Electric fields in matter
November 26, 2013
Suppose we apply a constant electric field to a block of material. Then the charges that make up the
matter are no longer in equilibrium: the electrons tend to move in the −E direction and the positive nuclei
tend to move in the direction of the field. In the simplest circumstances, this creates a dipole moment within
the material, and within the material this dipole field, Edipole generally opposes the applied field. The total
electric field is therefore reduced to E − Edipole within the material.
Similarly, an electron approaching a spherical atom will alter the atom. In the simplest model of what
happens, we have a sphere of negative charge around an equal but positively charged center. As the additional
electron approaches, it repels the electron sphere and attracts the nucleus, and since we imagine these to
move freely of one another, the sphere of charge move slightly away and the nucleus moves nearer, creating
a slight dipole. This electric field points in the direction from the nucleus toward the approaching electron,
and therefore attracts the approaching negative charge.
These are highly simplified pictures. What happens in realistic situations may be different in different
directions, and may depend nonlinearly on the applied electric field.
We divide materials into two types: conductors and insulators. Conductors have free charges that
rearrange to ofset any applied fields, but in insulators the charges stay in their respective atoms or molecules.
We call these latter materials dielectrics.
1
The potential produced by a polarized material
We will assume that the polarization of materials is linear in both magnitude and direction, so that each
atom in the material acquires a dipole moment proportional to the applied field p ∝ E. This produces a
dipole moment per unit volume, P. This is a net result of the applied field together with the counter field
produced by the charge separation. Since a single dipole at the origin produces a potential
V (r) =
1 p·r
1 p · r̂
=
4π0 r2
4π0 r3
we may shift the origin to see that a single dipole at postion r0 has a potential
V (r) =
1 p · (r − r0 )
4π0 |r − r0 |3
Now we can add up the combined contributions of small volumes, Pd3 x0 , over a finite volume,
ˆ
1
P · (r − r0 ) 3 0
V (r) =
3 d x
4π0
|r − r0 |
V
where V is the volume of dielectric.
Now, notice that
0 1 ∂
1
1
0
0 ∂
∇
= r̂ 0 + θ̂ 0 0 √
|r − r0 |
∂r
r ∂θ
r2 + r02 − 2rr0 cosθ0
1
r̂0
=
r0 − rcosθ0
(r2 + r02 − 2rr0 cosθ0 )
+ θ̂
3/2
0
1
r0 rsinθ0
r0 (r2 + r02 − 2rr0 cosθ0 )3/2
0
r0 r̂0 + θ̂ rsinθ0 − r̂0 rcosθ0
=
(r2 + r02 − 2rr0 cosθ0 )
r0 − r
=
3/2
3
|r − r0 |
This lets us rewrite the potential:
V (r)
=
1
4π0
=
1
4π0
ˆ
P · (r − r0 )
|r − r0 |
V
ˆ
P · ∇0
3
d3 x0
1
|r − r0 |
d3 x0
V
Now the identity ∇0 · (f v) = ∇0 f · v + f ∇0 · v lets us replace
1
P
1
0
P · ∇0
=
∇
·
−
∇0 · P
|r − r0 |
|r − r0 |
|r − r0 |
The potential becomes
V (r)
=
1
4π0
=
1
4π0
ˆ
0
∇ ·
P
|r − r0 |
d3 x0 −
V
˛
P · n̂ 2 0
1
d x −
|r − r0 |
4π0
S
ˆ
1
4π0
ˆ
∇0 · P 3 0
d x
|r − r0 |
V
∇0 · P 3 0
d x
|r − r0 |
V
The first term on the right has the form of the potentials due to a surface charge P · n̂, while the second has
the form of the potential due to a charge density ∇0 · P. We define the bound surface charge density and
the bound charge density,
σb
≡
P · n̂
ρb
≡
−∇0 · P (r0 )
The potential is now
V (r)
=
1
4π0
˛
σb
1
d2 x0 +
|r − r0 |
4π0
S
ˆ
ρb
d3 x0
|r − r0 |
V
Once we know the dipole moment per unit volume, we can compute σb and ρb and find the dipole moments
produce in the usual way.
2
Electric displacement
We could like to characterize the electric field inside a dielectric, or in regions which include dielectrics. We
write the charge density as the effective bound charge density due to the polarization, together with any
other charges (“free charges”) in the system, inside or outside the dielectric,
ρ = ρb + ρf
2
Then Gauss’s law is
0 ∇ · E = ρ
= ρf − ∇ · P
The electric field here is the total field. Rewriting, we again have the form of Gauss’s law
∇·D
=
ρf
with the electric field replaced by the electric displacement,
D ≡ 0 E + P
and the charge density given by the free charge density.
