# Chapter 1 Introduction to Electrostatics# 1.9 Uniqueness of the Solution With Dirichlet or Neumann Boundary Conditionappropriate B.C. on the closed surface S \mathbb{S} S ∇ 2 Φ = − ρ / ϵ 0 \nabla^2 \Phi = - \rho / \epsilon_0 ∇ 2 Φ = − ρ / ϵ 0 V \mathbb{V} V
Dirichlet B.C. : Φ Neumann B.C. : ∂ Φ ∂ n Cauchy B.C. : Φ & ∂ Φ ∂ n → over specification
\begin{aligned}
\text{Dirichlet B.C. :} &\quad \Phi \\[0.8em]
\text{Neumann B.C. :} &\quad \dfrac{\partial \Phi}{\partial n} \\[0.8em]
\text{Cauchy B.C. :} &\quad \Phi \;\&\; \dfrac{\partial \Phi}{\partial n} \rightarrow \text{over specification}
\end{aligned}
Dirichlet B.C. : Neumann B.C. : Cauchy B.C. : Φ ∂ n ∂ Φ Φ & ∂ n ∂ Φ → over specification prove : If ∃ Φ 1 , Φ 2 \exist \, \, \Phi_1 , \Phi_2 ∃ Φ 1 , Φ 2
{ Φ 1 = Φ 2 = Φ D on S ∂ Φ 1 ∂ n = ∂ Φ 1 ∂ n = Φ N ′
\left\{
\begin{aligned}
\displaystyle \Phi_1 = \Phi_2 = \Phi_D \text{ on } \mathbb{S} \\
\displaystyle \frac{\partial \Phi_1}{\partial n} = \frac{\partial \Phi_1}{\partial n} = \Phi_N'
\end{aligned}
\right.
⎩ ⎨ ⎧ Φ 1 = Φ 2 = Φ D on S ∂ n ∂ Φ 1 = ∂ n ∂ Φ 1 = Φ N ′ U = Φ 2 − Φ 1
U= \Phi_2 - \Phi_1
U = Φ 2 − Φ 1 ∇ 2 Φ 1 = ∇ 2 Φ 2 = − ρ / ϵ 0
\nabla^2 \Phi_1 = \nabla^2 \Phi_2 = - \rho / \epsilon_0
∇ 2 Φ 1 = ∇ 2 Φ 2 = − ρ / ϵ 0 From Green’s first identity, take ϕ = ψ = U \phi = \psi = U ϕ = ψ = U
∫ V ( U ∇ 2 U + ∣ ∇ U ∣ 2 ) d 3 x = ∮ S U ∂ ∂ n U d a
\int_{\mathbb{V}}(U \nabla^2 U +|\nabla U|^2)d^3x = \oint_{\mathbb{S}} U \frac{\partial }{\partial n} U da
∫ V ( U ∇ 2 U + ∣∇ U ∣ 2 ) d 3 x = ∮ S U ∂ n ∂ U d a If U U U ( ∇ 2 U = 0 ) (\nabla^2 U = 0) ( ∇ 2 U = 0 )
∫ V ∣ ∇ U ∣ 2 d 3 x = ∮ S U ∂ ∂ n U d a
\int_{\mathbb{V}}|\nabla U|^2d^3x = \oint_{\mathbb{S}} U \frac{\partial }{\partial n} U da
∫ V ∣∇ U ∣ 2 d 3 x = ∮ S U ∂ n ∂ U d a Use Dirichlet B.C. ( U = 0 on S ) (U=0 \text{ on } \mathbb{S}) ( U = 0 on S )
