6.1 Polarization by Superposition
The Gauss’ law electric field produced by a sphere with uniform charge density ρ centered at the origin is
E(r) =
An identical sphere, but with charge density−ρ displaced from the origin by δ, produces the negative of this field except that r→ r − δ. Moreover, to lowest order in δ,
|r − δ|−3 = [(r− δ) · (r − δ)]−3/2
Hence, the total field produced by the superposition of the two spheres is
E(r) =
This may be compared with the field produced by a sphere with volume V and polarization P:
6.2 How to Make a Uniformly Charged Sphere The equation to be solved is
∇ · P =
−ρP r < R, 0 r > R.
We solve this by analogy with the problem
∇ · E =
ρ/0 r < R, 0 r > R.
The Gauss’ law solution for the latter problem is
E(r) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ ρ 30
r r < R,
ρ 30
R3
r3r r > R.
Therefore, the desired polarization is
P(r) =
⎧⎪
⎪⎨
⎪⎪
⎩
−ρP
3 r r < R, ρP
3 R3
r3 r r > R.
Source: A.M. Portis, Electromagnetic Fields (Wiley, New York, 1978).
6.3 The Energy of a Polarized Ball
Choose P = P ˆz so the surface polarization charge density is σP(θ) = P· ˆn = P ˆz · r = P cos θ.
The volume polarization density is zero for this system. Therefore, the total energy is
UE = 1 2
dS σ(θ)ϕ(rS) = 1 8π0
dS
dS σ(θ)σ(θ)
|rS − rS| = P2 8π0
dS
dScos(θ) cos(θ)
|rS − rS| .
Now, cos θ =
$4π
3 Y10(θ, φ) and, since rS = rS, we can use 1
|rS − rS| = 1 R
∞ L = 0
L M =−L
4π
2L + 1YL M∗ (θ, φ)YL M(θ, φ).
Both integrals are now examples of the orthonormality relation for the spherical harmonics:
dΩ Ym∗ (Ω)Ym(Ω) = δδm m. Hence,
UE = 2P2R3 90
.
Source: A.M. Portis, Electromagnetic Fields (Wiley, New York, 1978).
6.4 A Hole in Radially Polarized Matter
The polarization is uniform in magnitude but always points in the radial direction outside an origin-centered sphere of radius R as shown below.
R P
The surface density of polarization charge is σ =−P·ˆr = −P at r = R. The volume density of polarization charge density is
ρP =−∇ · P = −∇ · [Pˆr].
Now,
∇ · ˆr = ∇ ·r r
= ∇ · r
r + r· ∇
1 r
= 3 r + r·
−ˆr r2
=2 r. Therefore,
ρP(r) =−2P
r r≥ R.
By Gauss’ law, both σP and ρP produce purely radial electric fields outside the hole. Specif-ically,
Eρ(r) = Q(r)
4π0r2ˆr with Q(r) =−4π
r
0
dss2ρP(s) = 4πP (R2− r2),
and
Eσ(r) =−4πP R2
4π0r2ˆr =−P R2
0r2ˆr.
Therefore, the total electric field everywhere is
E(r) = Eρ(r) + Eσ(r) =
⎧⎪
⎨
⎪⎩
−P
0ˆr r≥ R,
0 r≤ R.
6.5 The Field at the Center of a Polarized Cube
The volume polarization charge density is zero for a uniformly polarized object. The surface polarization σP = P· ˆn is P on the right (R) face of the cube and −P on the left (L) face of the cube.
R
Since we only have surface charge,
E(r) = P
At the origin,
E(0) =− P
By symmetry, the x and y components of these integrals are zero. Therefore, if the origin of the primed system is at the center of the cube,
Ez(0) = − P
The integral is the solid angle subtended by the right face at the center of the cube. By symmetry, this number must be 4π/6. Therefore, the electric field at the center of the cube is
E(0) =− P 30
.
This is exactly the same as the electric field at the center of a uniformly polarized sphere found in Application 6.1!
Source: Prof. H.B. Biritz, Georgia Institute of Technology (private communication).
