Chapter 3
The static electromagnetic ?eld
3.1 Static ?elds and steady currents
Perhaps the most carefully studied area of electromagnetics is that in which the ?elds
are time-invariant. This area, known generally as statics, o?ers (1)the most direct op-
portunities for solution of the governing equations, and (2)the clearest physical pictures
of the electromagnetic ?eld. We therefore devote the present chapter to a treatment
of static ?elds. We begin to seek and examine speci?c solutions to the ?eld equations;
however, our selection of examples is shaped by a search for insight into the behavior of
the ?eld itself, rather than a desire to catalog the solutions of numerous statics problems.
We note at the outset that a static ?eld is physically sensible only as a limiting case
of a time-varying ?eld as the latter approaches a time-invariant equilibrium, and then
only in local regions. The static ?eld equations we shall study thus represent an idealized
model of the physical ?elds.
If we examine the Maxwell–Minkowski equations (2.1)–(2.4) and set the time deriva-
tives to zero, we obtain the static ?eld Maxwell equations
?×E(r) = 0, (3.1)
?·D(r) = ρ(r), (3.2)
?×H(r) = J(r), (3.3)
?·B(r) = 0. (3.4)
We note that if the ?elds are to be everywhere time-invariant, then the sources J and
ρ must also be everywhere time-invariant. Under this condition the dynamic coupling
between the ?elds described by Maxwell’s equations disappears; any connection between
E, D, B, and H imposed by the time-varying nature of the ?eld is gone. For static ?elds
we also require that any dynamic coupling between ?elds in the constitutive relations
vanish. In this static ?eld limit we cannot derive the divergence equations from the curl
equations, since we can no longer use the initial condition argument that the ?elds were
identically zero prior to some time.
The static ?eld equations are useful for approximating many physical situations in
which the ?elds rapidly settle to a local, macroscopically-static state. This may occur
so rapidly and so completely that, in a practical sense, the static equations describe the
?elds within our ability to measure and to compute. Such is the case when a capacitor
is rapidly charged using a battery in series with a resistor; for example, a 1 pF capacitor
charging through a 1 Omega1 resistor reaches 99.99% of its total charge static limit within
10 ps.
3.1.1 Decoupling of the electric and magnetic ?elds
For the remainder of this chapter we shall assume that there is no coupling between
E and H or between D and B in the constitutive relations. Then the static equations
decouple into two independent sets of equations in terms of two independent sets of ?elds.
The static electric ?eld set (E,D)is described by the equations
?×E(r) = 0, (3.5)
?·D(r) = ρ(r). (3.6)
Integrating these over a stationary contour and surface, respectively, we have the large-
scale forms
contintegraldisplay
Gamma1
E · dl = 0, (3.7)
contintegraldisplay
S
D · dS =
integraldisplay
V
ρ dV. (3.8)
The static magnetic ?eld set (B,H)is described by
?×H(r) = J(r), (3.9)
?·B(r) = 0, (3.10)
or, in large-scale form,
contintegraldisplay
Gamma1
H · dl =
integraldisplay
S
J · dS, (3.11)
contintegraldisplay
S
B · dS = 0. (3.12)
We can also specialize the Maxwell–Bo? equations to static form. Assuming that the
?elds, sources, and equivalent sources are time-invariant, the electrostatic ?eld E(r) is
described by the point-form equations
?×E = 0, (3.13)
?·E =
1
epsilon1
0
(ρ ??·P), (3.14)
or the equivalent large-scale equations
contintegraldisplay
Gamma1
E · dl = 0, (3.15)
contintegraldisplay
S
E · dS =
1
epsilon1
0
integraldisplay
V
(ρ ??·P) dV. (3.16)
Similarly, the magnetostatic ?eld B is described by
?×B = μ
0
(J +?×M), (3.17)
?·B = 0, (3.18)
or
contintegraldisplay
Gamma1
B · dl = μ
0
integraldisplay
S
(J +?×M)· dS, (3.19)
contintegraldisplay
S
B · dS = 0. (3.20)
Figure 3.1: Positive point charge in the vicinity of an insulated, uncharged conductor.
It is important to note that any separation of the electromagnetic ?eld into independent
static electric and magnetic portions is illusory. As we mentioned in § 2.3.2, the electric
and magnetic components of the EM ?eld depend on the motion of the observer. An
observer stationary with respect to a single charge measures only a static electric ?eld,
while an observer in uniform motion with respect to the charge measures both electric
and magnetic ?elds.
3.1.2 Static ?eld equilibrium and conductors
Suppose we could arrange a group of electric charges into a static con?guration in free
space. The charges would produce an electric ?eld, resulting in a force on the distribution
via the Lorentz force law, and hence would begin to move. Regardless of how we arrange
the charges they cannot maintain their original static con?guration without the help
of some mechanical force to counterbalance the electrical force. This is a statement of
Earnshaw’s theorem, discussed in detail in § 3.4.2.
The situation is similar for charges within and on electric conductors. A conductor
is a material having many charges free to move under external in?uences, both electric
and non-electric. In a metallic conductor, electrons move against a background lattice
of positive charges. An uncharged conductor is neutral: the amount of negative charge
carried by the electrons is equal to the positive charge in the background lattice. The
distribution of charges in an uncharged conductor is such that the macroscopic electric
?eld is zero inside and outside the conductor. When the conductor is exposed to an addi-
tional electric ?eld, the electrons move under the in?uence of the Lorentz force, creating
a conduction current. Rather than accelerating inde?nitely, conduction electrons experi-
ence collisions with the lattice, thereby giving up their kinetic energy. Macroscopically,
the charge motion can be described in terms of a time-average velocity, hence a macro-
scopic current density can be assigned to the density of moving charge. The relationship
between the applied, or “impressed,” ?eld and the resulting current density is given by
Ohm’s law; in a linear, isotropic, nondispersive material this is
J(r,t) = σ(r)E(r,t). (3.21)
The conductivity σ describes the impediment to charge motion through the lattice: the
Figure 3.2: Positive point charge near a grounded conductor.
higher the conductivity, the farther an electron may move on average before undergoing
a collision.