∫ V ∣ ∇ ⃗ U ∣ 2 d 3 x = 0 ⇒ ∇ ⃗ U = 0 ⇒ U = Φ 2 − Φ 1 = 0
\int_{\mathbb{V}} |\vec{\nabla}U|^2 d^3x = 0 \Rightarrow \vec{\nabla}U = 0 \Rightarrow U = \Phi_2-\Phi_1 = 0
∫ V ∣ ∇ U ∣ 2 d 3 x = 0 ⇒ ∇ U = 0 ⇒ U = Φ 2 − Φ 1 = 0 Use Neumann B.C. ( ∂ U ∂ n = 0 on S ) (\frac{\partial U}{\partial n}=0 \text{ on } \mathbb{S}) ( ∂ n ∂ U = 0 on S )
∫ V ∣ ∇ ⃗ U ∣ 2 d 3 x = c o n s t ⇒ ∇ ⃗ U = c o n s t ⇒ U = Φ 2 − Φ 1 = c o n s t
\int_{\mathbb{V}} |\vec{\nabla}U|^2 d^3x = const \Rightarrow \vec{\nabla}U = const \Rightarrow U = \Phi_2-\Phi_1 = const
∫ V ∣ ∇ U ∣ 2 d 3 x = co n s t ⇒ ∇ U = co n s t ⇒ U = Φ 2 − Φ 1 = co n s t Exercise
∇ ⃗ ∣ x ⃗ − x ′ ⃗ ∣ n = n ∣ x ⃗ − x ′ ⃗ ∣ n − 2 ( x ⃗ − x ′ ⃗ )
\vec{\nabla}|\vec{x}-\vec{x'}|^n = n|\vec{x}-\vec{x'}|^{n-2}(\vec{x}-\vec{x'})
∇ ∣ x − x ′ ∣ n = n ∣ x − x ′ ∣ n − 2 ( x − x ′ ) solution
∇ ⃗ ∣ x ⃗ − x ′ ⃗ ∣ n = [ ∂ ∂ x e x ⃗ + ∂ ∂ y e y ⃗ + ∂ ∂ z e z ⃗ ] [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] n 2 = [ 2 ( x − x ′ ) e x ⃗ + 2 ( y − y ′ ) e y ⃗ + 2 ( z − z ′ ) e z ⃗ ] [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] n 2 − 1 = n ∣ x ⃗ − x ′ ⃗ ∣ n − 2 ( x ⃗ − x ′ ⃗ )
\begin{aligned}
\vec{\nabla}|\vec{x}-\vec{x'}|^n &=\left[
\frac{\partial}{\partial x} \vec{e_x} +
\frac{\partial}{\partial y} \vec{e_y} +
\frac{\partial}{\partial z} \vec{e_z}
\right] \left[
(x-x')^2 +
(y-y')^2 +
(z-z')^2
\right]^{\frac{n}{2}} \\
\displaystyle
&= \left[
2(x-x') \vec{e_x} +
2(y-y') \vec{e_y} +
2(z-z') \vec{e_z}
\right] \left[
(x-x')^2 +
(y-y')^2 +
(z-z')^2
\right]^{\frac{n}{2}-1} \\
\displaystyle
&= n|\vec{x}-\vec{x'}|^{n-2}(\vec{x}-\vec{x'})
\end{aligned}
∇ ∣ x − x ′ ∣ n = [ ∂ x ∂ e x + ∂ y ∂ e y + ∂ z ∂ e z ] [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 2 n = [ 2 ( x − x ′ ) e x + 2 ( y − y ′ ) e y + 2 ( z − z ′ ) e z ] [ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] 2 n − 1 = n ∣ x − x ′ ∣ n − 2 ( x − x ′ ) Note : ∇ ⃗ \vec{\nabla} ∇ x ⃗ \vec{x} x ∇ ′ ⃗ \vec{\nabla'} ∇ ′ x ′ ⃗ \vec{x'} x ′