6.6 Practice with Poisson’s Formula
A body with volume V and uniform charge density ρ0 produces an electric field E0(r). If we replace ρ0 in the body by a uniform polarization P0, Poisson’s relation asserts that the electrostatic potential produced by the polarized body is
ϕ(r) = P0· E0(r) ρ0 .
Let the plane z = 0 bisect the slab with uniform charge density ρ. From Gauss’ law in integral form, the electric field everywhere due to the slab is
E(z) =
⎧⎨
⎩
sgn(z)(ρt/0)ˆz |z| > t, (ρz/0) ˆz |z| < t.
As far as electrostatics is concerned, the slab with uniform polarization P is equivalent to a plane at z = t with uniform charge/area σ = P· ˆz and a plane at z = −t with uniform charge/area σ =−P·ˆz. The potential of a sheet of charge at z = 0 with uniform charge/area σ is ϕσ(z) =−(σ/20)|z| . Therefore, the potential of the polarized sheet is
ϕ(z) = P· ˆz
20 {|z + t| − |z − t|} .
By checking the four intervals z <−t, −t ≤ z < 0, 0 ≤ z < t, and t ≤ z separately, it is easy to confirm that ϕ(z) and E(z) do indeed satisfy the Poisson relation.
6.7 Isotropic Polarization
ϕ(r) = 1 4π0
d3rP(r)· ∇ 1
|r − r| =− 1 4π0
d3rP(r)· ∇ 1
|r − r|
= −
R
0
drP(r)· ∇
dS 4π0
1
|r − r|
.
The quantity in square brackets is the potential ϕ(r) of a sphere of radius r with uniform surface charge density σ = 1. This means that the charge of that sphere is Q = 4πr2 and
ϕ(r) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ r
0 r < r, r2
0r r > r.
(a) Substituting ϕ(r) for r > rabove gives the potential of the polarized sphere for r > R:
ϕ(r) =− 1 4π0
R
0
drr2P(r)· ∇1
r =− 1 4π0
d3rP(r)· ∇1 r. This is exactly the potential of a point dipole at the origin,
ϕ(r) =− 1
4π0p· ∇1 r, with electric dipole moment
p =
d3rP(r).
(b) ϕ(r) is independent of r when r < r so the potential of the polarized sphere is zero when r < R.
6.8 E and D for an Annular Dielectric
(a) We treat the geometry shown below as the superposition of a ball with radius b and uniform polarization P and a concentric ball with radius a and uniform polarization
−P.
b a
P
From the text, the field produced by an origin-centered polarized ball with volume V is
E(r) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
− P 30
r < R, V
4π0
3(r· P)r r5 −P
r3
r > R.
Therefore, the field in question is
E(r) =
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
0 r < a,
−P 30 − a3
30
3(r· P)r r5 −P
r3
a < r < b,
b3− a3 30
3(r· P)r r5 −P
r3
r > b.
(b) By symmetry, we must have D(r) = D(r)ˆr. Therefore, the choice of a spherical Gaussian surface of radius r gives
S
dS· D = D(r)4πr2= Qc,en cl = 0.
Therefore, D = 0 everywhere.
Source: A.M. Portis, Electromagnetic Fields (Wiley, New York, 1978).
6.9 The Correct Way to Define E
In the presence of a charge q, nearby conductors or dielectrics are polarized and create a field Eind at the position of the charge. Therefore, if E is the field of interest, the force measured when q is placed at a point is
Fq= q(E + Eind).
The linearity of electrostatics guarantees that the induced field changes sign when the charge that polarizes the conductor/dielectric changes sign.Therefore, the force measured when−q sits at the point in question is
F−q =−q(E − Eind).
Therefore, the electric field we seek is
E = Fq− F−q
2q .
Source: W.M. Saslow, Electricity, Magnetism, and Light (Academic, Amsterdam, 2002).
6.10 Charge and Polarizable Matter Coincident
(a) We will compute the polarization from
P = D− 0E = D−D
κ =κ− 1 κ D.
Gauss’ law in integral form applies to a volume V enclosed by a surface S:
S
dS· D =
V
d3r ρc.
This problem has spherical symmetry so D(r) = D(r)ˆr. Choosing a Gaussian sphere of radius r gives
D(r)4πr2 = ρc
4π 3 r3. Hence,
D(r) = ρc
3r ⇒ P = ρcκ− 1 3κ r.