Let us examine how a state of equilibrium is established in a conductor. We shall con-
sider several important situations. First, suppose we bring a positively charged particle
into the vicinity of a neutral, insulated conductor (we say that a conductor is “insulated”
if no means exists for depositing excess charge onto the conductor). The Lorentz force
on the free electrons in the conductor results in their motion toward the particle (Figure
3.1).Areactionforce F attractstheparticletotheconductor.Iftheparticleandthe
conductor are both held rigidly in space by an external mechanical force, the electrons
within the conductor continue to move toward the surface. In a metal, when these elec-
trons reach the surface and try to continue further they experience a rapid reversal in the
direction of the Lorentz force, drawing them back toward the surface. A su?ciently large
force (described by the work function of the metal)will be able to draw these charges
from the surface, but anything less will permit the establishment of a stable equilibrium
at the surface. If σ is large then equilibrium is established quickly, and a nonuniform
static charge distribution appears on the conductor surface. The electric ?eld within the
conductor must settle to zero at equilibrium, since a nonzero ?eld would be associated
with a current J = σE. In addition, the component of the ?eld tangential to the surface
must be zero or the charge would be forced to move along the surface. At equilibrium,
the ?eld within and tangential to a conductor must be zero. Note also that equilibrium
cannot be established without external forces to hold the conductor and particle in place.
Next, suppose we bring a positively charged particle into the vicinity of a grounded
(ratherthaninsulated)conducto rasinFigure3.2.Useoftheterm“grounded”means
that the conductor is attached via a ?lamentary conductor to a remote reservoir of charge
known as ground; in practical applications the earth acts as this charge reservoir. Charges
are drawn from or returned to the reservoir, without requiring any work, in response to
the Lorentz force on the charge within the conducting body. As the particle approaches,
negative charge is drawn to the body and then along the surface until a static equilibrium
is re-established. Unlike the insulated body, the grounded conductor in equilibrium has
excess negative charge, the amount of which depends on the proximity of the particle.
Again, both particle and conductor must be held in place by external mechanical forces,
and the total ?eld produced by both the static charge on the conductor and the particle
must be zero at points interior to the conductor.
Finally, consider the process whereby excess charge placed inside a conducting body
redistributes as equilibrium is established. We assume an isotropic, homogeneous con-
ducting body with permittivity epsilon1 and conductivity σ. An initially static charge with
density ρ
0
(r) is introduced at time t = 0. The charge density must obey the continuity
equation
?·J(r,t) =?
?ρ(r,t)
?t
;
since J = σE,wehave
σ?·E(r,t) =?
?ρ(r,t)
?t
.
By Gauss’s law, ?·E can be eliminated:
σ
epsilon1
ρ(r,t) =?
?ρ(r,t)
?t
.
Solving this di?erential equation for the unknown ρ(r,t) we have
ρ(r,t) = ρ
0
(r)e
?σt/epsilon1
. (3.22)
The charge density within a homogeneous, isotropic conducting body decreases exponen-
tially with time, regardless of the original charge distribution and shape of the body. Of
course, the total charge must be constant, and thus charge within the body travels to
the surface where it distributes itself in such a way that the ?eld internal to the body
approaches zero at equilibrium. The rate at which the volume charge dissipates is deter-
mined by the relaxation time epsilon1/σ; for copper (a good conductor)this is an astonishingly
small 10
?19
s. Even distilled water, a relatively poor conductor, has epsilon1/σ = 10
?6
s. Thus
we see how rapidly static equilibrium can be approached.
3.1.3 Steady current
Since time-invariant ?elds must arise from time-invariant sources, we have from the
continuity equation
?·J(r) = 0. (3.23)
In large-scale form this is
contintegraldisplay
S
J · dS = 0. (3.24)
A current with the property (3.23)is said to be a steady current. By (3.24), a steady
current must be completely lineal (and in?nite in extent)or must form closed loops.
However, if a current forms loops then the individual moving charges must undergo
acceleration (from the change in direction of velocity). Since a single accelerating particle
radiates energy in the form of an electromagnetic wave, we might expect a large steady
loop current to produce a great deal of radiation. In fact, if we superpose the ?elds
produced by the many particles comprising a steady current, we ?nd that a steady current
produces no radiation [91]. Remarkably, to obtain this result we must consider the exact
relativistic ?elds, and thus our ?nding is precise within the limits of our macroscopic
assumptions.
If we try to create a steady current in free space, the ?owing charges will tend to
disperse because of the Lorentz force from the ?eld set up by the charges, and the
resulting current will not form closed loops. A beam of electrons or ions will produce
both an electric ?eld (because of the nonzero net charge of the beam)and a magnetic ?eld
(because of the current). At nonrelativistic particle speeds, the electric ?eld produces
an outward force on the charges that is much greater than the inward (or pinch)force
produced by the magnetic ?eld. Application of an additional, external force will allow
the creation of a collimated beam of charge, as occurs in an electron tube where a series
of permanent magnets can be used to create a beam of steady current.
More typically, steady currents are created using wire conductors to guide the moving
charge. When an external force, such as the electric ?eld created by a battery, is applied
to an uncharged conductor, the free electrons will begin to move through the positive
lattice, forming a current. Each electron moves only a short distance before colliding with
the positive lattice, and if the wire is bent into a loop the resulting macroscopic current
will be steady in the sense that the temporally and spatially averaged microscopic current
will obey ?·J = 0. We note from the examples above that any charges attempting to
leave the surface of the wire are drawn back by the electrostatic force produced by the
resulting imbalance in electrical charge. For conductors, the “drift” velocity associated
with the moving electrons is proportional to the applied ?eld:
u
d
=?μ
e
E
where μ
e
is the electron mobility. The mobility of copper (3.2 × 10
?3
m
2
/V · s)is such
that an applied ?eld of 1 V/m results in a drift velocity of only a third of a centimeter
per second.