∇ ′ ⃗ ∣ x ⃗ − x ′ ⃗ ∣ n = − ∇ ⃗ ∣ x ⃗ − x ′ ⃗ ∣ n
\vec{\nabla'}|\vec{x}-\vec{x'}|^n = - \vec{\nabla}|\vec{x}-\vec{x'}|^n
∇ ′ ∣ x − x ′ ∣ n = − ∇ ∣ x − x ′ ∣ n e.g. : n = − 1 n = -1 n = − 1
∇ ⃗ 1 ∣ x ⃗ − x ′ ⃗ ∣ = x ⃗ − x ′ ⃗ ∣ x ⃗ − x ′ ⃗ ∣ 3 ( = − e r ⃗ r 2 )
\vec{\nabla} \frac{1}{|\vec{x}-\vec{x'}|} = \frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3} (=-\frac{\vec{e_r}}{r^2})
∇ ∣ x − x ′ ∣ 1 = ∣ x − x ′ ∣ 3 x − x ′ ( = − r 2 e r ) ⇒ E ⃗ = 1 4 π ϵ 0 ∫ ρ ( x ′ ⃗ ) ( x ⃗ − x ′ ⃗ ) ∣ x ⃗ − x ′ ⃗ ∣ 3 d 3 x ′ = ∇ ⃗ 1 4 π ϵ 0 ∫ ρ ( x ′ ⃗ ) ∣ x ⃗ − x ′ ⃗ ∣ d 3 x ′ = − ∇ ⃗ Φ
\Rightarrow \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho (\vec{x'})(\vec{x}-\vec{x'})}{|\vec{x}-\vec{x'}|^3} d^3x' = \vec{\nabla} \frac{1}{4 \pi \epsilon_0} \int \frac{\rho (\vec{x'})}{|\vec{x}-\vec{x'}|} d^3x' = -\vec{\nabla}\Phi
⇒ E = 4 π ϵ 0 1 ∫ ∣ x − x ′ ∣ 3 ρ ( x ′ ) ( x − x ′ ) d 3 x ′ = ∇ 4 π ϵ 0 1 ∫ ∣ x − x ′ ∣ ρ ( x ′ ) d 3 x ′ = − ∇ Φ Green Function : a class of function satisfying
∇ ′ 2 G ( x ⃗ , x ′ ⃗ ) = 4 π δ ( x ⃗ − x ′ ⃗ )
\nabla '^2 G(\vec{x},\vec{x'}) = 4 \pi \delta (\vec{x}-\vec{x'})
∇ ′2 G ( x , x ′ ) = 4 π δ ( x − x ′ ) where
G ( x ⃗ , x ′ ⃗ ) = 1 ∣ x ⃗ , x ′ ⃗ ∣ + F ( x ⃗ , x ′ ⃗ )
G(\vec{x},\vec{x'}) = \frac{1}{|\vec{x},\vec{x'}|} + F(\vec{x},\vec{x'})
G ( x , x ′ ) = ∣ x , x ′ ∣ 1 + F ( x , x ′ ) ∇ ′ 2 F ( x ⃗ , x ′ ⃗ ) = 0 → satisfying Laplace equation
\nabla'^2 F (\vec{x},\vec{x'}) = 0 \rightarrow \text{satisfying Laplace equation}
∇ ′2 F ( x , x ′ ) = 0 → satisfying Laplace equation Let
{ ψ = G ( x ⃗ , x ′ ⃗ ) ϕ = Φ
\left\{
\begin{array}{ll}
\psi = G(\vec{x},\vec{x'})\\
\phi = \Phi
\end{array}
\right.