(b) The volume density of polarization charge is ρP =−∇ · P = −ρc
κ− 1
3κ ∇ · r = ρc
1− κ κ . The surface density of polarization charge is
σP = P· ˆr = ρc
κ− 1 3κ R.
Therefore, the total polarization charge is
QP = ρPV + σPA = ρc1− κ
κ ×4πR3
3 + ρcκ− 1
3κ R× 4πR2 = 0.
This is the expected value because the dielectric is neutral. The free charge ρc is extraneous to the dielectric matter.
6.11 Cavity Field
The matching conditions at a dielectric interface are the continuity of Eand the continuity of D⊥ = [E]⊥. When h l, it is sufficient to enforce these conditions on the large flat surface with ˆn as its normal. All we need is the decomposition
E = E⊥+ E= (E· ˆn)ˆn + [E − (E · ˆn)ˆn] . Applying the matching conditions gives
Ecav =
0
(E0· ˆn)ˆn + [E0− (E0· ˆn)ˆn] .
6.12 Making External Fields Identical
Both spheres produce a dipole field outside of themselves. The dipole moment of the dielec-tric sphere is
p = 4πa30κ− 1 κ + 2E0.
The dipole moment of the conducting sphere is the κ→ ∞ limit: p = 4πb30E0. The fields will be identical for r > a if p = p or
b = a
κ− 1 κ + 2
1/3
.
6.13 The Capacitance Matrix for a Spherical Sandwich
Let ϕ2 and ϕ1 be the potentials of the shells. For r ≥ R2, the system acts like a point charge, so
ϕ2 = q1+ q2 4π0R2
.
From Gauss’ law, the electric field between the spheres is E(r) = ˆr E(r) = ˆr q1 4π0κr2. Therefore,
ϕ1− ϕ2=
R2
R1
ds· E(r) = q1 4π0κ
1 R1 − 1
R2
.
From the previous two equations, we deduce that
q1 = 4π0κ R1R2
R2− R1
(ϕ1− ϕ2)
q2 = 4π0κ R1R2
R1− R2
ϕ1+ 4π0R2
1 + κ R1
R2− R1
ϕ2.
By definition, the elements of the capacitance matrix satisfy
q1 = C11ϕ1+ C12ϕ2
q2 = C21ϕ1+ C22ϕ2. Therefore,
C11 =−C12 =−C21 = 4π0κ R1R2
R2− R1
C22 = 4π0R2
1 + κ R1
R2− R1
.
6.14 A Spherical Conductor Embedded in a Dielectric
R2 R1
κ
(a) Gauss’ law in integral form is
S
dS· D = Qf ,en cl. For this problem with spherical symmetry where D = κ0E, we find immediately that
E(r) =
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎩
0 r < R1, Q
4π
ˆr
r2 R1 < r < R2, Q
4π0
ˆr
r2 r > R2.
There is no free charge anywhere except on the conductor surface so the bulk polar-ization charge is zero everywhere:
ρP =−∇ · P = −(κ − 1)∇ · E = 0.
The surface density of polarization charge is
σP(R1) = −P · ˆr|r = R1 =−(κ − 1)E(R1)· ˆr = −κ− 1 κ
Q 4π0R1 σP(R2) = P· ˆr|r = R2 = (κ− 1)E(R2)· ˆr = κ− 1
κ Q 4π0R2
.
The dielectric is neutral so the total polarization charge vanishes, as it must:
r = R1
dSσP(R1) +
r = R2
dSσP(R2) = 0.
(b) There is no bulk polarization charge as in part (a). Besides E0, which imposes azimuthal symmetry on the problem, the only electric fields in the problem are produced by sur-face polarization charge densities at r = R1and r = R2. Therefore, the potential must have the form of an exterior azimuthal multipole expansion for r > R2 (supplemented by E0) and the form of a sum of interior and exterior azimuthal multipole expansions for R1< r < R2.
ϕout(r, θ) =−E0r cos θ +
∞
= 0
A
r+ 1P(cos θ) r > R2, ϕin(r, θ) =
∞
= 0
Br+ C
r+ 1
P(cos θ) R1 < r < R2.
