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,vecp}={"(Overscript[ϵ, ->])","(Overscript[x, ->])","((Overscript[x, ->]),t)","(Overscript[D, ->])","(Overscript[p, ->])"};
{vecP}={"(Overscript[P, ->])"};
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 3 - Boundary Value Problems with Dielectrics

Reading Assignment: Jackson Sections 4.4 and 4.5

Macroscopic Electrostatics

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

Here (Overscript[D, ->])((Overscript[x, ->])) =  (Overscript[E, ->])((Overscript[x, ->])) + 4π (Overscript[P, ->])((Overscript[x, ->])) =  where (Overscript[P, ->])((Overscript[x, ->]))  is the polarization in the dielectric materials.

The simplest model for polarization
Recall, (Overscript[P, ->])((Overscript[x, ->])) =  N((Overscript[x, ->])) ⟨ (Subscript[(Overscript[p, ->]), molecule])((Overscript[x, ->])) ⟩.
Assume (Overscript[P, ->])((Overscript[x, ->])) is proportional to (Overscript[E, ->])((Overscript[x, ->])).

Why does this seem reasonable?
But how can it fail?
Many ways.

Assuming a uniform material, write (Overscript[P, ->])((Overscript[x, ->])) = (Subscript[χ, e]) (Overscript[E, ->])((Overscript[x, ->])).

(Subscript[χ, e]) ≡ electric susceptibility;
dimensionless in Gaussian units.

Then (Overscript[D, ->])((Overscript[x, ->])) = ϵ (Overscript[E, ->])((Overscript[x, ->])) where ϵ = 1 + 4π (Subscript[χ, e]).

ϵ ≡ electric permittivity, or dielectric constant;
dimensionless in Gaussian units.

Boundary value problems with dielectrics

Example 1

First, draw a picture to define the problem.

scan31

The problem is to calculate the electric field (Overscript[E, ->])((Overscript[x, ->])). The equations are

StringJoin[div,vecD," = 4π (Subscript[ρ, free]) = 4π q (δ^3)((Overscript[x, ->])- (Subscript[(OverHat[e]), z])d)"]//TF
StringJoin[curl,vecE," = 0"]//TF

StringJoin[vecD," = (Subscript[ϵ, 1])",vecE, " in region R1 ( z > 0 )"] //TF
StringJoin[vecD," = (Subscript[ϵ, 2])",vecE, " in region R2 ( z < 0 )"]//TF

Solution

A point charge with a dielectric interface

First, because ∇× (Overscript[E, ->]) = 0 we can write (Overscript[E, ->])((Overscript[x, ->])) = –∇Φ((Overscript[x, ->]));
now the problem is to determine Φ((Overscript[x, ->])).

Note that (∇^2)Φ = – 4π q (δ^3)( (Overscript[x, ->]) – (Subscript[(OverHat[e]), z]) d )    in region R1,
and (∇^2)Φ = 0 in R2.

The problem has cylindrical symmetry, but we won’t use that to find the solution. Instead we will use a clever trick, called the method of images.

Region R1:

StringJoin["(Subscript[ϵ, 1]) Φ(",vecx,") = ",
"(q/(|(Overscript[x, ->]) - (Subscript[(OverHat[e]), z]) d|)) + Ψ '(",vecx,")"]//TF

where (∇^2)Ψ' = 0. This does not mean that Ψ ’ = 0! We can “guess” that the form of Ψ ’ = potential due to a fictitious “image” charge q’ at –(Subscript[(OverHat[e]), z]) d (over in region R2).  So

StringJoin["(Subscript[ϵ, 1]) Φ(",vecx,") = ",
"(q/(|(Overscript[x, ->]) - (Subscript[(OverHat[e]), z]) d|)) +\!\(\*FractionBox[\(\(\\ \)\(q'\)\), \(\(|\)\(\*OverscriptBox[\(x\), \(->\)] + \*SubscriptBox[OverscriptBox[\(e\), \(^\)], \(z\)] d\)\(|\)\)]\)","   for (Overscript[x, ->]) ∈ R1"]//TF

Region R2:

(∇^2)Φ = 0. This does not mean that Φ = 0! We can “guess” that the form of Φ = potential due to a fictitious “image” charge q’’ at +(Subscript[(OverHat[e]), z]) d (over in region R1). So

StringJoin["(Subscript[ϵ, 2]) Φ(",vecx,") = ",
"((q'')/(|(Overscript[x, ->]) - (Subscript[(OverHat[e]), z]) d|))","   for (Overscript[x, ->]) ∈ R2"]//TF

So now we have (∇^2)Φ = – 4π (Subscript[ρ, free]) .