{ ψ = G ( x , x ′ ) ϕ = Φ ⇒ ∫ V ( Φ ∇ ′ 2 G ⏞ − 4 π δ ( x ⃗ − x ′ ⃗ ) − G ∇ ′ 2 Φ ⏞ − ρ ( x ′ ⃗ ) ) d 3 x ′ = ∮ S ( Φ ∂ ∂ n ′ G − G ∂ ∂ n ′ Φ ) d a ′
\Rightarrow \int_{\mathbb{V}}(\Phi \overbrace{\nabla'^2 G}^{-4 \pi \delta(\vec{x}-\vec{x'})} - G \overbrace{\nabla'^2 \Phi}^{-\rho(\vec{x'})}) d^3x' = \oint_{\mathbb{S}}(\Phi \frac{\partial}{\partial n'}G-G\frac{\partial}{\partial n'}\Phi)da'
⇒ ∫ V ( Φ ∇ ′2 G − 4 π δ ( x − x ′ ) − G ∇ ′2 Φ − ρ ( x ′ ) ) d 3 x ′ = ∮ S ( Φ ∂ n ′ ∂ G − G ∂ n ′ ∂ Φ ) d a ′ ⇒ Φ ( x ⃗ ) = 1 4 π ϵ 0 ∫ V ρ ( x ′ ⃗ ) G ( x ⃗ , x ′ ⃗ ) d 3 x ′ + 1 4 π ∮ S [ G ( x ⃗ , x ′ ⃗ ) ∂ Φ ( x ′ ⃗ ) ∂ n ′ − Φ ( x ′ ⃗ ) ∂ G ( x ⃗ , x ′ ⃗ ) ∂ n ′ ] d a ′
\Rightarrow \Phi(\vec{x}) = \frac{1}{4 \pi \epsilon_0 }\int_{\mathbb{V}} \rho (\vec{x'}) G(\vec{x},\vec{x'}) d^3x' + \frac{1}{4 \pi} \oint_{\mathbb{S}}\left[
G(\vec{x},\vec{x'}) \frac{\partial \Phi (\vec{x'})}{\partial n'} - \Phi(\vec{x'}) \frac{\partial G (\vec{x},\vec{x'})}{\partial n'}
\right] da'
⇒ Φ ( x ) = 4 π ϵ 0 1 ∫ V ρ ( x ′ ) G ( x , x ′ ) d 3 x ′ + 4 π 1 ∮ S [ G ( x , x ′ ) ∂ n ′ ∂ Φ ( x ′ ) − Φ ( x ′ ) ∂ n ′ ∂ G ( x , x ′ ) ] d a ′ Using the additional freedom F ( x ⃗ , x ′ ⃗ ) F(\vec{x},\vec{x'}) F ( x , x ′ ) G ( x ⃗ , x ′ ⃗ ) G(\vec{x},\vec{x'}) G ( x , x ′ )
Dirichlet B.C. G D ( x ⃗ , x ′ ⃗ ) = 0 for x ′ ⃗ on S G_D(\vec{x},\vec{x'}) = 0 \text{ for } \vec{x'} \text{ on } \mathbb{S} G D ( x , x ′ ) = 0 for x ′ on S
Φ ( x ⃗ ) = 1 4 π ϵ 0 ∫ V ρ ( x ′ ⃗ ) G D ( x ⃗ , x ′ ⃗ ) d 3 x ′ − 1 4 π ∮ S [ Φ ( x ′ ⃗ ) ∂ G D ( x ⃗ , x ′ ⃗ ) ∂ n ′ ] d a ′
\Phi(\vec{x}) = \frac{1}{4 \pi \epsilon_0 }\int_{\mathbb{V}} \rho (\vec{x'}) G_D(\vec{x},\vec{x'}) d^3x' - \frac{1}{4 \pi} \oint_{\mathbb{S}}\left[
\Phi(\vec{x'}) \frac{\partial G_D (\vec{x},\vec{x'})}{\partial n'}
\right] da'
Φ ( x ) = 4 π ϵ 0 1 ∫ V ρ ( x ′ ) G D ( x , x ′ ) d 3 x ′ − 4 π 1 ∮ S [ Φ ( x ′ ) ∂ n ′ ∂ G D ( x , x ′ ) ] d a ′
