Classical Electrodynamics 1-2

# Chapter 1 Introduction to Electrostatics

# 1.5 Another Equation of Electrostatics and the Scalar Potential

Helmholt’s Theorem ( the fundamental theorem of rector calculus )
vector field : divergence + curl ( + boundary condition )

$$\begin{align*} \displaystyle \vec{\nabla} \times \vec{E}(\vec{x}) &= \frac{1}{4\pi \epsilon_0}\int \rho (\vec{x'}) \vec{\nabla} \times \left(\frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3} \right) d^3x' \\ &= \frac{1}{4\pi \epsilon_0}\int \rho (\vec{x'}) \underbrace {\vec{\nabla} \times \left( -\vec{\nabla} \frac{1}{|\vec{x}-\vec{x'}|} \right) }_{\textcolor{blue}{ \begin{array}{l} \textcolor{blue}{=0} \\ \textcolor{blue}{\text{curl free}} \\ \textcolor{blue}{\text{only when } \frac{\partial \vec{B}}{\partial t}=0} \end{array} }} =0 \end{align*}$$ $$\vec{E}(\vec{x}) = - \frac{1}{4\pi \epsilon_0} \vec{\nabla} \int \frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|}d^3x'$$

Scalar Potential or Electrostatic Potential $\Phi$

$$\vec{E}(\vec{x})=-\vec{\nabla}\Phi(\vec{x})$$ $$\underset{\textcolor{red}{\text{all charges in the universe}}} {\Phi(\vec{x})=\frac{1}{4 \pi \epsilon_0}\int \frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|}d^3x }$$

More generally

$$\vec{\nabla}|\vec{x}-\vec{x'}|^n = n|\vec{x}-\vec{x'}|^{n-2}(\vec{x}-{x'})$$ $$\therefore \vec{\nabla} \times \vec{E}=0$$

only depends on the “central force” nature of Coulomb’s law does not require $\frac{1}{r^2}$
Physical Interpretation of $\Phi(x)$
Work done by an external force to “counteract” the $\vec{E}$ field

$$\begin{align*} W^{A \to B}&= \int_A^B \vec{F} \cdot d \vec{l} = -q \int_A^B \vec{E} \cdot d \vec{l} \\ &= q \int_A^B \Phi \cdot d \vec{l} = q \int_A^B d \Phi \\ &=q (\Phi_B-\Phi_A) \quad \textcolor{blue}{\text{independent of the path}} \end{align*}$$

For close loop

$$\oint_C \vec{E} \cdot d \vec{l} = 0$$

Stock’s Theorem (Curl theorem)

$$\oint_C \vec{A} \cdot d \vec{l} = \int_S (\vec{\nabla} \times \vec{A} \cdot \vec{n} \, da)$$

# 1.6 Surface Distributions of Charges & Dipoles & Discontinuities in the Electric Field and Potential

Surface charge density

$$\sigma (\vec{x}) = \frac{dq}{da}$$

Tangential component of $\vec{E}$

$$\oint_C \vec{E} \cdot d \vec{l} = 0$$ $$\Rightarrow (\vec{E_2}-\vec{E_1}) \cdot \vec{e_\parallel} = 0 \quad \text{for static case}$$

Normal component of $\vec{E}$ :discontinuous across $\sigma(\vec{x})$

$$\oint_{\mathbb{P}} \vec{E} \cdot \vec{n} \, da = \frac{q}{\epsilon_0} = \int_{\mathbb{A}} \frac{\sigma(\vec{x})}{\epsilon_0} \, da$$ $$\Rightarrow (\vec{E_2}-\vec{E_1}) \cdot \vec{n} = \frac{\sigma(\vec{x})}{\epsilon_0} \ne 0 \quad \text{if } \sigma(\vec{x})= 0$$

c.f.

$$\Phi(\vec{x}) = \frac{1}{4\pi \epsilon_0}\int_V \frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|} d^3x' = \frac{1}{4\pi \epsilon_0} \int_{\mathbb{S}}\frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|} da'$$ $\Rightarrow$ continuous for surface / volume charge

