec{x})\equiv\lim_{q \to 0}\frac {\vec{F}} q \end{gathered} E(x)=q14πϵ0err2=q14πϵ0xx1xx13 \begin{gathered} \Rightarrow\vec{E}(\vec{x})=\frac {q_1} {4\pi\epsilon_0 }\frac {\vec{e_r}} {r^2}= \frac {q_1}{4\pi\epsilon_0} \frac {\vec{x}-\vec{x_1}}{|\vec{x}-\vec{x_1}|^3} \end{gathered}

principle of linear superposition

E(x)=14πϵ0i=1nqixxixxi3 \begin{gathered} \Rightarrow\vec{E}(\vec{x})= \frac 1 {4\pi\epsilon_0} \sum_{i=1}^{n}q_i \frac {\vec{x}-\vec{x_i}}{|\vec{x}-\vec{x_i}|^3} \end{gathered}

continuous charge distribution

E(x)=14πϵ0ρ(x)xxxx3d3x \begin{gathered} \Rightarrow\vec{E}(\vec{x})= \frac 1 {4\pi\epsilon_0} \int \rho (\vec{x'}) \frac {\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3} d^3x' \end{gathered}

discrete set of point charges

ρ(x)=i=1nqiδ(xxi) \begin{gathered} \rho (\vec{x'})= \sum_{i=1}^{n}q_i\delta(\vec{x'}-\vec{x_i}) \end{gathered} E(x)=14πϵ0iqiδ(xxi)xxxx3d3x \begin{gathered} \vec{E}(\vec{x})= \frac 1 {4\pi\epsilon_0} \int \sum_i q_i \delta(\vec{x'}-\vec{x_i}) \frac {\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3} d^3x' \end{gathered}

Dirac Delta Function
Paul Dirac 1927 “The Physical Interpretation of the Quantum Mechanics”

ψ>=f(x)x>dx \begin{gathered} *|\psi>=\int f(x)|x>dx \end{gathered} f(a)=<aψ>=f(x)<ax>δ(xa)dx \begin{gathered} f(a)=<a|\psi>=\int f(x)\underbrace{<a|x>}_ {\delta (x-a)}dx \end{gathered}

Kronecker Delta

i=mnfiδia={faif man, aZ0otherwise \sum_{i=m}^n f_i \delta_{ia} = \left\{ \begin{array}{ll} f_a & \text{if } m \le a \le n,\ a \in \mathbb{Z} \\ 0 & \text{otherwise} \end{array} \right. continuous analogy  \begin{gathered} \textcolor{blue}{\downarrow \text{continuous analogy }} \end{gathered}

Dirac Delta Function δ(xa) \delta (x-a)

a1a2f(x)δ(xa)dx={f(a)if a1<a<a2R0otherwise \int_{a_1}^{a_2} f(x) \delta (x-a) dx= \left\{ \begin{array}{ll} f(a) & \text{if } a_1 < a < a_2 \in \mathbb{R} \\ 0 & \text{otherwise} \end{array} \right.
  • δ(xa)=0 , for xa \displaystyle\delta (x-a) = 0 \text{ , for }x \ne a
  • a1a2δ(xa)dx=1 , if a(a1,a2) \displaystyle\int_{a_1}^{a_2} \delta (x-a)dx = 1 \text{ , if } a \in (a_1,a_2)
  • a1a2δ(xa)dx=f(x)δ(xa)a1a2a1a2f(x)δ(xa)dx=f(a) \displaystyle\int_{a_1}^{a_2} \delta'(x-a) dx = f(x) \delta (x-a) |_{a_1}^{a_2}- \int_{a_1}^{a_2}f'(x) \delta (x-a)dx =-f'(a)
  • δ(f(x))=i1dfdx(xi)δ(xxi) \displaystyle\delta (f(x))= \sum_{i} \frac{1}{|\frac{df}{dx}(x_i)|} \delta (x-x_i)

( f(x) f(x) has simple zeros at x=xi x=x_i )

# 1.3 Gauss’s Law (Gauss’s Flux Theorem)

Electric Flux 電通量 ϕ \phi

dϕ=Enda=q4πϵ0cosθr2da=q4πϵ0dΩ d \phi = \vec{E} \cdot \vec{n} \, da = \frac{q}{4 \pi \epsilon_0} \frac{cos \theta }{r^2}\,da = \frac{q}{4 \pi \epsilon_0}\,d\Omega ϕ=SEnda={qϵ0q inside S0q outside S \phi = \oint_S \vec{E} \cdot \vec{n} \, da = \left\{ \begin{array}{ll} \frac{q}{\epsilon_0} & q \text{ inside } S \\ 0 & q \text{ outside } S \\ \end{array} \right.

