10. Ch. 4, part 2

Lecture 10 - Dielectric Media

Almost always, we are interested in the electric fields in the presence of matter that is itself made up of charged particles. The total microscopic field can be written

E_{micro} = E_{matter}^{(0)} + E′

where E_{matter}^{(0)} is the electric field of the matter before the “external” or “true” charge density ρ is brought onto the scene. In this chapter we consider only the case when E_{matter}^{(0)} = 0 outside the regions occupied by material bodies (thus excluding ferroelectrics). We also assume that inside the material bodies E_{matter}^{(0)} vanishes on the average. What precisely we mean by average is a long story. For fluids, a quantum-statistical average or a time average are enough to make ⟨E_{matter}^{(0)}⟩ = 0. For solids this is not true, and to get a

zero ⟨E_{matter}^{(0)}⟩ we must average explicitly over regions larger than the interatomic spacing, with special attention to the vicinity of surfaces. In any case:

• Outside material bodies E_{micro} = E′: there is only one electric field to deal with and we call it simply E.

• Inside matter, we assume that our averaging makes ⟨E_{matter}^{(0)}⟩ = 0 and we define E ≡ ⟨E_{micro}⟩ = ⟨E′⟩ . However, in the theory of solids one often takes only the quantum-statistical average and the three average fields are all non-zero. In this case, the convenient quantity to deal with is usually the quantum-statistical average of E′.

10.1. The field equations

It can be shown explicitly (see Jackson, page 228) that Maxwell’s equations are preserved by the averaging process. Thus from

∇ × E_{micro} = 0
∇ · E_{micro} = 4π(ρ_{matter} + ρ)

(where ρ is the “external” charge density) and

∇ × E_{matter}^{(0)} = 0
∇ · E_{matter}^{(0)} = 4πρ_{matter}^{(0)}

we get

∇ × E′ = 0
∇ · E′ = 4π(ρ_{matter} − ρ_{matter}^{(0)} + ρ)

and upon averaging

∇ × E = 0
∇ · E = 4π(ρ_{ind} + ρ)

where

ρ_{ind} = ⟨ρ_{matter} − ρ_{matter}^{(0)}⟩

is the average charge density induced in matter by the presence of the “external” charge ρ.

To compute ρ_{ind} we must find the averaged current j_{ind} that is induced in the materials when ρ is brought on the scene. This is in general a difficult problem, but we can make simple models. From j_{ind} we can find ρ_{ind} by integrating its time derivative, given by the continuity equation

\frac{∂ρ_{ind}}{∂t} + ∇ · j_{ind} = 0

with the initial condition ρ_{ind} = 0. We deal first with materials in which j_{ind} comes to a halt after a short while and a final, static ρ_{ind} is reached. These are called insulators or dielectrics. It is convenient to define a vector P(x) such that

\frac{∂P}{∂t} = j

Integration of the continuity equation gives

ρ_{ind} = −∇ · P

The physical meaning of P can be seen if we suppose that the matter at hand consists of separate molecules, each containing charges q_{i} located at points x_{i}. The induced current in each molecule is

j_{mol} = ∑_{i} q_{i} \frac{dx_{i}}{dt} = \frac{d}{dt} ∑_{i} q_{i}x_{i} = \frac{dp}{dt}

where

p = ∑_{i} q_{i}x_{i}

is the dipole moment of the molecule. The total current in any volume V can be written in the two equivalent ways

J = ∫_{V} j_{ind}(x) d^{3}x = ∫_{V} \frac{∂P}{∂t} d^{3}x = \frac{d}{dt} ∫_{V} P(x) d^{3}x

J = ∑_{mol}^{in V} j_{mol} = ∑_{mol}^{in V} \frac{dp}{dt} = \frac{d}{dt} ∑_{mol}^{in V} p

Since this holds for any volume V, P(x) is the dipole moment density, better known as polarization.

The main advantage of introducing P(x) is that often there is (to a good approximation) a local relation between P and E. In the simplest case this relation is linear

P(x) = χ^{(e)}(x)E(x)

but in general χ^{(e)} will also contain terms proportional to E^{2} and higher non-linear terms. In any case, χ^{(e)} is called the electric susceptibility of the medium. It may vary from point to point and be nonlinear. It is also possible that P and E are not collinear, in which case χ^{(e)} is a tensor:

P_{i} = ∑_{j} χ_{ij}^{(e)}E_{j}

In any case, the field equations can now be written

∇ × E = 0
∇ · D = 4πρ

where we have introduced the electric displacement vector

D = E + 4πP

They are to be solved jointly with the constitutive relation between D and E, which in the simplest case takes the form

D = ϵE

where ϵ = 1 + 4πχ^{(e)} is called the dielectric constant (although it can depend on x, and also on E in a non-linear medium, and in time-dependent phenomena will also depend on frequency, etc.).