Neumann B.C. ∂ G N ( x ⃗ , x ′ ⃗ ) ∂ n ′ d a ′ = − 4 π A S for x ′ ⃗ on S \displaystyle \frac{\partial G_N(\vec{x},\vec{x'})}{\partial n'} da'= -\frac{4 \pi}{A_S} \text{ for } \vec{x'} \text{ on } \mathbb{S} ∂ n ′ ∂ G N ( x , x ′ ) d a ′ = − A S 4 π for x ′ on S
∵ \because ∵ Gauss's theorem
∮ S ∂ G N ( x ⃗ − x ′ ⃗ ) ∂ n ′ d a ′ = ∮ S ∇ ′ ⃗ G N ( x ⃗ , x ′ ⃗ ) ⋅ n ′ ⃗ d a ′
\oint_{\mathbb{S}} \frac{\partial G_N(\vec{x}-\vec{x'})}{\partial n'} da' = \oint_{\mathbb{S}} \vec{\nabla '} G_N (\vec{x},\vec{x'}) \cdot \vec{n'} da'
∮ S ∂ n ′ ∂ G N ( x − x ′ ) d a ′ = ∮ S ∇ ′ G N ( x , x ′ ) ⋅ n ′ d a ′ = ∫ − V − 4 π δ ( x ⃗ , x ′ ⃗ ) d 3 x ′ = − 4 π ≠ 0
= \int_{-\mathbb{V}} -4 \pi \delta (\vec{x},\vec{x'}) d^3x' = -4 \pi \ne 0
= ∫ − V − 4 π δ ( x , x ′ ) d 3 x ′ = − 4 π = 0 Φ ( x ⃗ ) = ⟨ Φ ⟩ S + 1 4 π ϵ 0 ∫ V ρ ( x ′ ⃗ ) G N ( x ⃗ , x ′ ⃗ ) d 3 x ′ + 1 4 π ∮ S ∂ Φ ( x ′ ⃗ ) ∂ n ′ G N ( x ⃗ , x ′ ⃗ ) d a ′
\Phi(\vec{x}) = \langle \Phi \rangle_{\mathbb{S}} + \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{V}} \rho (\vec{x'}) G_N (\vec{x},\vec{x'}) d^3x' + \frac{1}{4 \pi} \oint_{\mathbb{S}}\frac{\partial \Phi(\vec{x'})}{\partial n'}G_N (\vec{x},\vec{x'})da'
Φ ( x ) = ⟨ Φ ⟩ S + 4 π ϵ 0 1 ∫ V ρ ( x ′ ) G N ( x , x ′ ) d 3 x ′ + 4 π 1 ∮ S ∂ n ′ ∂ Φ ( x ′ ) G N ( x , x ′ ) d a ′ # 1.11 Electrostatic Potential Energy and Energy Density, Capacitance
Potential Energyq i q_i q i ( Φ ( ∞ ) = 0 ) (\Phi (\infty) = 0) ( Φ ( ∞ ) = 0 ) x i ⃗ \vec{x_i} x i E ⃗ \vec{E} E Φ \Phi Φ
⇒ W i = q i Φ ( x i ⃗ )
\Rightarrow W_i = q_i \Phi(\vec{x_i})
⇒ W i = q i Φ ( x i ) Adding charges ( q 1 , q 2 , . . . , q n ) (q_1,q_2,...,q_n) ( q 1 , q 2 , ... , q n ) ( x 1 ⃗ , x 2 ⃗ , . . . , x n ⃗ ) (\vec{x_1},\vec{x_2},...,\vec{x_n}) ( x 1 , x 2 , ... , x n )
Potential experienced by the charge q i q_i q i
Φ ( x i ⃗ ) = 1 4 π ϵ 0 ∑ j = 1 i − 1 q j ∣ x i ⃗ − x j ⃗ ∣
\Phi (\vec{x_i}) = \frac{1}{4 \pi \epsilon_0} \sum_{j=1}^{i-1} \frac{q_j}{|\vec{x_i}-\vec{x_j}|}
Φ ( x i ) = 4 π ϵ 0 1 j = 1 ∑ i − 1 ∣ x i − x j ∣ q j total potential energy