Notice that in general there is no potential for D since we may have ∇ × D = ∇ × P 6= 0.
Next, we assume a linear material so that the dipole moment per unit volume is proportional to the
electric field in that volume. It is convenient to write the proportionality as
P = 0 χe E
The constant χe is called the electric susceptibility. Clearly this lets us write the electric displacement as
D
= 0 (1 + χe ) E
= E
where = 0 (1 + χe ). These things all have names:
0
permitivity of f ree space
permitivity, dielectric constant
susceptibiliby
χe = 0 − 1
dielectric
constant
0
We can simplify these relations in linear materials. In a linear material,
P =
0 χe E
−0 χe ∇V
=
so that
≡
ρb
−∇ · P
= 0 χ e ∇2 V
1
= −0 χe ρ
0
= −χe (ρf + ρb )
Solving for ρb ,
ρb = −
χe
ρf
1 + χe
If there is no free charge suspended in the dielectric, ρf = 0 and therefore ρb = 0. The bound surface charge
is the normal derivative at the boundary,
σb
=
n̂ · P
=
−0 χe
3
∂V
∂n
3
Boundary conditions
Integrating the divergence of the displacement over a volume, and using the divergence theorem, we have
ˆ
ˆ
3
∇ · Dd x =
ρf d 3 x
V
V
˛
D · n̂d2 x =
Qf
S
and we can find the boundary condition at an interface by integrating over a small cylinder that pierces the
surface. Making the sides of the cylinder vanishingly small so that the charge density reduces to the surface
density and the surface integral is essentially the contribution of the top and bottom of the cylinder, we have
(Dtop · n̂ − Dbottom · n̂) A = σf A
and therefore
out
in
D⊥
− D⊥
= σf
For the tangential component of the displacement, we use
∇×D=∇×P
Integrating around an elongated rectangle with one long side parallel to the surface on the outside and one
long side parallel to the surface on the inside, and neglecting any contribution from the short sides leaves us
with
in
out
in
Dout
k − Dk = Pk − Pk
in
Recall that the same computation with ∇ × E = 0 tells us that Eout
k − Ek .
4
Summary: method of solution
The basic equations remain the same for linear materials in the absence of free charges. With free charges
absent, ρf = ρb = 0 and σf = 0, so for the interior and exterior regions we are solving only the Laplace
equation
∇2 Vin
2
∇ Vout
=
0
=
0
If we have azimuthal symmetry, the solutions may be written as
∞ X
Bl
Vin (r, θ) =
Al rl + l+1 Pl (cos θ)
r
l=0
∞
X
Bl
Vout (r, θ) =
Al rl + l+1 Pl (cos θ)
r
l=0
and we need only impose the boundary conditions. These are:
1. The boundary condition at infinity Vout (r = ∞)
2. The boundary condition at the origin, Vin (r = 0)
3. Continuity of the normal component of D across the boundary
out
in
D⊥
− D⊥
= σf = 0
4
4. Continuity of the tangential component of E acros the boundary
in
Eout
k − Ek = 0
If σf is nonzero, we keep it in condition 3; if ρf is nonzero, we solve the Poisson equation instead of the
Laplace equation. Assuming is constant,
5
5.1
∇·D
= ρj
∇ · (E)
= ρj
∇2 V
= −
ρj
Example
Dielectric sphere in a uniform field
A sphere of radius R, made of linear dielectric material is placed in an otherwise uniform electric field E0 .
Find the resulting field within the inside the sphere. The the susceptibility is χe .
First, choose the z-axis in the direction of the uniform electric field. Then the boundary condition at
infinity is
V∞ = −E0 r cos θ
Solving the Laplace equation for the potential outside the sphere, we have
Vout (r, θ) =
∞ X
Bl
Al rl + l+1 Pl (cos θ)
r
l=0
As r → ∞, the Bl terms become small but we must have
V∞ (r, θ) = −E0 r cos θ =
∞
X
Al rl Pl (cos θ)
l=0
so A1 = E0 , with all other Al = 0. Therefore, putting in the single nonzero Al term explicitly, we have
Vout (r, θ) = −E0 r cos θ +
∞
X
Bl
Pl (cos θ)
rl+1
l=0
for the exterior solution.