Integral properties of a steady current. Steady currents obey several useful inte-
gral properties. To develop these properties we need an integral identity. Let f (r) and
g(r) be scalar functions, continuous and with continuous derivatives in a volume region
V. Let J represent a steady current ?eld of ?nite extent, completely contained within
V. We begin by using (B.42)to expand
?·( fgJ) = fg(?·J)+ J ·?( fg).
Noting that ?·J = 0 and using (B.41), we get
?·( fgJ) = ( f J)·?g +(gJ)·?f.
Now let us integrate over V and employ the divergence theorem:
contintegraldisplay
S
( fg)J · dS =
integraldisplay
V
[( f J)·?g +(gJ)·?f ] dV.
Since J is contained entirely within S,wemusthave?n · J = 0 everywhere on S. Hence
integraldisplay
V
[( f J)·?g +(gJ)·?f ] dV = 0. (3.25)
We can obtain a useful relation by letting f = 1 and g = x
i
in (3.25), where (x, y, z) =
(x
1
, x
2
, x
3
). This gives
integraldisplay
V
J
i
(r)dV = 0, (3.26)
where J
1
= J
x
and so on. Hence the volume integral of any rectangular component of J
is zero. Similarly, letting f = g = x
i
we ?nd that
integraldisplay
V
x
i
J
i
(r)dV = 0. (3.27)
With f = x
i
and g = x
j
we obtain
integraldisplay
V
bracketleftbig
x
i
J
j
(r)+ x
j
J
i
(r)
bracketrightbig
dV = 0. (3.28)
3.2 Electrostatics
3.2.1 The electrostatic potential and work
The equation
contintegraldisplay
Gamma1
E · dl = 0 (3.29)
satis?ed by the electrostatic ?eld E(r) is particularly interesting. A ?eld with zero
circulation is said to be conservative. To see why, let us examine the work required to
move a particle of charge Q around a closed path in the presence of E(r). Since work is
the line integral of force and B = 0, the work expended by the external system moving
the charge against the Lorentz force is
W =?
contintegraldisplay
Gamma1
(QE + Qv × B)· dl =?Q
contintegraldisplay
Gamma1
E · dl = 0.
This property is analogous to the conservation property for a classical gravitational ?eld:
any potential energy gained by raising a point mass is lost when the mass is lowered.
Direct experimental veri?cation of the electrostatic conservative property is di?cult,
aside from the fact that the motion of Q may alter E by interacting with the sources of
E. By moving Q with nonuniform velocity (i.e., with acceleration at the beginning of the
loop, direction changes in transit, and deceleration at the end)we observe a radiative
loss of energy, and this energy cannot be regained by the mechanical system providing
the motion. To avoid this problem we may assume that the charge is moved so slowly,
or in such small increments, that it does not radiate. We shall use this concept later to
determine the “assembly energy” in a charge distribution.
The electrostatic potential. By the point form of (3.29),
?×E(r) = 0,
we can introduce a scalar ?eld Phi1 = Phi1(r) such that
E(r) =??Phi1(r). (3.30)
The function Phi1 carries units of volts and is known as the electrostatic potential. Let us
consider the work expended by an external agent in moving a charge between points P
1
at r
1
and P
2
at r
2
:
W
21
=?Q
integraldisplay
P
2
P
1
??Phi1(r)· dl = Q
integraldisplay
P
2
P
1
dPhi1(r) = Q [Phi1(r
2
)?Phi1(r
1
)].
The work W
21
is clearly independent of the path taken between P
1
and P
2
; the quantity
V
21
=
W
21
Q
= Phi1(r
2
)?Phi1(r
1
) =?
integraldisplay
P
2
P
1
E · dl, (3.31)
called the potential di?erence, has an obvious physical meaning as work per unit charge
required to move a particle against an electric ?eld between two points.
Figure 3.3: Demonstration of path independence of the electric ?eld line integral.
Of course, the large-scale form (3.29)also implies the path-independence of work in
the electrostatic ?eld. Indeed, we may pass an arbitrary closed contour Gamma1 through P
1
and P
2
andthensplititintotwopiecesGamma1
1
andGamma1
2
asshowninFigure3.3.Since
?Q
contintegraldisplay
Gamma1
1
?Gamma1
2
E · dl =?Q
integraldisplay
Gamma1
1
E · dl + Q
integraldisplay
Gamma1
2
E · dl = 0,
we have
?Q
integraldisplay
Gamma1
1
E · dl =?Q
integraldisplay
Gamma1
2
E · dl
as desired.
We sometimes refer to Phi1(r) as the absolute electrostatic potential. Choosing a suitable
reference point P
0
at location r
0
and writing the potential di?erence as
V
21
= [Phi1(r
2
)?Phi1(r
0
)] ? [Phi1(r
1
)?Phi1(r
0
)],
we can justify calling Phi1(r) the absolute potential referred to P
0
. Note that P
0
might
describe a locus of points, rather than a single point, since many points can be at the same
potential. Although we can choose any reference point without changing the resulting
value of E found from (3.30), for simplicity we often choose r
0
such that Phi1(r
0
) = 0.
Several properties of the electrostatic potential make it convenient for describing static
electric ?elds. We know that, at equilibrium, the electrostatic ?eld within a conducting
body must vanish. By (3.30)the potential at all points within the body must therefore
have the same constant value. It follows that the surface of a conductor is an equipotential
surface: a surface for which Phi1(r) is constant.
As an in?nite reservoir of charge that can be tapped through a ?lamentary conductor,
the entity we call “ground” must also be an equipotential object. If we connect a con-
ductor to ground, we have seen that charge may ?ow freely onto the conductor. Since no
work is expended, “grounding” a conductor obviously places the conductor at the same
absolute potential as ground. For this reason, ground is often assigned the role as the
potential reference with an absolute potential of zero volts. Later we shall see that for
sources of ?nite extent ground must be located at in?nity.