10.1.1. Boundary conditions

At the interface between two media ϵ(x) varies rapidly. Approximately, it has a step discontinuity. If z is the coordinate perpendicular to the interface, from medium 1 to medium 2, the equation ∇ · D = 4πρ implies that D_{z} is continuous at the interface, and therefore E_{z} is discontinuous:

ϵ^{(1)}E_{z}^{(1)} = ϵ^{(2)}E_{z}^{(2)}

On the other hand, the equation ∇ × E = 0 implies that E_{x} and E_{y} are continuous at the interface, and therefore D_{x} and D_{y} are discontinuous

\frac{D_{x}^{(1)}}{ϵ^{(1)}} = \frac{D_{x}^{(2)}}{ϵ^{(2)}}  and  \frac{D_{y}^{(1)}}{ϵ^{(1)}} = \frac{D_{y}^{(2)}}{ϵ^{(2)}}

Simple boundary-value problems with uniform dielectric media are usually solved by writing E = −∇Φ, solving for Φ in each medium, and then matching the parallel components of ∇Φ and ϵ∂Φ/∂n at each interface.

From the boundary conditions follows a physical interpretation of E and D :

• Inside a thin cavity aligned with the direction of E (either needle-like or lens-shaped), the electric field is equal to E in the surrounding medium

• Inside a needle aligned with the outside E, or a thin lens with the short axis perpendicular E, the electric field is equal to the outside E

• Inside a lens-shaped cavity perpendicular to E, the electric field is equal to D in the surrounding medium

• Inside a dielectric slab perpendicular to the outside electric field, D is equal to the outside electric field.

10.2. Boundary value problems

We can now revisit all the boundary value problems with conducting boundaries and replace the conductors with dielectric media. In many cases the solution is similar, but more difficult. The perfect conductor case is recovered formally in the limit ϵ → ∞. (We will see later that the proper limit for a metal is either ϵ → −∞ or ϵ → i∞).

10.2.1. Plane boundary by images

Assume ϵ = ϵ_{1} for z > 0, ϵ = ϵ_{2} for z < 0 and put a charge q at z = d on the z axis. Then;

• For z > 0, it is as if an image charge had appeared at z = −d in an infinite medium of dielectric constant ϵ_{1}. The strength of the image is

q′ = \frac{ϵ_{1} − ϵ_{2}}{ϵ_{1} + ϵ_{2}} q

A force F_{z} = qq′/4ϵ_{1}d^{2} acts on the charge and the energy of the configuration is W = qq′/4ϵ_{1}d. For ϵ_{2} >> ϵ_{1} the image charge q′ becomes −q, and F_{z} = −q^{2}/4ϵ_{1}d^{2} as expected. Note however that the charge q is repelled by the boundary when ϵ_{1} > ϵ_{2}.

• For z < 0, it is as if the charge q had been replaced by a charge

q′′ = \frac{2ϵ_{1}}{ϵ_{1} + ϵ_{2}} q

in an infinite medium of dielectric constant ϵ_{2}. As expected, q′′ → 0 when ϵ_{2} >> ϵ_{1}.

• A polarization-charge density σ_{pol} = (1/4π) (E_{2} − E_{1})_{z} = (P_{1} − P_{2})_{z} appears at the interface. It has the value

σ_{pol} = \frac{q}{2πϵ_{1}} \frac{ϵ_{1} − ϵ_{2}}{ϵ_{1} + ϵ_{2}} \frac{d}{(ρ^{2} + d^{2})^{3/2}}

It becomes 0 for ϵ_{1} → ϵ_{2} and for ϵ_{2} >> ϵ_{1} approaches

−\frac{q}{2πϵ_{1}} \frac{d}{\sqrt{ρ^{2} + d^{2}}}

• Let us take ϵ_{1} = 1 and think of q as an “external” charge that generates the field E_{ext} in the absence of medium 2. It is clear from a look at Jackson’s Fig. 4.8 that, when medium 2 is present, E differs from E_{ext} even on the vacuum side z > 0. Inside medium 2, E and D have the same direction as E_{ext} but different magnitudes:

E =\frac{2}{1 + ϵ_{2}} E_{ext} D =\frac{2ϵ_{2}}{1 + ϵ_{2}} E_{ext}

Any charge distribution near a flat boundary can be “imaged” by superposition of the source charges and their images. In particular, a wire parallel to the boundary has an “image wire”, or in other words the solution of the 2-d problem is entirely analogous to the 3-d case.