What next? Boundary Conditions

▮ (Subscript[D, norm.]) is continuous at z = 0  (no free charge on the surface) ⟹

StringJoin["(Subscript[ϵ, 2]) ((∂Φ)/(∂z))(x,y,0-) = (Subscript[ϵ, 1]) ((∂Φ)/(∂z))(x,y,0+)"] //TF
StringJoin["∴   q'' = q - q'" ]//TF

▮ (Subscript[E, tang.]) is continuous at z = 0   ⟹

StringJoin[" ((∂Φ)/(∂x))(x,y,0-) = ((∂Φ)/(∂x))(x,y,0+)"] //TF
StringJoin["∴  (1/(Subscript[ϵ, 2])) q'' =\!\(\*FractionBox[\(\(\\ \)\(1\)\), SubscriptBox[\(ϵ\), \(1\)]]\) (q + q')" ]//TF

Solve for q’ and q” ⟹

StringJoin["q' =(\!\(\*FractionBox[\(\*SubscriptBox[\(ϵ\), \(2\)]\(-\)\*SubscriptBox[\(ϵ\), \(1\)]\(\\ \)\), \(\*SubscriptBox[\(ϵ\), \(2\)] + \*SubscriptBox[\(ϵ\), \(1\)]\)]\))q  and  q''=(((2 (Subscript[ϵ, 2]))/((Subscript[ϵ, 2]) + (Subscript[ϵ, 1])))) q"]//TF

See Figure 4.7.

fig47

Exercise. Determine the surface charge density on the boundary surface. (It is not free charge !) Determine the force on q, and explain why it is attractive?

Example 2

A dielectric sphere with an applied electric field.

First, draw a picture to define the problem.

scan32

By cylindrical symmetry,

StringJoin["(Subscript[Φ, in]) = (Underoverscript[Σ, (l = 0), ∞]) (Subscript[A, l]) (r^l) (Subscript[P, l])(cosθ)","    ⟵   r <a"]//TF
StringJoin["(Subscript[Φ, out]) = (Underoverscript[Σ, (l = 0), ∞]) ((Subscript[B, l]) (r^l)+(Subscript[C, l])(r^(-(l + 1)))) (Subscript[P, l])(cosθ)","    ⟵   r >a"]//TF

Far from the sphere, i.e., r ⟶ ∞, Φ must approach the potential for a uniform field, Φ ⟶  –(Subscript[E, 0])z = -(Subscript[E, 0])r cosθ . Therefore (Subscript[B, 1]) = -(Subscript[E, 0]) and (Subscript[B, l]) = 0 for l  ≠ 1.

Now we have (∇^2)Φ = 0.

What’s next?  The boundary conditions.

(Subscript[E, tang.]) is continuous at r = a ⟹

StringJoin[" -\!\(\*FractionBox[\(\(1\)\(\\ \)\), \(a\)]\)[\!\(\*FractionBox[\(\(\\ \)\(∂\*SubscriptBox[\(Φ\), \(in\)]\)\), \(∂θ\)]\)]_(r = a)= -\!\(\*FractionBox[\(\(1\)\(\\ \)\), \(a\)]\)[\!\(\*FractionBox[\(\(\\ \)\(∂\*SubscriptBox[\(Φ\), \(out\)]\)\), \(∂θ\)]\)]_(r = a)   "]//TF
StringJoin["∴   (Subscript[A, 1]) = -(Subscript[E, 0]) +\!\(\*FractionBox[\(\*SubscriptBox[\(C\), \(1\)]\(\\ \)\), SuperscriptBox[\(a\), \(3\)]]\) "]//TF