3.2.2 Boundary conditions
Boundary conditions for the electrostatic ?eld. The boundary conditions found
for the dynamic electric ?eld remain valid in the electrostatic case. Thus
?n
12
×(E
1
? E
2
) = 0 (3.32)
and
?n
12
·(D
1
? D
2
) = ρ
s
. (3.33)
Here ?n
12
points into region 1 from region 2. Because the static curl and divergence
equations are independent, so are the boundary conditions (3.32)and (3.33).
For a linear and isotropic dielectric where D = epsilon1E, equation (3.33)becomes
?n
12
·(epsilon1
1
E
1
?epsilon1
2
E
2
) = ρ
s
. (3.34)
Alternatively, using D = epsilon1
0
E + P we can write (3.33)as
?n
12
·(E
1
? E
2
) =
1
epsilon1
0
(ρ
s
+ρ
Ps1
+ρ
Ps2
) (3.35)
where
ρ
Ps
= ?n · P
is the polarization surface charge with ?n pointing outward from the material body.
We can also write the boundary conditions in terms of the electrostatic potential. With
E =??Phi1, equation (3.32)becomes
Phi1
1
(r) = Phi1
2
(r) (3.36)
for all points r on the surface. Actually Phi1
1
and Phi1
2
may di?er by a constant; because
this constant is eliminated when the gradient is taken to ?nd E, it is generally ignored.
We can write (3.35)as
epsilon1
0
parenleftbigg
?Phi1
1
?n
?
?Phi1
2
?n
parenrightbigg
=?ρ
s
?ρ
Ps1
?ρ
Ps2
where the normal derivative is taken in the ?n
12
direction. For a linear, isotropic dielectric
(3.33)becomes
epsilon1
1
?Phi1
1
?n
?epsilon1
2
?Phi1
2
?n
=?ρ
s
. (3.37)
Again, we note that (3.36)and (3.37)are independent.
Boundary conditions for steady electric current. The boundary condition on the
normal component of current found in § 2.8.2 remains valid in the steady current case.
Assume that the boundary exists between two linear, isotropic conducting regions having
constitutive parameters (epsilon1
1
,σ
1
) and (epsilon1
2
,σ
2
), respectively. By (2.198) we have
?n
12
·(J
1
? J
2
) =??
s
· J
s
(3.38)
where ?n
12
points into region 1 from region 2. A surface current will not appear on the
boundary between two regions having ?nite conductivity, although a surface charge may
accumulate there during the transient period when the currents are established [31]. If
charge is in?uenced to move from the surface, it will move into the adjacent regions,
Figure 3.4: Refraction of steady current at a material interface.
rather than along the surface, and a new charge will replace it, supplied by the current.
Thus, for ?nite conducting regions (3.38)becomes
?n
12
·(J
1
? J
2
) = 0. (3.39)
A boundary condition on the tangential component of current can also be found.
Substituting E = J/σ into (3.32)we have
?n
12
×
parenleftbigg
J
1
σ
1
?
J
2
σ
2
parenrightbigg
= 0.
We can also write this as
J
1t
σ
1
=
J
2t
σ
2
(3.40)
where
J
1t
= ?n
12
× J
1
, J
2t
= ?n
12
× J
2
.
We may combine the boundary conditions for the normal components of current and
electric ?eld to better understand the behavior of current at a material boundary. Sub-
stituting E = J/σ into (3.34)we have
epsilon1
1
σ
1
J
1n
?
epsilon1
2
σ
2
J
2n
= ρ
s
(3.41)
where J
1n
= ?n
12
· J
1
and J
2n
= ?n
12
· J
2
. Combining (3.41)with (3.39), we have
ρ
s
= J
1n
parenleftbigg
epsilon1
1
σ
1
?
epsilon1
2
σ
2
parenrightbigg
= E
1n
parenleftbigg
epsilon1
1
?
σ
1
σ
2
epsilon1
2
parenrightbigg
= J
2n
parenleftbigg
epsilon1
1
σ
1
?
epsilon1
2
σ
2
parenrightbigg
= E
2n
parenleftbigg
epsilon1
1
σ
2
σ
1
?epsilon1
2
parenrightbigg
where
E
1n
= ?n
12
· E
1
, E
2n
= ?n
12
· E
2
.
Unless epsilon1
1
σ
2
? σ
1
epsilon1
2
= 0, a surface charge will exist on the interface between dissimilar
current-carrying conductors.
We may also combine the vector components of current on each side of the boundary
todeterminethee?ectsoftheboundaryoncurrentdirection(Figure3.4).Letθ
1,2
denote
the angle between J
1,2
and ?n
12
so that
J
1n
= J
1
cosθ
1
, J
1t
= J
1
sinθ
1
J
2n
= J
2
cosθ
2
, J
2t
= J
2
sinθ
2
.
Then J
1
cosθ
1
= J
2
cosθ
2
by (3.39), while σ
2
J
1
sinθ
1
= σ
1
J
2
sinθ
2
by (3.40). Hence
σ
2
tanθ
1
= σ
1
tanθ
2
. (3.42)
It is interesting to consider the case of current incident from a conducting material onto
an insulating material. If region 2 is an insulator, then J
2n
= J
2t
= 0; by (3.39)we have
J
1n
= 0. But (3.40)does not require J
1t
= 0; with σ
2
= 0 the right-hand side of (3.40)
is indeterminate and thus J
1t
may be nonzero. In other words, when current moving
through a conductor approaches an insulating surface, it bends and ?ows tangential to
the surface. This concept is useful in explaining how wires guide current.
Interestingly, (3.42)shows that when σ
2
lessmuch σ
1
we have θ
2
→ 0; current passing from a
conducting region into a slightly-conducting region does so normally.
3.2.3 Uniqueness of the electrostatic ?eld
In § 2.2.1 we found that the electromagnetic ?eld is unique within a region V when
the tangential component of E is speci?ed over the surrounding surface. Unfortunately,
this condition is not appropriate in the electrostatic case. We should remember that
an additional requirement for uniqueness of solution to Maxwell’s equations is that the
?eld be speci?ed throughout V at some time t
0
. For a static ?eld this would completely
determine E without need for the surface ?eld!