10.2.2. Sphere in a uniform field

Dielectric sphere problems cannot be solved by images, but are easily handled by spherical harmonics. For an uncharged sphere in a uniform field, only dipole fields (l = 1) are involved. (The solution for a charged dielectric sphere depends on the charge distribution on it). Jackson gives explicit formulas for a sphere of dielectric constant ϵ surrounded by vacuum. However, we can obtain the field ratios for a sphere of dielectric constant ϵ_{2} surrounded by a medium of dielectric constant ϵ_{1} by the simple trick: replace ϵ with ϵ_{2}/ϵ_{1}. Note that we could have applied the same trick to the plane-boundary problem. If the “external” field is E_{0} and the radius of the sphere is a :

• The field inside the sphere is

E =\frac{3ϵ_{1}}{ϵ_{2} + 2ϵ_{1}} E_{0}

• For ϵ_{2} > ϵ_{1} the inside field is smaller that E_{0}, as illustrated by Jackson in Fig. 4.7. The polarization is of course P = (ϵ_{2}−1)E/4π and the induced dipole moment is

p = (ϵ_{2}−1)Ea^{3}/3 = \frac{ϵ_{1} (ϵ_{2} − 1)}{ϵ_{2} + 2ϵ_{1}} a^{3}E_{0}  (10.1)

This reduces to ϵ_{1}a^{3}E_{0} in the case of the conducting sphere (ϵ_{2} >> ϵ_{1}).

• In particular, the field inside a spherical cavity (ϵ_{2} = 1) is

E = \frac{3ϵ_{1}}{1 + 2ϵ_{1}} E_{0}  (10.2)

For comparison, inside a thin cavity with the long axis parallel to E_{0} we have E_{needle} = E_{0} and inside a thin pillbox perpendicular to E_{0} we have E_{pillbox} = D_{0} = ϵ_{1}E_{0}. We see that the spherical cavity result is intermediate, as expected. For ϵ_{1} > 1, E_{needle} < E_{sphere} < E_{pillbox} or E_{0} < E_{sphere} < D_{0}.

• The field outside the sphere is E_{0} plus the field due to the “image dipole” p given in eq.(10.1).

• The polarization charge on the surface of a sphere immersed in vacuum (ϵ_{1} = 1) is

σ_{pol} = \frac{3}{4π} \frac{ϵ_{2} − 1}{ϵ_{2} + 2} E_{0} cos θ

10.3. Molecular polarizability

Gases can be regarded as a collection of molecules (including monatomic molecules, as in rare gases) to a good approximation. For many liquids and solids, this simple picture is also reasonable. Each molecule may have a permanent dipole moment p_{mol}. Symmetric molecules such as O_{2} have p_{mol} = 0, but otherwise p_{mol} is of order ea_{0}, where a_{0} is the Bohr radius, and is usually given in Debye units:

1 Debye ≡ 10^{−18} statcoul · cm = 3.336 · 10^{−30} C · m
ea_{0} = (4.803 · 10^{−10} statcoul) (0.529 · 10^{−8}cm) = 2.54 Debye

For example the molecular (gas-phase) dipole moment of CO is 0.112, of H_{2}O is 1.85, of NaCl is 9.00 Debye. In a solid these dipole moments may be regularly oriented at low temperature, but we will not consider this case here because we discuss in detail the analogous magnetic case later on. Then the polarization of molecular fluids and solids can arise in two ways:

• The applied field produces an induced dipole moment on each molecule;

• The applied field tends to line up the permanent dipole moments of the molecules, which are randomly oriented otherwise.

10.3.1. Induced dipole moment

We have seen that the induced dipole moment on a conducting sphere of radius a in a field E is simply a^{3}E. We can say that the polarizability of a conducting sphere is a^{3}. It turns out (Schiff, page 265) that the polarizability of a hydrogen atom is \frac{9}{2}a_{0}^{3}, where a_{0} is the Bohr radius h̄^{2}/me^{2} = 0.529 · 10^{−8} cm. Numerically, γ_{H} = 0.666 · 10^{−24} cm^{3}. Typical molecular polarizabilities γ_{mol} are of this order of magnitude, 10^{−24} cm^{3}. In general γ_{mol} ≃ v_{mol}, where v_{mol} is the volume of the atom or molecule (for H, note that 4.5 is close to 4π/3 = 4.1888).