(Subscript[D, norm.]) is continuous ⟹

StringJoin[" -ϵ[\!\(\*FractionBox[\(\(\\ \)\(∂\*SubscriptBox[\(Φ\), \(in\)]\)\), \(∂r\)]\)]_(r = a)= -[\!\(\*FractionBox[\(\(\\ \)\(∂\*SubscriptBox[\(Φ\), \(out\)]\)\), \(∂r\)]\)]_(r = a)  "]//TF
StringJoin[" ∴  ϵ (Subscript[A, 1]) = -(Subscript[E, 0]) - \!\(\*FractionBox[\(\(2\) \*SubscriptBox[\(C\), \(1\)]\(\\ \)\), SuperscriptBox[\(a\), \(3\)]]\) "]//TF

Exercise : Show that  (Subscript[A, l])=0 and (Subscript[C, l])= 0  for l ≠ 1;
this is sort of obvious.

Thus,

StringJoin["(Subscript[A, 1]) = - ((3/(ϵ + 2))) \!\(\*SubscriptBox[\(E\), \(\(0\)\(\\ \)\)]\)","   and    ","(Subscript[C, 1]) = (\!\(\*FractionBox[\(\(\\ \)\(ϵ - 1\)\), \(ϵ + 2\)]\)) (a^3) (Subscript[E, 0]) "]//TF

The potential inside the sphere corresponds to a constant electric field,

StringJoin["(Subscript[Φ, in])=–(3/(ϵ + 2))(Subscript[E, 0])r cosθ ;  ","∴ (Subscript[(Overscript[E, ->]), in]) =\!\(\*FractionBox[\(\(3\)\(\\ \)\), \(ϵ + 2\)]\) (Subscript[E, 0]) (Subscript[(OverHat[e]), z]) "]//TF

The potential outside the sphere corresponds to a constant electric field plus the field of a point-like electric dipole

StringJoin["(Subscript[Φ, out])=– (Subscript[E, 0]) r cosθ + ((ϵ - 1)/(ϵ + 2))(Subscript[E, 0])\!\(\*FractionBox[\(\*SuperscriptBox[\(a\), \(3\)]\(\\ \)\), SuperscriptBox[\(r\), \(2\)]]\) cosθ"," ; "]//TF

StringJoin["∴ (Subscript[(Overscript[E, ->]), out]) = (Subscript[E, 0]) (Subscript[(OverHat[e]), z]) - ∇(\!\(\*FractionBox[\(\(x•p\)\(\\ \)\), SuperscriptBox[\(r\), \(3\)]]\))"]//TF
StringJoin[vecp," = ","(((ϵ - 1)/(ϵ + 2))) (a^3) (Subscript[E, 0]) (Subscript[(OverHat[e]), z])"]//TF

"fig4.9"

Example 3

A spherical cavity in a dielectric medium with an applied electric field

First, draw a picture to define the problem.

"fig4.10"

This is the same as Example 2, except for these replacements...
(Subscript[ϵ, inside]) = ϵ   ⟶   1
(Subscript[ϵ, outside]) = 1   ⟶   ϵ

So in the final equations from Example 2, replace
ϵ  =  (Subscript[ϵ, inside])/(Subscript[ϵ, outside])  by 1/ϵ .

The electric field inside the cavity is

StringJoin["\!\(\*SubscriptBox[OverscriptBox[\(E\), \(->\)], \(in\\ cavity\)]\) = ((3  ϵ)/(2  ϵ + 1)) (Subscript[(Overscript[E, ->]), 0])"]//TF

Outside the cavity there is a dipole field, in addition to the applied field. The dipole moment of the cavity points opposite to the applied field because the bound surface charge density is negative on the down-field side and positive on the up-field side.