Let us determine conditions for uniqueness beginning with the static ?eld equations.
Consider a region V surrounded by a surface S. Static charge may be located entirely
or partially within V, or entirely outside V, and produces a ?eld within V. The region
may also contain any arrangement of conductors or other materials. Suppose (D
1
,E
1
)
and (D
2
,E
2
) represent solutions to the static ?eld equations within V with source ρ(r).
We wish to ?nd conditions that guarantee both E
1
= E
2
and D
1
= D
2
.
Since ?·D
1
= ρ and ?·D
2
= ρ, the di?erence ?eld D
0
= D
2
? D
1
obeys the
homogeneous equation
?·D
0
= 0. (3.43)
Consider the quantity
?·(D
0
Phi1
0
) = Phi1
0
(?·D
0
)+ D
0
·(?Phi1
0
)
where E
0
= E
2
? E
1
=??Phi1
0
=??(Phi1
2
? Phi1
1
). We integrate over V and use the
divergence theorem and (3.43)to obtain
contintegraldisplay
S
Phi1
0
[D
0
· ?n] dS =
integraldisplay
V
D
0
·(?Phi1
0
)dV =?
integraldisplay
V
D
0
· E
0
dV. (3.44)
Now suppose that Phi1
0
= 0 everywhere on S, or that ?n · D
0
= 0 everywhere on S, or that
Phi1
0
= 0 over part of S and ?n · D
0
= 0 elsewhere on S. Then
integraldisplay
V
D
0
· E
0
dV = 0. (3.45)
Since V is arbitrary, either D
0
= 0 or E
0
= 0. Assuming E and D are linked by the
constitutive relations, we have E
1
= E
2
and D
1
= D
2
.
Hence the ?elds within V are unique provided that either Phi1, the normal component
of D, or some combination of the two, is speci?ed over S. We often use a multiply-
connected surface to exclude conductors. By (3.33)we see that speci?cation of the
normal component of D on a conductor is equivalent to speci?cation of the surface
charge density. Thus we must specify the potential or surface charge density over all
conducting surfaces.
One other condition results in zero on the left-hand side of (3.44). If S recedes to
in?nity and Phi1
0
and D
0
decrease su?ciently fast, then (3.45)still holds and uniqueness
is guaranteed. If D,E ～ 1/r
2
as r →∞, then Phi1 ～ 1/r and the surface integral in (3.44)
tends to zero since the area of an expanding sphere increases only as r
2
. We shall ?nd
later in this section that for sources of ?nite extent the ?elds do indeed vary inversely
with distance squared from the source, hence we may allow S to expand and encompass
all space.
For the case in which conducting bodies are immersed in an in?nite homogeneous
medium and the static ?elds must be determined throughout all space, a multiply-
connected surface is used with one part receding to in?nity and the remaining parts
surrounding the conductors. Here uniqueness is guaranteed by specifying the potentials
or charges on the surfaces of the conducting bodies.
3.2.4 Poisson’s and Laplace’s equations
For computational purposes it is often convenient to deal with the di?erential versions
?×E(r) = 0, (3.46)
?·D(r) = ρ(r), (3.47)
of the electrostatic ?eld equations. We must supplement these with constitutive relations
between E and D; at this point we focus our attention on linear, isotropic materials for
which
D(r) = epsilon1(r)E(r).
Using this in (3.47)along with E =??Phi1 (justi?ed by (3.46)), we can write
?·[epsilon1(r)?Phi1(r)] =?ρ(r). (3.48)
This is Poisson’s equation. The corresponding homogeneous equation
?·[epsilon1(r)?Phi1(r)] = 0, (3.49)
holding at points r where ρ(r) = 0,isLaplace’s equation. Equations (3.48)and (3.49)
are valid for inhomogeneous media. By (B.42)we can write
?Phi1(r)·?epsilon1(r)+epsilon1(r)?·[?Phi1(r)] =?ρ(r).
For a homogeneous medium, ?epsilon1 = 0; since ?·(?Phi1) ≡?
2
Phi1,wehave
?
2
Phi1(r) =?ρ(r)/epsilon1 (3.50)
in such a medium. Correspondingly,
?
2
Phi1(r) = 0
at points where ρ(r) = 0.
Poisson’s and Laplace’s equations can be solved by separation of variables, Fourier
transformation, conformal mapping, and numerical techniques such as the ?nite di?erence
and moment methods. In Appendix A we consider the separation of variables solution
to Laplace’s equation in three major coordinate systems for a variety of problems. For
an introduction to numerical techniques the reader is referred to the books by Sadiku
[162], Harrington [82], and Peterson et al. [146]. Solution to Poisson’s equation is often
undertaken using the method of Green’s functions, which we shall address later in this
section. We shall also consider the solution to Laplace’s equation for bodies immersed in
an applied, or “impressed,” ?eld.
Uniqueness of solution to Poisson’s equation. Before attempting any solutions,
we must ask two very important questions. How do we know that solving the second-order
di?erential equation produces the same values for E =??Phi1 as solving the ?rst-order
equations directly for E? And, if these solutions are the same, what are the conditions
for uniqueness of solution to Poisson’s and Laplace’s equations? To answer the ?rst
question, a su?cient condition is to have Phi1 twice di?erentiable. We shall not attempt to
prove this, but shall instead show that the condition for uniqueness of the second-order
equations is the same as that for the ?rst-order equations.
Consider a region of space V surrounded by a surface S. Static charge may be located
entirely or partially within V, or entirely outside V, and produces a ?eld within V. This
region may also contain any arrangement of conductors or other materials. Now, assume
that Phi1
1
and Phi1
2
represent solutions to the static ?eld equations within V with source
ρ(r). We wish to ?nd conditions under which Phi1
1
= Phi1
2
.