For a non spherical molecule, γ_{mol} is a tensor, and the principal axes of (γ_{mol})_{ij} correspond to symmetry axes of the molecule, if any. According to γ ≈ a^{3} and to simple intuition, a molecule is most polarizable along its longest axis. Usually, however, the average over the random orientation of the molecules reduces (γ_{mol})_{ij} to its isotropic part γ_{mol}δ_{ij}, where

γ_{mol} = \frac{1}{3} ∑_{i} (γ_{mol})_{ii}

In effect, then, the polarizability is again a scalar γ_{mol}.

Another way to picture polarizability is to say that it results from an elastic deformation of the charge distribution in the molecule. When the molecule is subject to a periodic field, it resonates at certain angular frequencies ω_{j} that can be measured; when it is subject to a constant field, it just deforms. In the simplest picture, only one electron is responsible for the deformation and there is only one resonant frequency ω_{0}. Then

• the effective spring constant is mω_{0}^{2}, where m is the electron mass and h̄ω_{0} is of the order of the ionization energy, typically a few eV, which corresponds to ω_{0} ≃ 10^{16} sec^{−1}.

• the effective electron displacement when subject to the force eE is x = eE/mω_{0}^{2}, where e is the electron charge

• the induced dipole moment is ex = e^{2}E/mω_{0}^{2} and the polarizability is

γ_{mol} = \frac{e^{2}}{mω_{0}^{2}}

This formula gives γ_{mol} = 4a_{0}^{3} for the H atom, when we take ω_{0} = e^{2}/2a_{0}h̄. As we already said, the exact result for the H atom is γ_{mol} = 4.5 a_{0}^{3}.

The electric susceptibility χ^{(e)} is of the order of Nγ_{mol}, where N is the number of molecules per unit volume (see next section for the more accurate Clausius-Mossotti formula). Then ϵ = 1 + 4πχ^{(e)} ≃ 1 + 10 Nv_{mol} is close to unity for a gas and can be 10 or more for a condensed body. Actual numbers are 1.00054 for dry air at STP and 11.7 for Silicon.

Larger values of ϵ are common for “polar” liquids or solids where the molecules have permanent dipole moments. Water for instance has a large dielectric constant, ϵ = 78.5 at 25^{∘}C, which is very important to “wet” chemistry because it weakens the forces between ions in solutions.

10.3.2. Alignment of permanent moments

The energy of a dipole p in the field E is −p · E. The thermal average of p_{mol} has a non-zero component ⟨p_{mol}⟩ in the direction of E. Classically, at temperature T,

⟨p_{mol}⟩ = \frac{1}{Z} ∫_{0}^{π} p_{mol} cos θ exp (βp_{mol}E cos θ) sin θ dθ

where β = 1/k_{B}T and

Z = ∫_{0}^{π} exp (βp_{mol}E cos θ) sin θ dθ

We can easily compute

Z = −\frac{1}{ξ} exp (ξ cos θ)|_{0}^{π} = \frac{2 sinh ξ}{ξ}

where

ξ = βp_{mol}E = \frac{p_{mol}E}{k_{B}T}

and we find

⟨p_{mol}⟩ = p_{mol} \frac{∂ ln Z}{∂ξ} = p_{mol} (coth ξ − \frac{1}{ξ})

We can estimate that p_{mol}E ≈ ea_{0}E ≈ 10^{−8}E eV, where E is expressed in eV/cm. Typically then ξ << 1, because p_{mol}E can hardly be as large as 0.01eV, while k_{B}T = 0.025eV at room temperature. To get the small-ξ limit, it is easier to expand ln Z ≃ ln 2 + ln(1 + \frac{1}{6}ξ^{2}) ≃ ln 2 + \frac{1}{6}ξ^{2}. Thus

⟨p_{mol}⟩ = p_{mol} \frac{∂ ln Z}{∂ξ} ≃ p_{mol} \frac{ξ}{3} = \frac{p_{mol}^{2}E}{3k_{B}T}

On the other hand, the induced polarizability is temperature independent as long as the molecules stay in their ground state, or as long as the classical harmonic oscillator model is valid (see Jackson). At ordinary temperatures then

γ_{mol} ≃ v_{mol} + \frac{p_{mol}^{2}}{3k_{B}T}

10.4. The local field

The field acting on a molecule in a medium is called the local field E_{local}. It is neither the microscopic field E_{micro} nor the averaged field E. It is the part of E_{micro} that is due to the external sources and to all the molecules except the one under consideration.