( also : $\vec{E}$ discontinuous across $\sigma(\vec{x})$ but bounded )

e.g. uniform spherical shell of charges

$$\sigma = \frac{Q}{4\pi a^2}$$ $$\vec{E}= \frac{q}{4 \pi \epsilon_0} \frac{\vec{e_r}}{r^2}$$

Potential :

$$\Phi (r \leq a) = \frac{Q}{4 \pi \epsilon_0 a}$$ $$\Phi (r > a) = \frac{Q}{4 \pi \epsilon_0 r}$$

Dipole Layer

Method 1

$$\Phi(\vec{x})= \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{S}} \frac{\sigma(\vec{x})}{|\vec{x}-\vec{x'}|} \, da' + \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{S}'} \frac{\overbrace{-\sigma (\vec{x'})}^{\sigma (\vec{x''})}}{\vec{x} \underbrace{-\vec{x'}+n \, d(\vec{x'})}_{-\vec{x''}}}\, \overbrace{da'}^{da''}$$

Taylor expansion

$$\frac{1}{|\vec{x}-\vec{x'}+\vec{n}d|} \underset{|\vec{x}-\vec{x'}|>>d}{\sim}\frac{1}{|\vec{x}-\vec{x'}|}+ \vec{n} d \cdot \underbrace{\vec{\nabla}\left( \frac{1}{|\vec{x}-\vec{x'}|}\right)}_{-\vec{\nabla}(\frac{1}{|\vec{x}-\vec{x'}|}) } +\Theta(d^2)$$ $$\therefore \Phi(\vec{x}) = \frac{1}{4 \pi \epsilon_0 }\int_{\mathbb{S}} \sigma(\vec{x'}) d (\vec{x'}) \vec{n} \cdot \vec{\nabla '}\left( \frac{1}{|\vec{x}-\vec{x'}|} \right) \, da'$$

Dipole Layer Strength

$$\lim_{d \to 0} \sigma(\vec{x'})d(\vec{x'})= D(\vec{x'})$$ $$\vec{n} \cdot \vec{\nabla '}\left( \frac{1}{|\vec{x}-\vec{x'}|} \right) \, da' = \vec{n} \cdot \left( \frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3} \right)\, da' = - \frac{\cos{\theta}}{r^2} da' =-d \Omega$$ $$\Rightarrow \Phi (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{S}} D (\vec{x}) \, d \Omega$$

Method 2

$\vec{p}$ : dipole moment $$\Phi (\vec{x}) = \frac{1}{4 \pi \epsilon_0}\left( \frac{q}{r_+}-\frac{q}{r_-} \right) \approxeq \frac{1}{4\pi \epsilon_0}\frac{ \overbrace{q\,d\vec{n}}^{\textcolor{blue}{\vec{p}}} \cdot (\vec{x}-\vec{x'}) }{|\vec{x}-\vec{x'}|^3}$$ $$\begin{align*} \Phi(\vec{x})& = \frac{1}{4 \pi \epsilon_0} \int da' \sigma (\vec{x'}) d (\vec{x'}) \vec{n} \cdot \frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3} \\ &= - \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{S} } D(\vec{x'})d \Omega \end{align*}$$

Discontinuity of $\Phi$ across the dipole layer
observation point : infinitesimally close to the dipole layer

$\sigma$ uniform sheet of charge $\rightarrow$ uniform flat disc with constant $D= \sigma d$ $$\Phi_2= \frac{D}{2\epsilon_0} \quad \quad \Phi_1= -\frac{D}{2\epsilon_0}$$ $$\Rightarrow \Phi_2- \Phi_1 = \frac{D}{\epsilon_0}$$

Electric Field

$$2EA = \frac{\sigma A}{\epsilon_0}$$ $$\Phi_2 - \Phi_1 = - \int \vec{E} \cdot d \vec{l} = \frac{\sigma d}{\epsilon_0}$$

# 1.7 Poisson and Laplace Equations

$$\left\{ \begin{array}{ll} \displaystyle \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0} \\ \vec{\nabla} \times \vec{E} = 0 \quad \Leftrightarrow \quad \vec{E} = - \vec{\nabla} \Phi \end{array} \right.$$