Since we have
?·[epsilon1(r)?Phi1
1
(r)] =?ρ(r), ?·[epsilon1(r)?Phi1
2
(r)] =?ρ(r),
the di?erence ?eld Phi1
0
= Phi1
2
?Phi1
1
obeys
?·[epsilon1(r)?Phi1
0
(r)] = 0. (3.51)
That is, Phi1
0
obeys Laplace’s equation. Now consider the quantity
?·(epsilon1Phi1
0
?Phi1
0
) = epsilon1|?Phi1
0
|
2
+Phi1
0
?·(epsilon1?Phi1
0
).
Integration over V and use of the divergence theorem and (3.51)gives
contintegraldisplay
S
Phi1
0
(r)[epsilon1(r)?Phi1
0
(r)] · dS =
integraldisplay
V
epsilon1(r)|?Phi1
0
(r)|
2
dV.
As with the ?rst order equations, we see that specifying either Phi1(r) or epsilon1(r)?Phi1(r)· ?n over
S results in Phi1
0
(r) = 0 throughout V, hence Phi1
1
= Phi1
2
. As before, specifying epsilon1(r)?Phi1(r)· ?n
for a conducting surface is equivalent to specifying the surface charge on S.
Integral solution to Poisson’s equation: the static Green’s function. The
method of Green’s functions is one of the most useful techniques for solving Poisson’s
equation. We seek a solution for a single point source, then use Green’s second identity
to write the solution for an arbitrary charge distribution in terms of a superposition
integral.
We seek the solution to Poisson’s equation for a region of space V as shown in Figure
3.5.Theregionisassumedhomogeneouswithpermittivityepsilon1,anditssurfaceismultiply-
connected, consisting of a bounding surface S
B
and any number of closed surfaces internal
to V. We denote by S the composite surface consisting of S
B
and the N internal surfaces
S
n
, n = 1,...,N. The internal surfaces are used to exclude material bodies, such as the
Figure 3.5: Computation of potential from known sources and values on bounding sur-
faces.
plates of a capacitor, which may be charged and on which the potential is assumed
to be known. To solve for Phi1(r) within V we must know the potential produced by a
point source. This potential, called the Green’s function, is denoted G(r|r
prime
); it has two
arguments because it satis?es Poisson’s equation at r when the source is located at r
prime
:
?
2
G(r|r
prime
) =?δ(r ? r
prime
). (3.52)
Later we shall demonstrate that in all cases of interest to us the Green’s function is
symmetric in its arguments:
G(r
prime
|r) = G(r|r
prime
). (3.53)
This property of G is known as reciprocity.
Our development rests on the mathematical result (B.30)known as Green’s second
identity. We can derive this by subtracting the identities
?·(φ?ψ)= φ?·(?ψ)+(?φ)·(?ψ),
?·(ψ?φ) = ψ?·(?φ)+(?ψ)·(?φ),
to obtain
?·(φ?ψ ?ψ?φ)= φ?
2
ψ ?ψ?
2
φ.
Integrating this over a volume region V with respect to the dummy variable r
prime
and using
the divergence theorem, we obtain
integraldisplay
V
[φ(r
prime
)?
prime
2
ψ(r
prime
)?ψ(r
prime
)?
prime
2
φ(r
prime
)] dV
prime
=?
contintegraldisplay
S
[φ(r
prime
)?
prime
ψ(r
prime
)?ψ(r
prime
)?
prime
φ(r
prime
)] · dS
prime
.
The negative sign on the right-hand side occurs because ?n is an inward normal to V.
Finally, since ?ψ(r
prime
)/?n
prime
= ?n
prime
·?
prime
ψ(r
prime
),wehave
integraldisplay
V
[φ(r
prime
)?
prime
2
ψ(r
prime
)?ψ(r
prime
)?
prime
2
φ(r
prime
)] dV
prime
=?
contintegraldisplay
S
bracketleftbigg
φ(r
prime
)
?ψ(r
prime
)
?n
prime
?ψ(r
prime
)
?φ(r
prime
)
?n
prime
bracketrightbigg
dS
prime
as desired.
To solve for Phi1 in V we shall make some seemingly unmotivated substitutions into this
identity. First note that by (3.52)and (3.53)we can write
?
prime
2
G(r|r
prime
) =?δ(r
prime
? r). (3.54)
We now set φ(r
prime
) = Phi1(r
prime
) and ψ(r
prime
) = G(r|r
prime
) to obtain
integraldisplay
V
[Phi1(r
prime
)?
prime
2
G(r|r
prime
)? G(r|r
prime
)?
prime
2
Phi1(r
prime
)] dV
prime
=
?
contintegraldisplay
S
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
, (3.55)
hence
integraldisplay
V
bracketleftbigg
Phi1(r
prime
)δ(r
prime
? r)? G(r|r
prime
)
ρ(r
prime
)
epsilon1
bracketrightbigg
dV
prime
=
contintegraldisplay
S
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
.
By the sifting property of the Dirac delta
Phi1(r) =
integraldisplay
V
G(r|r
prime
)
ρ(r
prime
)
epsilon1
dV
prime
+
contintegraldisplay
S
B
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
+
+
N
summationdisplay
n=1
contintegraldisplay
S
n
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
. (3.56)
With this we may compute the potential anywhere within V in terms of the charge
density within V and the values of the potential and its normal derivative over S.We
must simply determine G(r|r
prime
) ?rst.
Let us take a moment to specialize (3.56)to the case of unbounded space. Provided
that the sources are of ?nite extent, as S
B
→∞we shall ?nd that
Phi1(r) =
integraldisplay
V
G(r|r
prime
)
ρ(r
prime
)
epsilon1
dV
prime
+
N
summationdisplay
n=1
contintegraldisplay
S
n
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
.
Ausefulderivativeidentity. Many di?erential operations on the displacement vector
R = r ? r
prime
occur in the study of electromagnetics. The identities
?R =??
prime
R =
?
R, ?
parenleftbigg
1
R
parenrightbigg
=??
prime
parenleftbigg
1
R
parenrightbigg
=?
?
R
R
2
, (3.57)
for example, follow from direct di?erentiation of the rectangular coordinate representa-
tion
R = ?x(x ? x
prime
)+ ?y(y ? y
prime
)+ ?z(z ? z
prime
).
The identity
?
