Remove["Global`*"]
{grad,div,curl}={"∇", "∇•","∇×"};
{tdE,tdB}={"((∂(Overscript[E, ->]))/(∂t))","((∂(Overscript[B, ->]))/(∂t))"};
{vecE,vecB,vecJ,vecF,vecv}={"(Overscript[E, ->])","(Overscript[B, ->])","(Overscript[J, ->])","(Overscript[F, ->])","(Overscript[v, ->])"};
{vecϵ,vecx,vecxt,vecD}={"(Overscript[ϵ, ->])","(Overscript[x, ->])","((Overscript[x, ->]),t)","(Overscript[D, ->])"};
M1=StringJoin[div,vecE,"=","4π ρ"];
M2=StringJoin[curl,vecE,"= –\!\(\*FractionBox[\(\(1\)\(\\\)\), \(c\)]\)",tdB];
M3=StringJoin[div,vecB,"=","0"];
M4=StringJoin[curl,vecB,"=","((4  π)/c)", vecJ ," + ",tdE];
potn="Φ((Overscript[x, ->]))";
TF=TraditionalForm;

Lecture 2 - Macroscopic Electrostatics

Reading Assignment : Jackson, Chapter 4.

The field equations of microscopic electrostatics are

StringJoin[div ,vecϵ," = 4π ρ'", "   and   ", curl,vecϵ ," =0."]//TF

Here ρ'((Overscript[x, ->])) is the total charge density that creates the field
(Overscript[ϵ, ->])((Overscript[x, ->])). These functions are independent of time .

In principle the solution is

StringJoin[vecϵ,"(",vecx,")"," = ∫((((Overscript[x, ->]) - (Overscript[x, ->])'))/(|(Overscript[x, ->]) - (Overscript[x, ->])'((|)^3)))ρ'((Overscript[x, ->])') (d^3)x' "]//TF

This is OK if ρ’ consists of a few isolated charges or simple charge densities. But what if ρ’ includes charges from a macroscopic sample of matter? All electric charge comes from electrons and ions. The number of electrons and ions is terribly large, like Avogadro’s number! We cannot handle a calculation that treats each particle individually.

Instead, we average over small volumes ΔV that contain large numbers of particles. It' s a subtle limiting process,

"molecule" ≪ ΔV ≪ (Subscript[V, sample])//TF

which is possible because the size of a molecule is so tiny.

The macroscopic density and the macroscopic field

StringJoin["⟨ρ'(",vecx,")⟩ = ","(1/(ΔV)) (∫_(ΔV)) ρ'((Overscript[x, ->])+(Overscript[ξ, ->])) (d^3)ξ"]//TF
StringJoin["⟨(Overscript[ϵ, ->])(",vecx,")⟩ = ","(1/(ΔV)) (∫_(ΔV)) (Overscript[ϵ, ->])((Overscript[x, ->])+(Overscript[ξ, ->])) (d^3)ξ"]//TF

You need to draw a picture to understand this!

Free charge and bound charge

Next, we divide ⟨ ρ’((Overscript[x, ->])) ⟩ into two parts,

StringJoin["⟨ ρ'((Overscript[x, ->])) ⟩ = (Subscript[ρ, free])((Overscript[x, ->])) + (Subscript[ρ, bound])((Overscript[x, ->])) "]//TF

Make sure you understand the meanings of these two charge densities!

For insulators ( “dielectrics” ):
 (Subscript[ρ, free]) is the "excess charge" added to the system
 (Subscript[ρ, bound]) is the charge of molecules in the material

For metals ( “conductors” ):
 (Subscript[ρ, free]) is the "excess charge" plus conduction electrons
 (Subscript[ρ, bound]) is the charge of ions in the material

Derivation of (Subscript[ρ, bound])( (Overscript[x, ->]))

Suppose the molecular density at the point (Overscript[x, ->]) is N((Overscript[x, ->])). You might think we could write
(Subscript[ρ, bound])( (Overscript[x, ->])) = N((Overscript[x, ->])) ⟨(Subscript[ρ, molecule])( (Overscript[x, ->]) )⟩ where

StringJoin["(Subscript[ρ, mol])((Overscript[x, ->])) = \!\(\*UnderscriptBox[\(Σ\), \(i\\  ∈ \\ mol\)]\) (Subscript[e, i]) (δ^3)((Overscript[x, ->])–(Subscript[(Overscript[x, ->]), i])) ;"]//TF

but that is certainly wrong, at least for dielectrics, because  
⟨(Subscript[ρ, mol])⟩ = 0 ;  molecules are electrically neutral ; (Underscript[Σ, i]) (Subscript[e, i]) = 0.

ρmol is not zero but ⟨(Subscript[ρ, mol])⟩ is zero;
the averaging over ΔV  gives 0.

(Metals are a little different, because the ions are charged.)