Integral Form of Gauss’s Law

superpositonSEnda=iqiϵ0 \textcolor{red}{\xrightarrow{\text{superpositon}}} \oint_S \vec{E} \cdot \vec{n} \, da = \sum_i \frac{q_i}{\epsilon_0} continuous limitSEnda=1ϵ0Vρ(x) \textcolor{blue}{\xrightarrow{\text{continuous limit}}} \oint_S \vec{E} \cdot \vec{n} \, da = \frac{1}{\epsilon_0} \int_{\mathbb{V}} \rho (\vec{x} )

# 1.4 Differential Form of Gauss’s Law

高斯散度定理 (Gauss’s theorem)

SEnda=1ϵ0Vρ(x)d3x=VEfor arbitrary volumed3x \oint_S \vec{E} \cdot \vec{n} \, da = \frac{1}{\epsilon_0} \int_{\mathbb{V}} \rho (\vec{x}) \, d^3 x = \int_{\mathbb{V}} \underbrace{\vec{\nabla} \cdot \vec{E}}_{\textcolor{blue}{\text{for arbitrary volume}}} \, d^3 x

(Using the divergence theorem)

SAnda=VAd3x \oint_S \vec{A} \cdot \vec{n} \, da = \int_{\mathbb{V}} \vec{\nabla} \cdot \vec{A} \, d^3 x E=ρϵ0Mathematically EquivalentDifferential Form of Gauss’s Law \Leftrightarrow \underset{\text{Mathematically Equivalent}}{ \boxed{ \vec{\nabla} \cdot \vec{E}= \frac{\rho}{\epsilon_0} }} \textcolor{red}{\text{Differential Form of Gauss's Law}}

Note

  1. Gauss’s law also applies to moving charges(while Coulomb’s law doesn’t)
  2. Newton’s gravitational force field : similar form(inverse square, central force, superposition)
  3. 1r2E \displaystyle\frac{1}{r^2} \rightarrow \vec{\nabla} \cdot \vec{E} does not depend on the shape of the surface, only charge density

Accuracy of the Inverse Square Law

  1. 1r2+ϵ \displaystyle\frac{1}{r^{2+\epsilon}}
  2. Φ(r)=14πϵ0qreμr \displaystyle \Phi(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \, e^{-\mu r} where μ=mr/c \displaystyle \mu = \frac{m_r/c}{\hbar}

Concentric Shell Experiment
Inverse square law => Shell Theorems :

  1. No E \vec{E} inside an uniform spherical shell of charges
  2. Uniform spherical shell of charges exerts E \vec{E} as if its charges were concentrated at the center point

Q : Gauss’s Law \Rightarrow Coulomb’s Law ?

No ! Need special assumptions : electrostatic -> E-field symmetric with point charge
Mathematically : Need E , ×E \vec{\nabla} \cdot \vec{E} \text{ , } \vec{\nabla} \times \vec{E} + Boundary condition to find E \vec{E}

# Delta Function in 3D

  1. Cartesian Coordinates
x=(x1,x2,x3)=(x,y,z) \vec{x}=(x_1,x_2,x_3)= (x,y,z) δ(xx0)δ(xx0)δ(yy0)δ(zz0) \delta(\vec{x}-\vec{x_0})\equiv \delta(x-x_0)\delta(y-y_0)\delta(z-z_0) d3x=dxdydz d^3x=dxdydz free spaceδ(xx0)d3x=δ(xx0)dxδ(yy0)dyδ(zz0)dz \int_{\text{free space}} \delta (\vec{x}-\vec{x_0})d^3x=\int_{-\infty}^{\infty} \delta(x-x_0)dx\int_{-\infty}^{\infty} \delta(y-y_0)dy\int_{-\infty}^{\infty} \delta(z-z_0)dz
  1. Cylindrical Coordinates
x=(ρ,ϕ,z) \vec{x}= (\rho,\phi,z) δ(xx0)1ρδ(ρρ0)δ(ϕϕ0)δ(zz0) \delta(\vec{x}-\vec{x_0}) \equiv \frac{1}{\rho} \delta(\rho- \rho_0)\delta(\phi- \phi_0)\delta(z-z_0) d3x=ρdρdϕdz d^3x=\rho d\rho d\phi dz free spaceδ(xx0)d3x=0δ(ρρ0)dρ02πδ(ϕϕ0)dϕδ(zz0)dz \int_{\text{free space}} \delta (\vec{x}-\vec{x_0})d^3x=\int_{0}^{\infty} \delta(\rho- \rho_0)d \rho \int_{0}^{2\pi} \delta(\phi-\phi_0)d\phi\int_{-\infty}^{\infty} \delta(z-z_0)dz
  1. Spherical Coordinates
x=(r,θ,ϕ) \vec{x}= (r,\theta,\phi) δ(xx0){1r2sinθδ(rr0)δ(θθ0)δ(ϕϕ0)1r2δ(rr0)δ(cosθcosθ0)δ(ϕϕ0) \delta(\vec{x}-\vec{x_0}) \equiv \left\{ \begin{array}{ll} \displaystyle\frac{1}{r^2\sin{\theta}}\delta (r-r_0) \delta(\theta-\theta_0) \delta(\phi-\phi_0) \\ \displaystyle\frac{1}{r^2}\delta (r-r_0) \delta(\cos{\theta}-cos{\theta_0}) \delta(\phi-\phi_0) \end{array} \right. d3x={r2sinθdrdθdϕr2drd(cosθ)dϕ d^3x= \left\{ \begin{array}{ll} \displaystyle r^2\sin{\theta}drd\theta d\phi \\ \displaystyle r^2drd(\cos{\theta})d\phi \end{array} \right. free spaceδ(xx0)d3x={0δ(rr0)dr0πδ(θθ0)dθ02πδ(ϕϕ0)dϕ0δ(rr0)dr11δ(cosθcosθ0)dcosθ02πδ(ϕϕ0)dϕ \int_{\text{free space}} \delta (\vec{x}-\vec{x_0})d^3x= \left\{ \begin{array}{ll} \displaystyle \int_{0}^{\infty} \delta(r-r_0)dr \int_{0}^{\pi} \delta(\theta- \theta_0)d \theta \int_{0}^{2\pi} \delta(\phi-\phi_0)d \phi \\ \displaystyle \int_{0}^{\infty} \delta(r-r_0)dr \int_{-1}^{1} \delta(\cos{\theta}- cos{\theta_0})d cos{\theta} \int_{0}^{2\pi} \delta(\phi- \phi_0)d \phi \end{array} \right.