W = 1 4 π ϵ 0 ∑ i = 1 n ∑ j < i q i q j ∣ x i ⃗ − x j ⃗ ∣ = 1 8 π ϵ 0 ∑ i ∑ j ≠ i q i q j ∣ x i ⃗ − x j ⃗ ∣
W= \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^{n} \sum_{j<i} \frac{q_i q_j}{|\vec{x_i}-\vec{x_j}|}= \frac{1}{8 \pi \epsilon_0} \sum_{i} \sum_{j \ne i} \frac{q_i q_j}{|\vec{x_i}-\vec{x_j}|}
W = 4 π ϵ 0 1 i = 1 ∑ n j < i ∑ ∣ x i − x j ∣ q i q j = 8 π ϵ 0 1 i ∑ j = i ∑ ∣ x i − x j ∣ q i q j energy can be negative (due to forces acting between the charges)
generalize to continuous charge
⇒ W = 1 8 π ϵ 0 ∬ ρ ( x ⃗ ) ρ ( x ′ ⃗ ) ∣ x ⃗ − x ′ ⃗ ∣ d 3 x d 3 x ′ = 1 2 ρ ( x ⃗ ) Φ ( x ⃗ ) d 3 x
\Rightarrow W = \frac{1}{8 \pi \epsilon_0} \iint \frac{\rho(\vec{x})\rho(\vec{x'})}{|\vec{x}-\vec{x'}|} d^3x \,d^3x' = \frac{1}{2} \rho (\vec{x}) \Phi (\vec{x}) d^3x
⇒ W = 8 π ϵ 0 1 ∬ ∣ x − x ′ ∣ ρ ( x ) ρ ( x ′ ) d 3 x d 3 x ′ = 2 1 ρ ( x ) Φ ( x ) d 3 x Alternatively , in terms of E ⃗ \vec{E} E
∵ ∇ ⃗ ⋅ ( E ⃗ Φ ) = ( ∇ ⃗ ⋅ E ⃗ ) Φ + E ⃗ ⋅ ( ∇ ⃗ Φ )
\because \vec{\nabla} \cdot (\vec{E} \Phi) = (\vec{\nabla}\cdot \vec{E}) \Phi + \vec{E} \cdot (\vec{\nabla}\Phi)
∵ ∇ ⋅ ( E Φ ) = ( ∇ ⋅ E ) Φ + E ⋅ ( ∇ Φ ) ⇒ W = − ϵ 0 ∫ Φ ∇ 2 Φ d 3 x = ϵ 0 2 { − ∫ E ⃗ ⋅ ∇ ⃗ Φ d 3 x + ∫ ∇ ⃗ ⋅ ( E ⃗ Φ ) d 3 x }
\Rightarrow W = - \epsilon_0 \int \Phi \nabla^2 \Phi d^3x = \frac{\epsilon_0}{2}\left\{
-\int \vec{E} \cdot \vec{\nabla} \Phi d^3 x + \int \vec{\nabla} \cdot (\vec{E} \Phi) d^3x
\right\}
⇒ W = − ϵ 0 ∫ Φ ∇ 2 Φ d 3 x = 2 ϵ 0 { − ∫ E ⋅ ∇ Φ d 3 x + ∫ ∇ ⋅ ( E Φ ) d 3 x } ϵ 0 2 ∫ V ∣ E ⃗ ∣ 2 d 3 x + ϵ 0 2 ∮ S Φ E ⃗ ⋅ n ⃗ d a = ϵ 0 2 ∫ all space ∣ E ⃗ ∣ 2 d 3 x
\frac{\epsilon_0}{2}\int_{\mathbb{V}}|\vec{E}|^2 d^3x +\frac{\epsilon_0}{2} \oint_{\mathbb{S}} \Phi \vec{E} \cdot \vec{n} \, da = \frac{\epsilon_0}{2} \int_{\text{all space}}|\vec{E}|^2 \, d^3x
2 ϵ 0 ∫ V ∣ E ∣ 2 d 3 x + 2 ϵ 0 ∮ S Φ E ⋅ n d a = 2 ϵ 0 ∫ all space ∣ E ∣ 2 d 3 x ∴ W = 1 2 ∫ ρ ( x ⃗ ) Φ ( x ⃗ ) d 3 x = ϵ 0 2 ∫ ∣ E ⃗ ∣ 2 d 3 x = ∫ W E d 3 x