Lorentz showed how to compute E_{local} in a cubic crystal made up of identical molecules of one kind. The Lorentz formula

E_{local} = E + \frac{4π}{3} P  (10.3)

is exact when E is uniform (see below for a derivation). It is also valid for fluids and is widely used as a good approximation when E is not uniform. Corrections to the Lorentz formula for non-cubic crystals are available. When several types of molecules or ions are present, the local field on each type of molecule may be different.

The induced dipole on a molecule is γ_{mol}E_{local} and the polarization is

P = Nγ_{mol}E_{local}  (10.4)

where N is the number of molecules per unit volume. This is an exact formula, to the extent that separate molecular dipoles can be defined. From (10.3) and (10.4)

E_{local} = \frac{E}{1 − \frac{4π}{3} Nγ_{mol}} P = \frac{Nγ_{mol}E}{1 − \frac{4π}{3} Nγ_{mol}}

Hence, recalling that P = χ^{(e)}E and ϵ = 1 + 4πχ^{(e)}, we obtain

χ^{(e)} = \frac{Nγ_{mol}}{1 − \frac{4π}{3} Nγ_{mol}} ϵ = \frac{1 + \frac{8π}{3} Nγ_{mol}}{1 − \frac{4π}{3} Nγ_{mol}}

Since ϵ (or χ^{(e)}) is easily measured macroscopically, either formula can be used backwards to determine the microscopic quantity γ_{mol}. For instance

γ_{mol} = \frac{3}{4πN} \frac{ϵ − 1}{ϵ + 2}

This is called the Clausius-Mossotti relation and was known empirically before Lorentz. The simple relation χ^{(e)} = Nγ_{mol} is good only for dilute media (gases) where the difference between E and E_{local} is negligible.

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Proof of the Lorentz local field equation.

In a uniform medium with polarization P, the average field is E. To compute the local field correction to E, cut a spherical hole out of the uniform medium and replace its contents by the actual distribution of molecules. Then E_{local} = E − E_{P} + E_{near} where

• E_{P} is the field of the polarization charge density P cos θ that appears on the surface of the sphere. E_{P} is actually uniform within the sphere. We know by our spherical harmonics trickery that it is enough to compute the potential on the z axis. We find

Φ_{P} = ∑_{l} \frac{r^{l}}{a^{l+1}} M_{l}P_{l}(cos θ)

where M_{l} are the multipoles of the surface charge density P cos θ. Using

2πP ∫_{0}^{π} cos θ′ P_{l}(cos θ′) a^{2} sin θ′ dθ′ = \frac{4πP a^{2}}{3} δ_{l1}

we obtain Φ_{P} = (4π/3)P z and E_{P} = −4πP/3.

• E_{near} is the sum of the dipolar fields

\frac{3 (p · r) r − r^{2}p}{r^{5}}

of all the other molecules within the sphere, computed at the origin. E_{near} turns out to be zero if the molecules are on a simple cubic lattice, i.e. if they are located at the points

r_{ijk} = (i, j, k)a i, j, k non-zero integers

If the molecular dipoles are oriented along z, it is obvious that E_{near} can only have a z component. We are left with

(E_{near})_{z} = ∑_{ijk} \frac{3pk^{2} − p (i^{2} + j^{2} + k^{2})}{a^{3} (i^{2} + j^{2} + k^{2})^{5/2}}

which vanishes because

∑_{ijk} \frac{i^{2}}{(i^{2} + j^{2} + k^{2})^{5/2}} = ∑_{ijk} \frac{j^{2}}{(i^{2} + j^{2} + k^{2})^{5/2}} = ∑_{ijk} \frac{k^{2}}{(i^{2} + j^{2} + k^{2})^{5/2}}

The argument also applies to face-centered and to body-centered cubic lattices, and to any arrangement where the x, y, z directions are equivalent. One can show directly that E_{near} = 0 for a spherically symmetric distribution of dipoles, because then the numerator of (E_{near})_{z} is proportional to P_{2}(cos θ), which integrates to zero.