From the fact that ϕin(R1, θ) = 0, we conclude that
ϕin(r, θ) =
∞
= 0
B
&
r−R2+ 11 r+ 1
'
P(cos θ) R1 < r < R2.
We also have the two matching conditions at r = R2. One is ϕout = ϕin, which tells us that
B1
R32− R31
=−E0R32+ A1
and
A= B[R2+ 12 + R2+ 11 ] = 1. (1)
The other matching condition at r = R2 is
κ∂ϕin
∂r = ∂ϕout
∂r . This tells us that
κB1[R32+ 2R31] =−E0R32− 2A1
and
A =−κB
+ 1R2+ 12 + R2+ 11
= 1. (2)
Equations (1) and (2) are not compatible unless A = B = 0 for = 0. Therefore,
ϕin(r, θ) = B1
r−R31
r2
cos θ
ϕout(r, θ) =
−E0r + A1
r2
cos θ, where
A1 = −E0R32(R31− R32) 2(R32− R31) + κ(R31+ R32)
B1 = −3E0
2[1− (R1/R2)3] + κ[1 + 2(R1/R2)3].
The charge density on the conductor surface is σ(θ) = −0(∂ϕin/∂r)|r = R1 ∝ cos θ. This integrates to zero; no charge is drawn up from ground. The external field polarizes both the neutral metal and the neutral dielectric, but there is no impetus to attract charge from ground.
6.15 A Parallel-Plate Capacitor with an Air Gap
The figure below shows the capacitor with the dielectric slab inserted. The potential differ-ence is maintained at V and a charge per unit area ±σf develops on the inner surface of the conducting plates. A polarization charge develops on the surfaces of the dielectric, but we will not need this information to solve the problem.
E2
D1 = ε0κE1 d
t V z
0
−σc
+σc
(a) We use the dark dashed Gaussian surface to find the electric field E2 in the air:
E2 =σf
0
. (1)
We use the white dashed Gaussian surface to find the D-field in the dielectric:
D1 = σf = κ0E1. Finally, since the potential difference is maintained at V ,
V =
t
0
E1dz +
d
t
E2dz = E1t + (d− t)E2 = σf κ0
t + σf
0
(d− t). (2)
The capacitance C is defined, so
Q = σfA = CV.
Therefore, using (2) to find σf, we find
C = 0κA κd− (κ − 1)t.
(b) Electric breakdown occurs when the electric field exceeds a critical value. Therefore, the criterion that V be the breakdown voltage is
E2 = V0
d .
Using this and (1) with the results of (a) gives the desired result,
V = V0
κd[t + κ(d− t)] = V0
1− t
d
1−1
κ
.
Source: O.D. Jefimenko, Electricity and Magnetism (Appleton-Century-Crofts, New York, 1966).
6.16 Helmholtz Theorem for D(r)
The Maxwell equations for dielectric matter are
∇ · D = ρf ∇ × E = 0.
To exploit the Helmholtz theorem, we use D = 0E + P to write these in the form
∇ · D = ρf ∇ × D = ∇ × P.
The Helmholtz theorem gives
D(r) = − 1 4π∇
d3r∇· D(r)
|r − r| + 1 4π∇ ×
d3r ∇× D(r)
|r − r|
= − 1 4π∇
d3r ρf(r)
|r − r|+ 1 4π∇ ×
d3r∇× P(r)
|r − r| .
For simple dielectric matter, P = 0χeE. Therefore,∇×P = 0χe∇×E = 0. Consequently,
D(r) =− 1 4π∇
d3r ρf(r)
|r − r| = 1 4π
d3rρf(r) r− r
|r − r|3.
6.17 Electrostatics of a Doped Semiconductor
(a) Gauss’ law is ∇ · D = ρf. Therefore, the electric field produced by the ions satisfies (d/dz)E+ = eND for 0 < z≤ d and (d/dz)E+ = 0 outside this region. The latter means that the field is constant outside the doping region. Moreover, E+(d/2) = 0 by symmetry, so
E+(z) = eND
z−d
2
0≤ z ≤ d.