So what is (Subscript[ρ, bound])?
For a dielectric, the answer is

StringJoin["(Subscript[ρ, bound])((Overscript[x, ->])) = – ",div,"( N((Overscript[x, ->])) ⟨\!(Subscript[(Overscript[p, ->]), mol])((Overscript[x, ->]))⟩ )"]//TF

where (Subscript[(Overscript[p, ->]), mol])((Overscript[x, ->])) = the mean electric dipole moment of a molecule at (Overscript[x, ->]).

This introduces a vector function (Overscript[P, ->])((Overscript[x, ->])) called the polarization;
 (Overscript[P, ->])((Overscript[x, ->])) = N((Overscript[x, ->])) ⟨ (Subscript[p, mol])((Overscript[x, ->])) ⟩ = the dipole moment density.

PROOF

Start with the electric field created by the charges in a single molecule,

StringJoin[vecϵ,"\!\(\*SubscriptBox[\(\), \(1\)]\)","(",vecx,")"," = ∫((((Overscript[x, ->]) - (Overscript[ξ, ->])))/(|(Overscript[x, ->]) - (Overscript[ξ, ->])((|)^3)))(Subscript[ρ, mol])((Overscript[ξ, ->])) (d^3)ξ "]//TF
StringJoin["= -",grad,"∫(1/(|(Overscript[x, ->]) - (Overscript[ξ, ->])|))(Subscript[ρ, mol])((Overscript[ξ, ->])) (d^3)ξ"]//TF

We will eventually average over ΔV which is large compared to the molecule. So we can approximate (Subscript[(Overscript[ϵ, ->]), 1]) by taking r ≫ ξ ; 	⟹multipole expansion

StringJoin["(1/(|(Overscript[x, ->]) - (Overscript[ξ, ->])|)) ≃ \!\(\*FractionBox[\(\(1\)\(\\ \)\), \(r\)]\) + \!\(\*FractionBox[\(\*OverscriptBox[\(r\), \(^\)]\(\\ \)\(•\)\(\\ \)\*OverscriptBox[\(ξ\), \(->\)]\(\\ \)\), SuperscriptBox[\(r\), \(2\)]]\) + O (\!\(\*FractionBox[\(\(\\ \)\*SuperscriptBox[\(ξ\), \(2\)]\), SuperscriptBox[\(r\), \(3\)]]\) )"]//TF

StringJoin[" ∴ (Subscript[Φ, 1])((Overscript[x, ->])) ≃ \!\(\*FractionBox[\(\*SubscriptBox[\(q\), \(1\)]\(\\ \)\), \(r\)]\) + \!\(\*FractionBox[\(\*OverscriptBox[\(r\), \(^\)]\(\\ \)\(•\)\(\\ \)\*SubscriptBox[OverscriptBox[\(p\), \(->\)], \(1\)]\(\\ \)\), SuperscriptBox[\(r\), \(2\)]]\) + neglect"]//TF

StringJoin["(Subscript[q, 1]) = ∫ (Subscript[ρ, mol])\!\(\*Cell[TextData[{
"(",
Cell[BoxData[FormBox[
OverscriptBox["ξ", "->"], TraditionalForm]],ExpressionUUID->"6e31694a-1bd0-4aed-b922-abeffa5cb170"],
") ",
Cell[BoxData[FormBox[RowBox[{
SuperscriptBox["d", "3"], "ξ"}], TraditionalForm]],ExpressionUUID->"79b34082-38fb-4453-974d-6daff8bd970c"]
}],ExpressionUUID->"81acc084-4fa6-4d3f-91cd-5ccf6ec3067f"]\) = Σ (Subscript[e, i]) = 0 for a dielectric;" ]//TF
StringJoin["(Subscript[(Overscript[p, ->]), 1]) = ∫ (Subscript[ρ, mol])\!\(\*Cell[TextData[{
"(",
Cell[BoxData[FormBox[
OverscriptBox["ξ", "->"], TraditionalForm]],ExpressionUUID->"e38e36a1-3a5f-44cd-9120-8af37164a56e"],
") ",
Cell[BoxData[FormBox[
OverscriptBox["ξ", "->"], TraditionalForm]],FormatType->"TraditionalForm",ExpressionUUID->"453ba448-2578-4c37-9f53-2093da062d90"],
" ",
Cell[BoxData[FormBox[RowBox[{
SuperscriptBox["d", "3"], "ξ"}], TraditionalForm]],ExpressionUUID->"46921846-256d-48f6-886f-c83eb2a9f20b"]
}],ExpressionUUID->"1733e499-8284-427d-b3d0-7a691a90deca"]\) = dipole moment"]//TF
StringJoin[" = Σ (Subscript[e, i]) (Subscript[(Overscript[ξ, ->]), i])"]//TF