Collecting results, we have obtained E_{local} = E+4πP/3 = E (2 + ϵ) /3. This is different from the field inside an actual spherical cavity, which according to (10.4) is 3ϵE/(2 + ϵ).

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10.5. Energy in dielectrics

The formula used in most applications is

W = \frac{1}{8π} ∫ E · D d^{3}x  (10.5)

which can be transformed into

W = \frac{1}{2} ∫ ρ(x)Φ(x) d^{3}x

by using E = −∇Φ and ∇ · D = 4πρ. However, these formulas are valid only for linear media, in which χ^{(e)} and ϵ do not depend on E. The general formula for a scalar ϵ is

W = \frac{1}{4π} ∫ d^{3}x (∫_{0}^{D} E′ dD′)  (10.6)

where of course

dD = (ϵ + E \frac{dϵ}{dE} ) dE

The evaluation of (10.6) is not simple in general, because one must solve a different boundary value problem for each ϵ. However, in the case of charges embedded in a uniform infinite medium, D is the same as the free-space solution for E, independent of ϵ.

To obtain (10.6), start from the infinitesimal work

δW = ∫ δρ(x)Φ(x) d^{3}x

use δρ = 4π∇·δD and integrate by parts to obtain

δW = −\frac{1}{4π} ∫ ∇Φ·δD d^{3}x = \frac{1}{4π} ∫ E·δD d^{3}x

When the relation between E and D is linear the integration is trivial and leads to (10.5).

A a consequence of (10.5), or directly from computing E and D, it is seen that the capacitance coefficients C_{ij} in a medium can be obtained as follows: compute them in vacuum and multiply by ϵ. The assumption is of course that all the charges are embedded in the medium of dielectric constant ϵ. Thus the capacitance of an isolated sphere of radius a is ϵa, and the (internal) capacitance of a large parallel plate capacitor of area A is ϵA/d, where d is the spacing between the plates.

If the dielectrics are bounded by equipotential surfaces, or by surfaces of zero flux, the capacitance is also easily obtained from the free-space capacitance.

Forces acting on dielectrics are obtained by differentiating W with respect to the coordinates of the dielectric. Just as in the-free space case, both these formulas are correct for linear dielectrics:

F_{ξ} = − (\frac{∂W}{∂ξ} )_{Q} and in particular F_{ξ} = −\frac{Q^{2}}{2} (\frac{∂(1/C)}{∂ξ})

F_{ξ} = (\frac{∂W}{∂ξ} )_{V} and in particular F_{ξ} = \frac{V^{2}}{2} (\frac{∂C}{∂ξ} )

and both show that dielectrics with ϵ > 1 are attracted to the high-field region, such as the space between capacitor plates.

10.5.1. Green’s function of dielectric sphere

Consider first r′ > a. For a < r′ < r we want the usual Green’s function, plus an arbitrary sum of exterior multipoles

∑_{l} (\frac{r^{′ l}}{r^{l+1}} + \frac{C_{l}a^{l}}{r^{l+1}} ) P_{l}(cos θ)

For r < a we have an arbitrary sum of interior multipoles

∑_{l} (\frac{A_{l}r^{l}}{a^{l+1}} ) P_{l}(cos θ)

The usual matching of solutions at r = a gives

A_{l} = (\frac{a}{r′} )^{l+1} + C_{l}
ϵ lA_{l} = l (\frac{a}{r′} )^{l+1} − (l + 1)C_{l}

From these equations:

A_{l} = \frac{ϵ + 2}{ϵ(l + 1) + 1} (\frac{a}{r′} )^{l+1}
C_{l} = − \frac{l(ϵ − 1)}{(l + 1)ϵ + 1} (\frac{a}{r′} )^{l+1}

and the matching of solutions at r = r′ gives

C_{l} = B_{l} + B′_{l} (\frac{a}{r′} )^{2l+1}
lC_{l} = lB_{l} − (l + 1)B′_{l} (\frac{a}{r′} )^{2l+1}

Manifestly C_{l} = B_{l} and B′_{l} = 0.

In the intermediate region a < r < r′ we want the usual Green’s function, plus an arbitrary sum of both interior and exterior multipoles

∑_{l} (\frac{r^{l}}{r^{′ l+1}} + \frac{B_{l}a^{l}}{r^{l+1}} + \frac{B′_{l}r^{l}}{a^{l+1}} ) P_{l}(cos θ)