In the interior of the sphere, we have the free and bound charge densities related by
ρb =
χe
ρf
1 − χe
but there is no free charge so both vanish, and we again solve the Laplace equation,
Vin (r, θ) =
∞ X
A0l rl
l=0
Bl0
+ l+1
r
Pl (cos θ)
This time, regularity at the origin tells us that Bl0 = 0 for all l, so
Vin (r, θ) =
∞
X
l=0
5
A0l rl Pl (cos θ)
The boundary conditions are:
out
in
D⊥
− D⊥
=
σf
Dout
k
=
in
Pout
k − Pk
−
Din
k
It is often convenient to mix these for linear dielectrics. The normal component of D has a discontinuity of
σf , while the parallel components satisfy:
in
Dout
k − Dk
in
= Pout
k − Pk
in
out Eout
k − in Ek
in
= 0 χe Eout
k − 0 χe Ek
in
(out − 0 χe ) Eout
k − (in − 0 χe ) Ek
out
in
out − 0
−1
Eout
−1
Ein
k − in − 0
k
0
0
in
0 Eout
k − 0 Ek
=
0
=
0
=
0
so this just reflects the continuity of the parallel component of the electric field across the boundary.
Now, compute the electric field:
Eout
=
=
=
−∇Vout (r, θ)
!
∞
X
∂
1 ∂
Bl
+ θ̂
Pl (cos θ)
− r̂
−E0 r cos θ +
∂r
r ∂θ
rl+1
l=0
!
!
∞
∞
X
X
(l + 1) Bl
Bl ∂
r̂ E0 cos θ +
Pl (cos θ) − θ̂ E0 sin θ +
Pl (cos θ)
rl+2
rl+2 ∂θ
l=0
Ein
=
=
l=0
−∇Vin (r, θ)
∞
1 ∂ X 0 l
∂
Al r Pl (cos θ)
+ θ̂
− r̂
∂r
r ∂θ
l=0
=
−r̂
∞
X
lA0l rl−1 Pl (cos θ) − θ̂
l=0
∞
X
A0l rl−1
l=0
∂
Pl (cos θ)
∂θ
For the parallel components at r = R
∞
X
Bl ∂
Pl (cos θ)
−θ̂ E0 sin θ +
Rl+2 ∂θ
!
= −θ̂
∞
X
l=0
l=0
∞
X
A0l Rl−1
∂
Pl (cos θ)
∂θ
∞
X
B2 ∂
B1
Bl ∂
∂
∂
0
0
sin
θ
+
P
(cos
θ)
+
P
(cos
θ)
=
−A
sin
θ
+
A
R
P
(cos
θ)
+
A0l Rl−1 Pl (cos θ)
2
l
2
1
2
R3
R4 ∂θ
Rl+2 ∂θ
∂θ
∂θ
l=3
l=0
∞ X
B1
B2
∂
Bl
∂
E0 − 3 + A01 sin θ +
− A02 R
P2 (cos θ) =
A0l Rl−1 − l+2
Pl (cos θ)
R
R4
∂θ
R
∂θ
E0 sin θ −
l=0
The derivative terms are all independent (see below), and independent of sin θ, so each coefficient must
vanish. For each l ≥ 2, this implies
A0l
Bl
R2l+1
=
while for the l = 1 term,
E0 −
B1
+ A01 = 0
R3
6
Now we look at the normal components of D. Again, there is no free surface charge, so we have continuity
of the normal components,
!
∞
∞
X
X
(l + 1) Bl
Pl (cos θ) + in r̂
out r̂ E0 cos θ +
lA0l Rl−1 Pl (cos θ) =
Rl+2
l=0
∞
X
l=0
∞
X
2B1
(l + 1) Bl
B0
+
cos
θ
+
Pl (cos θ) + in A01 cos θ + in
lA0l Rl−1 Pl (cos θ)
out
out
2
3
R
R
Rl+2
l=2
l=2
∞ X
2B1
(l + 1) out Bl
B0
0 l−1
+
l
A
R
Pl (cos θ)
out 2 + out E0 + out 3 + in A01 cos θ +
in l
R
R
Rl+2
out E0 cos θ + out
l=2
Again, each coefficient must vanish separately,
B0
=
0
2B1
+ in A01
R3
=
0
(l + 1) out Bl
+ lin A0l Rl−1
Rl+2
=
0 l≥2
out E0 + out
Now we solve the system. For l ≥ 2 we must have
A0l
=
Bl
R2l+1
(l + 1) out Bl
+ lin A0l Rl−1
Rl+2
=
0
so combining,
(l + 1) out Bl
Bl
+ lin 2l+1 Rl−1 =
Rl+2 R
(l + 1) out + lin
Bl
=
Rl+2
0
0
and therefore Bl = 0 for all l ≥ 2.