Now average over  ΔV  ⟹  
	⟨ (Overscript[p, ->]) ⟩ = ∫ over many molecules/ΔV;
and integrate over (Overscript[x, ->])’ to get the full potential,

StringJoin[potn," = ","∫ (d^3)x'\!\(\*FractionBox[\(\((\\ \*OverscriptBox[\(x\), \(->\)]\\  - \\ \*OverscriptBox[\(x\), \(->\)]')\) •\\ \*OverscriptBox[\(P\), \(->\)] \((\*OverscriptBox[\(x\), \(->\)]')\)\), \(\(|\)\(\\ \)\(\*OverscriptBox[\(x\), \(->\)] - \\ \*OverscriptBox[\(x\), \(->\)]'\)\*SuperscriptBox[\(|\), \(3\)]\)]\) "]//TF

The result is sort of obvious, isn’t it?

And now a bit of calculus ...

scan

Because of the polarization, i.e., because the molecules are polarized...
• there is a surface charge density (Subscript[σ, bound]) on the surface of V
• there is a volume charge density due to divergence of the  polarization

StringJoin["(Subscript[σ, bound]) = (OverHat[n]) • (Overscript[P, ->])","   and   ","(Subscript[ρ, bound]) = -∇ • P"]//TF

Comments

Study this figure from Jackson

fig4.X

Polarization does not produce a volume charge density.
Divergence of polarization produces a volume charge density.

The macroscopic field equations in electrostatics

The microscopic field equations are

StringJoin[div ,vecϵ," = 4π ρ'", "   and   ", curl,vecϵ ," =0."]//TF

Now derive the macroscopic field equations.
▮ Average over small volume elements ΔV throughout macroscopic materials.
▮ Replace ⟨ ρ’((Overscript[x, ->])) ⟩  by   (Subscript[ρ, free ])((Overscript[x, ->])) + (Subscript[ρ, bound])((Overscript[x, ->])) = (Subscript[ρ, free])((Overscript[x, ->])) – ∇•(Overscript[P, ->])((Overscript[x, ->])).
▮ Replace ⟨ (Overscript[ϵ, ->])((Overscript[x, ->])) ⟩ by (Overscript[E, ->])((Overscript[x, ->])).
▮ Define the displacement field (Overscript[D, ->])((Overscript[x, ->])) by (Overscript[D, ->])  =  (Overscript[E, ->])  +  4π (Overscript[P, ->])  . 
Then

StringJoin[div ,vecD," = 4π (Subscript[ρ, free])", "   and   ", curl,vecE ," =0."]//TF

Boundary conditions on (Overscript[E, ->])((Overscript[x, ->])) and (Overscript[D, ->])((Overscript[x, ->]))

Let S be any open surface in an electrostatic system of free charges and materials.

We are mainly thinking of dielectrics; i.e., insulators.
Metals are different.

The tangential components of (Overscript[E, ->]) are continuous across S;
	(OverHat[ n]) × (Overscript[E, ->]) is continuous;
	(OverHat[ n]) × (Overscript[E, ->])(1) = (OverHat[ n]) × (Overscript[E, ->])(2) .
PROOF:
Let Γ be an arbitrary loop that cuts through S, and let Σ be the surface enclosed by Γ; then
 (Subscript[∫ , Σ])(∇ × (Overscript[E, ->])) • d(Overscript[A, ->]) = (∮_(   Γ)) (Overscript[E, ->]) • d(Overscript[s, ->])  by Stokes’ theorem
 left = 0;  right ⟶ (Subscript[(OverHat[e]), tangential]) • ( (Overscript[E, ->])(2) – (Overscript[E, ->])(1) ) L(Γ)
 in the limit where Γ shrinks to the surface. This must be 0 for any Γ. Therefore (Subscript[(Overscript[E, ->]), tang.])(2) = (Subscript[(Overscript[E, ->]), tang.])(1).   QED

The normal component of (Overscript[D, ->]) is discontinuous if there is a free surface charge density on S, otherwise continuous;
	(OverHat[ n]) • (Overscript[D, ->])(2) – (OverHat[ n]) • (Overscript[D, ->])(1) = 4π (Subscript[σ, free]) .
PROOF:
Let Σ be an arbitrary closed surface that cuts through S, and let B be the volume enclosed by Σ; then
 (Subscript[∫ , B])(∇ • (Overscript[D, ->])) (d^3)x =  (∮_(  Σ)) (OverHat[n]) • (Overscript[D, ->]) dA   by Gauss’s theorem.
In the limit where Σ shrinks to the surface, 
  ∫ 4π (Subscript[σ, free]) dA   = ∫  [(Subscript[D, n])(2)–(Subscript[D, n])(1) ] dA .
This must be true for any Σ.
Therefore (Subscript[D, norm.])(2) – (Subscript[D, norm.])(1) = 4π (Subscript[σ, free]).   QED