Poisson Equation

$$\Rightarrow \vec{\nabla} \cdot (\vec{\nabla}\Phi) = \boxed{\nabla^2\Phi = - \frac{\rho }{\epsilon_0}}$$

Laplace Eqution
Special case ( $\rho = 0$ )

$$\boxed{\nabla^2\Phi = 0}$$

Poisson Equation ( differential form ) + Boundary Condition $\rightarrow$ One can solve for $\Phi$ and thus $\vec{E}$

$$\Phi = \frac{1}{4\pi \epsilon_0 }\int \frac{\rho(\vec{x'})d^3x'}{|\vec{x}-\vec{x'}|}$$

which satisfies Laplace Eqution

$$\nabla^2\Phi = 0$$

start with a point charge $q$ at $\vec{x'}$

$$\Phi = \frac{1}{4\pi \epsilon_0 }\int \frac{q}{|\vec{x}-\vec{x'}|}$$ $$\rho(\vec{x}) = q \delta(\vec{x}-\vec{x'})$$ $$\nabla^2\Phi(\vec{x}) = \frac{q}{4\pi \epsilon_0 } \nabla^2 \frac{1}{|\vec{x}-\vec{x'}|} \overset{?}{=} -\frac{q}{\epsilon_0}\delta(\vec{x}-\vec{x'})$$

prove:

$$\boxed{\nabla^2 \frac{1}{|\vec{x}-\vec{x'}|}= -4 \pi \delta (\vec{x}-\vec{x'})}$$

The above statement is equivalent to proving

$$\left\{ \begin{array}{ll} \displaystyle \nabla^2 \frac{1}{|\vec{x}-\vec{x'}|}=0 \quad \text{if } \vec{x} \ne \vec{x'} \\ \displaystyle \int \nabla^2 \frac{1}{|\vec{x}-\vec{x'}|} d^3x = - 4 \pi \int \delta (\vec{x}-\vec{x'}) d^3x = - 4 \pi \end{array} \right.$$

Switch to spherical coordinate with $r \equiv |\vec{x}-\vec{x'}|$

$$\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r}\left( r^2\frac{\partial}{\partial r} \right) + \frac{1}{r^2 \sin{\theta}} \frac{\partial}{\partial \theta}\left( \sin{\theta}\frac{\partial}{\partial \theta} \right) + \frac{1}{r^2 \sin{\theta}}\frac{\partial^2}{\partial \phi^2}$$ $$\Rightarrow \nabla^2 \frac{1}{|\vec{x}-\vec{x'}|} = \nabla^2 \frac{1}{r} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( - \frac{r^2}{r^2} \right) = 0 \quad \text{if } r \ne 0$$ $$\int_{\mathbb{V}} \nabla^2 \left( \frac{1}{r} \right) d^3x = \int_{\mathbb{V}} \vec{\nabla} \cdot \left( \vec{\nabla} \frac{1}{r} \right) d^3x \overset{\text{divergence theorem}}{\Longrightarrow} \oint_{\mathbb{S}}\left( \vec{\nabla} \frac{1}{r} \right) \cdot \vec{e_r} \, da$$ $$\oint_{\mathbb{S}} \overbrace{- \frac{1}{r^2} \vec{e_r}}^{\vec{\nabla} \frac{1}{r}} \cdot \vec{e_r} \overbrace{r^2 \sin{\theta} \, d \theta \, d \phi} = -4 \pi$$

where $r \ne 0$ on the surface

Generalize to continuous charge distribution

$$\nabla^2 \overbrace{\frac{1}{4\pi \epsilon_0}}^{\Phi} \frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|} d^3x' = - \frac{4 \pi}{4 \pi \epsilon_0} \int \rho (\vec{x'}) \delta(\vec{x}-\vec{x'}) d^3x = - \frac{\rho (\vec{x'})}{\epsilon_0}$$

# 1.8 Green’s Theorem

George Green “An Essay on the Application of Mathematical
Analysis to the Theories of Electricity and Magnetism” (1828)

$\rightarrow$ Mathematical tool to handle boundary conditions

Recall divergence throrem:

$$\int_{\mathbb{V}} \vec{\nabla} \cdot \vec{A} \,d^3x = \oint_{\mathbb{S}} \vec{A} \cdot \vec{n} \, da$$