# Volume Charge Density with Delta Functions

  1. point charge at (x0,y0,z0) (x_0,y_0,z_0)
ρ(x)=qδ(xx0)δ(yy0)δ(zz0) \rho (\vec{x'}) = q \delta (x'-x_0) \delta (y'-y_0) \delta (z'-z_0) E(x)=14πϵ0ρ(x)(xx)xx3d3x=q4πϵ0(xx0)xx031r2 \vec{E}(\vec{x}) = \frac{1}{4\pi \epsilon_0} \int \frac{\rho (\vec{x'})(\vec{x}-\vec{x'})}{|\vec{x}-\vec{x'}|^3 }d^3x' = \frac{q}{4\pi \epsilon_{0}}\frac{(\vec{x}-\vec{x_0})}{|\vec{x}-\vec{x_0}|^3} \textcolor{red}{\propto \frac{1}{r^2} }
  1. line charge (infinite wire along the z-axis)
ρ(x)=λδ(x)δ(y) \rho (\vec{x'}) = \lambda \delta (x') \delta (y') E(x)=λ4πϵ0cexzezcexzez3dz=λ4πϵ0excdz(c2+z2)32=λ4πϵ0ex[z/c(c2+z2)12]z==14πϵ02λcex1r \begin{align*} \displaystyle E(\vec{x}) &=\int_{-\infty}^{\infty} \frac{\lambda}{4\pi \epsilon_0} \frac{c \vec{e_x} - z' \vec{e_z}}{|c \vec{e_x} - z' \vec{e_z}|^3}dz' \\ &=\frac{\lambda}{4\pi \epsilon_0} \int_{-\infty}^{\infty} \frac{\vec{e_x}c \, dz'}{(c^2+z'^2)^{\frac{3}{2}}} \\ &=\frac{\lambda}{4\pi \epsilon_0}\vec{e_x}\left[\frac{z'/c}{(c^2+z'^2)^{\frac{1}{2}}}\right]_{z'=-\infty}^{\infty} \\ &=\frac{1}{4\pi \epsilon_0} \frac{2\lambda}{c}\vec{e_x}\quad \textcolor{red}{\propto \frac{1}{r}} \end{align*}
  1. surface charge (infinite plate in the x-y plane)
ρ(x)=σδ(z) \rho (\vec{x'}) = \sigma \delta (z) E(x)=σ4πϵ0(x)ex+(y)ey+cez(x)ex+(y)ey+cez3dxdy=σcez4πϵ0rdrdθ(r2+c2)32=σ4πϵ002πdθ[1r2+c2]r=0={σ2ϵ0ezc>0σ2ϵ0ezc<01r0non-divergent \begin{align*} \displaystyle \vec{E}(\vec{x})&=\frac{\sigma}{4\pi \epsilon_0} \int \frac{(-\vec{x'})\vec{e_x}+(-\vec{y'})\vec{e_y}+c\vec{e_z}}{|(-\vec{x'})\vec{e_x}+(-\vec{y'})\vec{e_y}+c\vec{e_z}|^3} \, dx' \, dy'\\ &= \frac{\sigma c \vec{e_z}}{4\pi \epsilon_0} \int \frac{r' \, dr \, d\theta}{(r'^2+c^2)^{\frac{3}{2}}} \\ &= \frac{\sigma}{4 \pi \epsilon_0} \int_{0}^{2\pi}d\theta ' \left[ \frac{-1}{\sqrt{r'^2+c^2}} \right]_{r'=0}^{\infty} \\ &= \left\{ \begin{align*} \displaystyle &\frac{\sigma}{2 \epsilon_0} \vec{e_z} \quad c>0 \\ \displaystyle -&\frac{\sigma}{2 \epsilon_0} \vec{e_z} \quad c<0 \end{align*} \right. \quad \textcolor{red}{\propto \frac{1}{r_0} \quad \text{non-divergent}} \end{align*}

ref: 古典電力學一 王喬萱教授 上課講義