2
parenleftbigg
1
R
parenrightbigg
=?4πδ(r ? r
prime
), (3.58)
crucial to potential theory, is more di?cult to establish. We shall prove the equivalent
version
?
prime
2
parenleftbigg
1
R
parenrightbigg
=?4πδ(r
prime
? r)
Figure 3.6: Geometry for establishing the singular property of ?
2
(1/R).
by showing that
integraldisplay
V
f (r
prime
)?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
=
braceleftBigg
?4π f (r), r ∈ V,
0, r /∈ V,
(3.59)
holds for any continuous function f (r). By direct di?erentiation we have
?
prime
2
parenleftbigg
1
R
parenrightbigg
= 0 for r
prime
negationslash= r,
hence the second part of (3.59)is established. This also shows that if r ∈ V then the
domain of integration in (3.59)can be restricted to a sphere of arbitrarily small radius
εcenteredat r (Figure3.6).Theresultweseekisfoundinthelimitasε→ 0.Thuswe
are interested in computing
integraldisplay
V
f (r
prime
)?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
= lim
ε→0
integraldisplay
V
ε
f (r
prime
)?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
.
Since f is continuous at r
prime
= r, we have by the mean value theorem
integraldisplay
V
f (r
prime
)?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
= f (r) lim
ε→0
integraldisplay
V
ε
?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
.
The integral over V
ε
can be computed using ?
prime
2
(1/R) =?
prime
·?
prime
(1/R) and the divergence
theorem:
integraldisplay
V
ε
?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
=
integraldisplay
S
ε
?n
prime
·?
prime
parenleftbigg
1
R
parenrightbigg
dS
prime
,
where S
ε
bounds V
ε
. Noting that ?n
prime
=?
?
R, using (57), and writing the integral in
spherical coordinates (ε,θ,φ)centered at the point r,wehave
integraldisplay
V
f (r
prime
)?
prime
2
parenleftbigg
1
R
parenrightbigg
dV
prime
= f (r) lim
ε→0
integraldisplay
2π
0
integraldisplay
π
0
?
?
R ·
parenleftBigg
?
R
ε
2
parenrightBigg
ε
2
sinθ dθ dφ =?4π f (r).
Hence the ?rst part of (3.59)is also established.
The Green’s function for unbounded space. In view of (3.58), one solution to
(3.52)is
G(r|r
prime
) =
1
4π|r ? r
prime
|
. (3.60)
This simple Green’s function is generally used to ?nd the potential produced by charge
in unbounded space. Here N = 0 (no internal surfaces)and S
B
→∞.Thus
Phi1(r) =
integraldisplay
V
G(r|r
prime
)
ρ(r
prime
)
epsilon1
dV
prime
+ lim
S
B
→∞
contintegraldisplay
S
B
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
.
We have seen that the Green’s function varies inversely with distance from the source,
and thus expect that, as a superposition of point-source potentials, Phi1(r) will also vary
inversely with distance from a source of ?nite extent as that distance becomes large with
respect to the size of the source. The normal derivatives then vary inversely with distance
squared. Thus, each term in the surface integrand will vary inversely with distance cubed,
while the surface area itself varies with distance squared. The result is that the surface
integral vanishes as the surface recedes to in?nity, giving
Phi1(r) =
integraldisplay
V
G(r|r
prime
)
ρ(r
prime
)
epsilon1
dV
prime
.
By (3.60)we then have
Phi1(r) =
1
4πepsilon1
integraldisplay
V
ρ(r
prime
)
|r ? r
prime
|
dV
prime
(3.61)
where the integration is performed over all of space. Since
lim
r→∞
Phi1(r) = 0,
points at in?nity are a convenient reference for the absolute potential.
Later we shall need to know the amount of work required to move a charge Q from
in?nity to a point P located at r. If a potential ?eld is produced by charge located in
unbounded space, moving an additional charge into position requires the work
W
21
=?Q
integraldisplay
P
∞
E · dl = Q[Phi1(r)?Phi1(∞)] = QPhi1(r). (3.62)
Coulomb’s law. We can obtain E from (61)by direct di?erentiation. We have
E(r) =?
1
4πepsilon1
?
integraldisplay
V
ρ(r
prime
)
|r ? r
prime
|
dV
prime
=?
1
4πepsilon1
integraldisplay
V
ρ(r
prime
)?
parenleftbigg
1
|r ? r
prime
|
parenrightbigg
dV
prime
,
hence
E(r) =
1
4πepsilon1
integraldisplay
V
ρ(r
prime
)
r ? r
prime
|r ? r
prime
|
3
dV
prime
(3.63)
by (3.57). So Coulomb’s law follows from the two fundamental postulates of electrostatics
(3.5)and (3.6).
Green’s function for unbounded space: two dimensions. We de?ne the two-
dimensional Green’s function as the potential at a point r = ρ + ?zz produced by a
z-directed line source of constant density located at r
prime
= ρ
prime
. Perhaps the simplest way
to compute this is to ?rst ?nd E produced by a line source on the z-axis. By (3.63)we
have
E(r) =
1
4πepsilon1
integraldisplay
Gamma1
ρ
l
(z
prime
)
r ? r
prime
|r ? r
prime
|
3
dl
prime
.
Then, since r = ?zz + ?ρρ, r
prime
= ?zz
prime
, and dl
prime
= dz
prime
,wehave
E(ρ) =
ρ
l
4πepsilon1
integraldisplay
∞
?∞
?ρρ + ?z(z ? z
prime
)
bracketleftbig
ρ
2
+(z ? z
prime
)
2
bracketrightbig
3/2
dz
prime
.
Carrying out the integration we ?nd that E has only a ρ-component which varies only
with ρ:
E(ρ) = ?ρ
ρ
l
2πepsilon1ρ
. (3.64)
The absolute potential referred to a radius ρ
0
can be found by computing the line integral
of E from ρ to ρ
0
:
Phi1(ρ) =?