Finally we need
B1
+ A01
R3
=
0
2B1
+ in A01
R3
=
0
E0 −
out E0 + out
Eliminating A01 ,
2B1
B1
out E0 + out 3 + in −E0 + 3
= 0
R
R
2B1
in B1
out E0 − in E0 + out 3 +
= 0
R
R3
2out
in
+ 3 B1 = (in − out ) E0
R3
R
in − out
B1 =
R3 E0
2out + in
7
0
=
0
=
0
We therefore have only two nonvanishing coefficients,
3out
A01 = −
E0
2out + in
in − out
B1 =
R3 E0
2out + in
The potential is
Vout (r, θ)
=
Vin (r, θ)
=
in − out R3
E0 cos θ
−E0 r cos θ +
2out + in r2
3out
−
E0 r cos θ
2out + in
At the boundary, we check continuity:
Vout (R, θ)
=
−E0 R cos θ +
=
−
in − out
2out + in
RE0 cos θ
3out
RE0 cos θ
2out + in
which is just Vin (R, θ).
The final electric fields are
2R3 in − out
in − out R3
Eout = r̂ 1 + 3
E0 cos θ − θ̂ 1 −
E0 sin θ
r
2out + in
2out + in r3
3out
3out
E0 cos θ − θ̂
E0 sin θ
Ein = r̂
2out + in
2out + in
A note on the independence of the derivative terms
In solving this, we had the equality
E0 sin θ−
∞
∞
X
X
B2 ∂
Bl ∂
∂
B1
∂
0
0
sin
θ+
P
(cos
θ)+
P
(cos
θ)
=
−A
sin
θ+A
R
P
(cos
θ)+
A0l Rl−1 Pl (cos θ)
2
l
2
1
2
R3
R4 ∂θ
Rl+2 ∂θ
∂θ
∂θ
l=3
l=0
It’s not hard to guess that
E0 sin θ = −A01 sin θ
but let’s look in detail at more general derivative terms. For P2 =
∂
P2 (cos θ)
∂θ
1
2
3x2 − 1 we have
∂ 1
3 cos2 θ − 1
∂θ 2
= −3 cos θ sin θ
=
Notice two things:
1. The derivative is of the same order as P2 in trig functions
2. The expression involves sin θ, which is indepenent of cos θ
These derivatives must be expressed in terms the associated Legendre polynomials, which give the θ-dependence
of the spherical harmonics (which also depend on ϕ). All of these are orthogonal to the rest. To show the
independence of two of the derivative terms,
∂
∂
Pl (cos θ) , Pl0 (cos θ)
∂θ
∂θ
8
we may integrate,
ˆπ
∂
∂
Pl (cos θ) Pl0 (cos θ) sin θdθ
∂θ
∂θ
0
To change to x = cos θ, we need
∂
∂θ
=
∂x ∂
∂θ ∂x
= − sin θ
∂
∂x
and the integral becomes
ˆπ
∂
∂
Pl (cos θ) Pl0 (cos θ) sin θdθ
∂θ
∂θ
ˆ1
sin2 θ
=
∂
∂
Pl (x)
Pl0 (x) dx
∂x
∂x
−1
0
ˆ1
=
∂
∂
Pl (x)
Pl0 (x) 1 − x2 dx
∂x
∂x
−1
Without loss of generality, suppose l0 < l. Then we may integrate by parts
ˆ1
∂Pl ∂Pl0
1 − x2 dx
∂x ∂x
ˆ1
=
−1
But
∂Pl0
∂x
∂
∂x
ˆ1
∂Pl0
∂ ∂Pl0
2
Pl
1−x
dx − Pl
1 − x2 dx
∂x
∂x ∂x
−1
0
−1
is a polynomial of order xl −1 so that
∂
∂x
∂Pl0
∂x
1 − x2
0
is a polynomial of order xl −1+2−1 = xl
0
and therefore expressible in terms of Legendre polynomials of order l0 or less. Since this linear combination,
0
∂
∂x
∂Pl0
1 − x2
∂x
=
l
X
αk Pk (x)
k=0
is multiplied by Pl with l > k, each term in the sum is orthogonal and the integral vanishes.
∂
∂
Pl (cos θ) , ∂θ
Pl0 (cos θ) are orthogonal if l 6= l0 .
This proves that ∂θ
You may assume this when working these problems.
9