E+(z) is continuous everywhere because the charge density is not singular. The re-sulting field is plotted as the solid line in the left figure below.
(b) The free charge in the doping layer has volume density eND. Each ion polarizes the dielectric medium so the total charge in the doping layer has volume density eND/κ.
Therefore, by conservation of charge, the surface charge density (composed of ionized electrons and positive surface polarization charge) must be σ =−eNDd/κ. This layer of charge produces a field
E0(z) = sgn(z) σ 20
=−sgn(z)eNDd 2 . This is plotted as the dashed line in the left figure below.
(c) The total field E = E+ + E− is plotted in the right figure below. The system of semiconductor plus dopant atoms has net zero charge. Therefore, from Gauss’ law for the total field,
∞
−∞
dzdE dz = 1
0
∞
−∞
dzρ = 0.
On the other hand, the integral on the far left is E(∞) − E(−∞). This is consistent with our graph where E(∞) = E(−∞) = 0.
eNDd E+
E−
E z
d ε d
6.18 Surface Polarization Charge
Let S be the dividing surface between the dielectrics. The dashed surfaces in the diagram below are Sout and Sin. Each lies entirely in the κout or κin material, respectively.
If Qf =%
qk, Gauss’ law tells us that
0
So u t
dS· Eout= Qf + Qpol
and
0
Si n
dS· Ein = Qf.
Therefore,
0
⎡
⎣
So u t
dS· Eout−
Si n
dS· Ein
⎤
⎦ = Qpol.
But D = E, so we can change the integration range in both cases to the surface S when we write
S
dS· D
0
out − 0
in
= Qpol.
The surface integral of D over S is Qf. Therefore,
Qp ol= Q
1 κout − 1
κin
.
Source: T.P. Doerr and Y.-K. Yu, American Journal of Physics 72, 190 (2004).
6.19 An Elastic Dielectric
(a) The energy is
U (q, d) = 1
2k(d− d0)2+ q2 2C(d)= 1
2k(d− d0)2+q2 2
d
A. The equilibrium is reached when ∂U/∂d = 0, i.e., for
d(q) = d0− q2 2kA. (b) At equilibrium the potential difference is
V (q) = q
C(q)= qd(q)
A = q C0
1− q2
3q02
, C0 = A
d0, 3q02 = 2kd0A.
Therefore, the capacitance is
Cd(q) = dq/dV = C0 1− q2/q02, which diverges at q = q0.
2
6.20 A Dielectric Inclusion
Let Vin be the volume with permittivity in and Vout be the complementary volume with permittivity out. The dipole moment of the entire system is
p =
6.21 A Classical Meson
(a) If we orient p along the z-axis, the dipole makes a contribution (p/4π0r2) cos θ to the interior potential, ϕin. Our experience with matching conditions and interior and exterior multipole expansions tells us that the potential produced by the medium must vary as cos θ also. Therefore, the most general potential for this problem is
ϕ(r, θ) =
There is no free charge at the cavity boundary, so the matching conditions are
ϕin(R, θ) = ϕout(R, θ) ∂ϕin
Solving these for A and B gives
A =− 2p 4π0a3
κ− 1
2κ + 1 and B = p
4π0 3 2κ + 1.
The corresponding electric field is
E(r, θ) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ p 4π0
3 cos θˆr− ˆz
r3 − Aˆz r < R, B3 cos θˆr− ˆz
r3 r > R.
Finally, Din = 0Ein and Dout= κ0Eout.
(b) We confirm immediately that Dout= 0 when κ = 0. Otherwise,
Uin = 1 2
a≤r≤R
d3r Ein· Din = 1 2
dΩ
R
a
drr20|Ein|2.
The various contributions to Uin behave like 1
R3, 1
a3, R3
a6 , ln R
a3 , and ln a a3 .
Only the first of these is independent of the cutoff and competes with the surface energy (∝ R2) to determine the size of the cavity.
6.22 An Application of the Dielectric Stress Tensor
κ Q
Q′
There is a radial inward force per unit area fin which acts on the inner surface of the shell and a radial outward force per unit area fout which acts on the outer surface of the shell.
The hemispheres will stay together if fin≥ fout.