StringJoin["(Subscript[(Overscript[p, ->]), cavity]) = -  ((ϵ - 1)/(2  ϵ + 1)) (a^3) (Subscript[(Overscript[E, ->]), 0]) (Subscript[(OverHat[e]), z])"]//TF

Molecular Polarizability and Electric Susceptibility

We' ll derive the Clausius-Mossotti equation (Section 4.5)

StringJoin["(Subscript[χ, e]) =\!\(\*FractionBox[\(\(N\)\(\\ \)\*SubscriptBox[\(γ\), \(mol\)]\(\\ \)\), \(1 - \((4  π/3)\)\\ N\\ \*SubscriptBox[\(γ\), \(mol\)]\)]\)"]//TF

where

StringJoin["(Subscript[γ, mol]) =\!\(\*FractionBox[\(\(3\)\(\\ \)\), \(4  π\\ N\)]\)(\!\(\*FractionBox[\(\(ϵ\)\(-\)\(1\)\(\\ \)\), \(ϵ + 2\)]\))"]//TF

We have defined (Subscript[χ, e]) by these equations

(Overscript[P, ->])((Overscript[x, ->])) = N((Overscript[x, ->]))  ⟨ (Subscript[(Overscript[p, ->]), mol]) ((Overscript[x, ->]) ) ⟩ = (Subscript[χ, e])  (Overscript[E, ->])((Overscript[x, ->])) .

Here (Overscript[E, ->])((Overscript[x, ->])) is the macroscopic field, i.e., averaged over ΔV ≫ molecule. But that is not quite right for a dense material.

We need a better calculation of (Subscript[(Overscript[p, ->]), mol])((Overscript[x, ->])).

The macroscopic field may be called an “external” field. It comes from outside the single molecule we are considering, by averaging over a large number ( ~ (10^23) ) of molecules in ΔV.

But there is also an “internal” field, coming from nearby polarized molecules. And that will also contribute to (Subscript[(Overscript[p, ->]), mol])((Overscript[x, ->])). So, we should write

StringJoin[vecp," = (Subscript[γ, mol]) (",vecE," + ",vecE,"(Subscript[, int]))"]//TF

where (Subscript[γ, mol]) = the molecular polarizability , a property of a single molecule.

Estimate the internal field

Imagine a sphere enclosing the molecule.  It is similar to  the example of a cavity in a dielectric. The molecules outside the sphere are polarized so there is a surface charge density (of bound charge!)  = – (Overscript[n, ⋀]) • (Overscript[P, ->]) on the surface of the sphere. The field inside the sphere due to that surface charge is

(OverHat[n])  is the outward normal

StringJoin[vecE,"\!\(\*SubscriptBox[\(\), \(int\)]\)"," = ",
" \!\(\*SubscriptBox[\(∮\), \(\(\\ \\ \)\(sphere\)\)]\)  (r^2) dΩ \!\(\*FractionBox[\(\*OverscriptBox[\(-P\), \(->\)] •\\ \*OverscriptBox[\(n\), \(^\)]\), SuperscriptBox[\(r\), \(2\)]]\) = \!\(\*FractionBox[\(\(4\)  \(π\)\(\\ \)\), \(3\)]\)",vecP]//TF

check the direction!

Susceptibility or permattivity

StringJoin[vecP," = N ⟨(Subscript[(Overscript[p, ->]), mol])⟩ = N (Subscript[γ, mol]) (",vecE,"+(Subscript[(Overscript[E, ->]), int]) )"]//TF
StringJoin[" = N (Subscript[γ, mol]) ( " ,vecE," + ((4  π)/3) ",vecP," )"," = (Subscript[χ, e])",vecE]//TF

StringJoin["(Subscript[χ, e]) =\!\(\*FractionBox[\(\*SubscriptBox[\(Nγ\), \(mol\)]\(\\ \)\), \(1 - \*FractionBox[\(4  π\), \(3\)] \*SubscriptBox[\(Nγ\), \(mol\)]\)]\)"]//TF

Or, in terms of the permittivity ϵ = 1 + 4π (Subscript[χ, e]) ,

StringJoin["(Subscript[γ, mol]) =\!\(\*FractionBox[\(\(3\)\(\\ \)\), \(4  π\\ N\)]\)(\!\(\*FractionBox[\(\(ϵ\)\(-\)\(1\)\(\\ \)\), \(ϵ + 2\)]\))"]//TF

The Clausius-Mossotti relation relates a molecular parameter (polarizability (Subscript[γ, mol])) to macroscopic parameters (density N and permittivity ϵ).
This provides a way to estimate (Subscript[γ, mol]) by laboratory measurements. It is accurate for gases (small N), but less so for liquids and solids (large N).

The ultimate theory of molecular polarizability would be a quantum theory.