\therefore W = \frac{1}{2} \int \rho (\vec{x}) \Phi (\vec{x}) \, d^3x = \frac{\epsilon_0}{2} \int |\vec{E}|^2 \, d^3x = \int W_E \, d^3x
∴ W = 2 1 ∫ ρ ( x ) Φ ( x ) d 3 x = 2 ϵ 0 ∫ ∣ E ∣ 2 d 3 x = ∫ W E d 3 x Energy density of E ⃗ \vec{E} E W E = ϵ 0 2 ∣ E ⃗ ∣ 2 W_E = \frac{\epsilon_0}{2} |\vec{E}|^2 W E = 2 ϵ 0 ∣ E ∣ 2
Self-Energy Contribution to the Energy Densityρ ( x ⃗ ) \rho (\vec{x}) ρ ( x ) ρ ( x ⃗ ) \rho (\vec{x}) ρ ( x )
κ ρ ( x ⃗ ) κ Φ ( x ⃗ ) κ = 0 ⟶ + ρ ( x ⃗ ) d κ κ ρ ( x ⃗ ) κ Φ ( x ⃗ ) κ = d κ ⟶ + ρ ( x ⃗ ) d κ κ ρ ( x ⃗ ) κ Φ ( x ⃗ ) κ = 2 d κ ⟶ . . . ⟶ ρ ( x ⃗ ) Φ ( x ⃗ ) κ = 1
\underset{\textcolor{blue}{\kappa = 0}}{
\begin{aligned}
\kappa \rho (\vec{x}) \\ \kappa \Phi (\vec{x})
\end{aligned}}
\overset{+ \rho (\vec{x})d\kappa }{\longrightarrow}
\underset{\textcolor{blue}{\kappa = d \kappa}}{
\begin{aligned}
\kappa \rho (\vec{x}) \\ \kappa \Phi (\vec{x})
\end{aligned}}
\overset{+ \rho (\vec{x})d\kappa }{\longrightarrow}
\underset{\textcolor{blue}{\kappa = 2d \kappa}}{
\begin{aligned}
\kappa \rho (\vec{x}) \\ \kappa \Phi (\vec{x})
\end{aligned}}
\longrightarrow ...\longrightarrow
\underset{\textcolor{blue}{\kappa = 1}}{
\begin{aligned}
\rho (\vec{x}) \\ \Phi (\vec{x})
\end{aligned}}
κ = 0 κ ρ ( x ) κ Φ ( x ) ⟶ + ρ ( x ) d κ κ = d κ κ ρ ( x ) κ Φ ( x ) ⟶ + ρ ( x ) d κ κ = 2 d κ κ ρ ( x ) κ Φ ( x ) ⟶ ... ⟶ κ = 1 ρ ( x ) Φ ( x ) d W = ∫ V κ Φ ( x ⃗ ) ρ ( x ⃗ ) d κ d 3 x
dW= \int_{\mathbb{V}} \kappa \Phi (\vec{x}) \rho (\vec{x}) d \kappa \, d^3x
d W = ∫ V κ Φ ( x ) ρ ( x ) d κ d 3 x ∴ W = ∫ κ = 0 κ = 1 d W = ∫ V ∫ κ = 0 1 κ d κ ρ ( x ⃗ ) Φ ( x ⃗ ) d 3 x = 1 2 ∫ Φ ( x ⃗ ) ρ ( x ⃗ ) d 3 x
\therefore W = \int_{\kappa = 0}^{\kappa = 1} dW = \int_{\mathbb{V}} \int _{\kappa = 0}^{1} \kappa \, d \kappa \rho (\vec{x}) \Phi (\vec{x}) d^3x = \frac{1}{2} \int \Phi (\vec{x}) \rho (\vec{x}) d^3x
∴ W = ∫ κ = 0 κ = 1 d W = ∫ V ∫ κ = 0 1 κ d κ ρ ( x ) Φ ( x ) d 3 x = 2 1 ∫ Φ ( x ) ρ ( x ) d 3 x Using Poisson Equation