The stress tensor formalism gives the force exerted on a sub-volume enclosed by a surface S as
F =
S
dS f =
S
dS[(ˆn· D)E −12n(Eˆ · D)].
To find fin, we use the Gauss’ law fields in the dielectric just inside the inner surface of the shell:
Din = 0κEin = Q 4π
ˆ r r2. If the shell has radius R, this gives
fin = Din2
20κ = Q2/κ 320R4π2.
To find fout, we use the Gauss’ law fields in the vacuum just outside the outer surface of the shell:
Dout= 0Eout= Q + Q 4π
ˆr r2. This gives
fout=D2out 20
=(Q + Q)2 320R4π2. Therefore, the shell will not separate if Q2/κ≥ (Q + Q)2 or
0≥ Q2(1−1
κ) + Q2+ 2QQ.
Since κ > 1, this shows that Q and Q must have opposite sign. If we put Q→ −Q and let x = Q/Q, the no-separation condition reads
y = x2− 2x + 1 −1
κ≤ 0. (1)
This function is positive at x = 0 and has positive curvature. Therefore (1) is satisfied for values of x that lie between the zeroes of y(x). From the quadratic equation, these occur at
x = Q
Q = 1± 1
√κ.
Source: V.C.A. Ferraro, Electromagnetic Theory (Athlone Press, London, 1954).
6.23 Two Dielectric Interfaces
The dashed lines in each figure below show a surface S which snugly encloses an interface.
The force on the interface is F =
S
dS
(ˆn· E)D −1
2n(Eˆ · D)
.
Let the z-axis point upward so the electric field in each medium is E1 = E1ˆz and E2 = E2ˆz.
κ2
κ2
z κ1 x
κ1
Horizontal Interface: The force per unit area f = f ˆz is
f = 1
2(E2D2− E1D1) = 1
20(κ2E22− κ1E12).
But D = Dˆz is continuous at the interface, so
D = 0κ1E1= 0κ2E2.
If the thickness of each dielectric layer is d/2, we have E0 = V /d and
E1
Vertical Interface: The force per unit area f = f ˆx is f = E2D2− E1D1. But E1= E2 = E0 = V /d, so
f = ˆx0E02(κ2− κ1).
6.24 The Force on an Isolated Dielectric
We set ρf = 0 because this piece requires no comment. Otherwise, we use (∂kPk)Ei= ∂k(PkEi) + (∂kEi)Pk
to eliminate the∇ · P contribution to get
Fi=−
Transforming the first term on the right into a surface integral gives
Fi=−
Now, the matching conditions at the surface of a polarized dielectric are
ˆ
Therefore,
Eout− Ein = σP
0 n =ˆ nˆ· P
0 n.ˆ Using dS = dS ˆn and restoring the free charge gives the final result:
F =
d3r [ρf(r) + P(r)· ∇ ]E(r) + 1 20
dS [ˆn(rS)· P(rS)]2.
6.25 Minimizing the Total Energy Functional Using the hint, we seek a minimum of the functional
F [ D ] = 1 2
V
d3r|D|2
−
V
d3r ϕ(r)(∇ · D − ρf).
The factor of 12 and the minus sign are inserted for convenience. Operationally, we compute δF = F [D + δD ]− F [D] and look for the conditions that make δ F = 0 to first order in δD . This extremum is a minimum if δ F > 0 to second order in δD .
The first step is to integrate by parts to get
F [ D ] = 1 2
V
d3r|D|2
+
V
d3r [D · ∇ϕ + ρfϕ] −
S
dS· D ϕ.
A direct calculation of δF to first order in δD gives
δ F =
V
d3r
D
+∇ϕ
· δD −
S
dS· δDϕ.
Finally, since the variation δD is arbitrary, δF vanishes if D (r) =−∇ϕ(r) and ˆn·δD|S = 0.
The second of these is true if we specify the normal component of D on the boundary surface.
The first implies that∇ × D = 0. Together with the divergence constraint, this guarantees that D and E =−∇ϕ satisfy Maxwell’s electrostatic equations. The second-order term in the variation of F [D] is 12
d3r|δD|2
. This is a positive quantity, so the extremum we have found does indeed correspond to a minimum of the total electrostatic energy.