Let $\vec{A} = \phi \vec{\nabla} \psi$

$$\text{Left side : } \vec{\nabla} \cdot \vec{A} = \vec{\nabla} \cdot (\phi \vec{\nabla} \psi ) \phi \nabla^2 \psi + \vec{\nabla}\phi \cdot \vec{\nabla}\psi$$ $$\text{Right side : } \vec{A} \cdot \vec{n} = \phi \vec{\nabla} \psi \cdot n =\phi \frac{\partial}{\partial n }\psi$$

Green’s first identity

$$\int_{\mathbb{V}} (\phi \nabla^2 \psi + \vec{\nabla \phi} \cdot \vec{\nabla} {\psi} ) \, d^3x = \oint_{\mathbb{S}} \phi \frac{\partial}{\partial n} \psi \, da \quad \text{\textcircled{1}}$$

Interchange $\psi \leftrightarrow \phi$

$$\int_{\mathbb{V}} (\psi \nabla^2 \phi + \vec{\nabla \psi} \cdot \vec{\nabla} {\phi} ) \, d^3x = \oint_{\mathbb{S}} \psi \frac{\partial}{\partial n} \phi \, da \quad \text{\textcircled{2}}$$ $\text{\textcircled{1} - \textcircled{2}}$

Green’s Second identity / Green’s theorem

$$\int_{\mathbb{V}} (\phi \nabla^2 \psi - \psi \nabla^2 \phi ) \, d^3x = \oint_{\mathbb{S}} \phi \frac{\partial}{\partial n} \psi - \psi \frac{\partial}{\partial n} \phi \, da \quad \text{\textcircled{3}}$$

switch to the prime coordinate $\vec{x} \rightarrow \vec{x'}$

$$\int_{\mathbb{V}} (\phi {\nabla'}^2 \psi - \psi {\nabla'}^2 \phi ) \, d^3x' = \oint_{\mathbb{S}} \phi \frac{\partial}{\partial n'} \psi - \psi \frac{\partial}{\partial n'} \phi \, da'$$

Let $\displaystyle \psi (\vec{x'}) = \frac{1}{|\vec{x}-\vec{x'}|}$

$${\nabla'}^2\frac{1}{|\vec{x}-\vec{x'}|} = {\nabla}^2\frac{1}{|\vec{x}-\vec{x'}|} = -4 \pi \delta (\vec{x}-\vec{x'})$$

Let $\phi (\vec{x'}) = \Phi (\vec{x'})$

$${\nabla'}^2 \Phi (\vec{x'}) = - \frac{\rho (\vec{x'})}{\epsilon_0}$$ $$\text{\textcircled{3}} \rightarrow \int_{\mathbb{V}}\left\{ \Phi (\vec{x'}) [ -4 \pi \delta( \vec{x}-\vec{x'} ) ] + \frac{\rho({\vec{x'}})}{ \epsilon_0 |\vec{x}-\vec{x'}|} \right\} \, d^3x'$$ $$= \oint_{\mathbb{S}} \left( \Phi (\vec{x'} ) \frac{\partial}{\partial n'}\frac{1}{|\vec{x}-\vec{x'}|}-\frac{1}{|\vec{x}-\vec{x'}|}\frac{\partial}{\partial n'}\Phi (\vec{x'} ) \right) \, da'$$

If $\vec{x}$ lies within $\mathbb{V}$ :

$$\int_{\mathbb{V}} \Phi (\vec{x'} )\delta(\vec{x}-\vec{x'}) \, d^3x' = \Phi (\vec{x})$$

Integral Statement of $\Phi (\vec{x})$

$$\therefore \Phi (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{V}} \frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|} d^3x' + \overbrace{\frac{1}{4 \pi} \oint_{\mathbb{S}} \left[ \frac{1}{|\vec{x}-\vec{x'}|}\frac{\partial \Phi}{\partial n'}-\Phi \frac{\partial}{\partial n'}\frac{1}{|\vec{x}-\vec{x'}|} \right]\, da' }^{ \rightarrow 0 \text{ for } \mathbb{S} = \text{ infinite space }}$$

not a solution “over specification”

ref: 古典電力學一 王喬萱教授 上課講義