⇒ − ϵ 0 2 ∫ Φ ( x ⃗ ) ∇ 2 Φ ( x ⃗ ) d 3 x
\Rightarrow - \frac{\epsilon_0}{2} \int \Phi(\vec{x}) \nabla^2 \Phi(\vec{x}) d^3x
⇒ − 2 ϵ 0 ∫ Φ ( x ) ∇ 2 Φ ( x ) d 3 x Using Green’s first identity with ϕ = ψ = Φ \phi = \psi =\Phi ϕ = ψ = Φ
⇒ ϵ 0 2 ∫ V ( ∇ ⃗ Φ ) ( ∇ ⃗ Φ ) d 3 x − ϵ 0 2 ∮ S Φ ∂ Φ ∂ n d a ⏟ = 0 , taking all space
\Rightarrow \frac{\epsilon_0}{2} \int_{\mathbb{V}}(\vec{\nabla} \Phi)(\vec{\nabla} \Phi) d^3x \underbrace{-\frac{\epsilon_0}{2} \oint_{\mathbb{S}} \Phi \frac{\partial \Phi}{\partial n} da }_{=0 \text{ , taking all space}}
⇒ 2 ϵ 0 ∫ V ( ∇ Φ ) ( ∇ Φ ) d 3 x = 0 , taking all space − 2 ϵ 0 ∮ S Φ ∂ n ∂ Φ d a
CapacitanceV i V_i V i Q i Q_i Q i
V i = ∑ j = 1 n P i j Q j ⟹ inversion Q i = C i j Q j
V_i = \sum_{j=1}^{n} P_{ij} Q_j \overset{\text{inversion}}{\Longrightarrow} Q_i = C_{ij} Q_j
V i = j = 1 ∑ n P ij Q j ⟹ inversion Q i = C ij Q j P i j ↔ C i j P_{ij} \leftrightarrow C_{ij} P ij ↔ C ij : geometric quantities
C i i \\ C_{ii} C ii : capacities / capacitances
C i j ( i ≠ j ) \\ C_{ij} (i \ne j) C ij ( i = j ) : coefficients of induction / mutual cross capacitance
Electrostatic potential energy of the n-conductor system
→ W i = 1 2 ∫ ρ i ( x ⃗ ) Φ i ( x ⃗ ) d 3 x = 1 2 Q i V i
\rightarrow W_i = \frac{1}{2} \int \rho_i (\vec{x}) \Phi_i (\vec{x}) d^3x = \frac{1}{2} Q_i V_i
→ W i = 2 1 ∫ ρ i ( x ) Φ i ( x ) d 3 x = 2 1 Q i V i W = 1 2 ∑ i = 1 n Q i V i = 1 2 ∑ i = 1 n ∑ j = 1 n C i j V i V j
W = \frac{1}{2} \sum_{i=1}^{n} Q_i V_i = \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} C_{ij} V_i V_j
W = 2 1 i = 1 ∑ n Q i V i = 2 1 i = 1 ∑ n j = 1 ∑ n C ij V i V j
Calculate force via change in the potential energyσ ( x ⃗ ) \sigma (\vec{x}) σ ( x )
( E 2 ⃗ − E 1 ⃗ ) ⋅ n ⃗ = σ ϵ 0
(\vec{E_2}-\vec{E_1}) \cdot \vec{n} = \frac{\sigma}{\epsilon_0}
( E 2 − E 1 ) ⋅ n = ϵ 0 σ immediately outside a conductor
E ⃗ = σ ( x ⃗ ) ϵ 0 n ⃗
\vec{E} = \frac{\sigma(\vec{x})}{\epsilon_0} \vec{n}
E = ϵ 0 σ ( x ) n ref: 古典電力學一 王喬萱教授 上課講義