ρ
l
2πepsilon1
integraldisplay
ρ
ρ
0
dρ
prime
ρ
prime
=
ρ
l
2πepsilon1
ln
parenleftbigg
ρ
0
ρ
parenrightbigg
.
We may choose any reference point ρ
0
except ρ
0
= 0 or ρ
0
=∞. This choice is equivalent
to the addition of an arbitrary constant, hence we can also write
Phi1(ρ) =
ρ
l
2πepsilon1
ln
parenleftbigg
1
ρ
parenrightbigg
+ C. (3.65)
The potential for a general two-dimensional charge distribution in unbounded space is
by superposition
Phi1(ρ) =
integraldisplay
S
T
ρ
T
(ρ
prime
)
epsilon1
G(ρ|ρ
prime
)dS
prime
, (3.66)
where the Green’s function is the potential of a unit line source located at ρ
prime
:
G(ρ|ρ
prime
) =
1
2π
ln
parenleftbigg
ρ
0
|ρ?ρ
prime
|
parenrightbigg
. (3.67)
Here S
T
denotes the transverse (xy)plane, and ρ
T
denotes the two-dimensional charge
distribution (C/m
2
)within that plane.
We note that the potential ?eld (3.66)of a two-dimensional source decreases logarith-
mically with distance. Only the potential produced by a source of ?nite extent decreases
inversely with distance.
Dirichlet and Neumann Green’s functions. The unbounded space Green’s func-
tion may be inconvenient for expressing the potential in a region having internal surfaces.
In fact, (3.56)shows that to use this function we would be forced to specify both Phi1 and its
normal derivative over all surfaces. This, of course, would exceed the actual requirements
for uniqueness.
Many functions can satisfy (3.52). For instance,
G(r|r
prime
) =
A
|r ? r
prime
|
+
B
|r ? r
i
|
(3.68)
satis?es (3.52)if r
i
/∈ V. Evaluation of (3.55)with the Green’s function (3.68)repro-
duces the general formulation (3.56)since the Laplacian of the second term in (3.68)is
identically zero in V. In fact, we can add any function to the free-space Green’s function,
provided that the additional term obeys Laplace’s equation within V:
G(r|r
prime
) =
A
|r ? r
prime
|
+ F(r|r
prime
), ?
prime
2
F(r|r
prime
) = 0. (3.69)
A good choice for G(r|r
prime
) will minimize the e?ort required to evaluate Phi1(r). Examining
(3.56)we notice two possibilities. If we demand that
G(r|r
prime
) = 0 for all r
prime
∈ S (3.70)
then the surface integral terms in (3.56)involving ?Phi1/?n
prime
will vanish. The Green’s
function satisfying (3.70)is known as the Dirichlet Green’s function. Let us designate it
by G
D
and use reciprocity to write (3.70)as
G
D
(r|r
prime
) = 0 for all r ∈ S.
The resulting specialization of (3.56),
Phi1(r) =
integraldisplay
V
G
D
(r|r
prime
)
ρ(r
prime
)
epsilon1
dV
prime
+
contintegraldisplay
S
B
Phi1(r
prime
)
?G
D
(r|r
prime
)
?n
prime
dS
prime
+
+
N
summationdisplay
n=1
contintegraldisplay
S
n
Phi1(r
prime
)
?G
D
(r|r
prime
)
?n
prime
dS
prime
, (3.71)
requires the speci?cation of Phi1 (but not its normal derivative)over the boundary surfaces.
In case S
B
and S
n
surround and are adjacent to perfect conductors, the Dirichlet bound-
ary condition has an important physical meaning. The corresponding Green’s function is
the potential at point r produced by a point source at r
prime
in the presence of the conductors
when the conductors are grounded — i.e., held at zero potential. Then we must specify
the actual constant potentials on the conductors to determine Phi1 everywhere within V
using (3.71). The additional term F(r|r
prime
) in (3.69)accounts for the potential produced
by surface charges on the grounded conductors.
By analogy with (3.70)it is tempting to try to de?ne another electrostatic Green’s
function according to
?G(r|r
prime
)
?n
prime
= 0 for all r
prime
∈ S. (3.72)
But this choice is not permissible if V is a ?nite-sized region. Let us integrate (3.54)over
V and employ the divergence theorem and the sifting property to get
contintegraldisplay
S
?G(r|r
prime
)
?n
prime
dS
prime
=?1; (3.73)
in conjunction with this, equation (3.72)would imply the false statement 0 =?1. Sup-
pose instead that we introduce a Green’s function according to
?G(r|r
prime
)
?n
prime
=?
1
A
for all r
prime
∈ S. (3.74)
where A is the total area of S. This choice avoids a contradiction in (3.73); it does not
nullify any terms in (3.56), but does reduce the surface integral terms involving Phi1 to
constants. Taken together, these terms all comprise a single additive constant on the
right-hand side; although the corresponding potential Phi1(r) is thereby determined only
to within this additive constant, the value of E(r) =??Phi1(r) will be una?ected. By
reciprocity we can rewrite (3.74)as
?G
N
(r|r
prime
)
?n
=?
1
A
for all r ∈ S. (3.75)
The Green’s function G
N
so de?ned is known as the Neumann Green’s function. Observe
that if V is not ?nite-sized then A →∞and according to (3.74)the choice (3.72)becomes
allowable.
Finding the Green’s function that obeys one of the boundary conditions for a given
geometry is often a di?cult task. Nevertheless, certain canonical geometries make the
Green’s function approach straightforward and simple. Such is the case in image theory,
when a charge is located near a simple conducting body such as a ground screen or
a sphere. In these cases the function F(r|r
prime
) consists of a single correction term as in
(3.68). We shall consider these simple cases in examples to follow.
Reciprocity of the static Green’s function. It remains to show that
G(r|r
prime
) = G(r
prime
|r)
for any of the Green’s functions introduced above. The unbounded-space Green’s function
is reciprocal by inspection; |r ? r
prime
| is una?ected by interchanging r and r
prime
. However, we
can give a more general treatment covering this case as well as the Dirichlet and Neumann
cases. We begin with
?
2
G(r|r
prime
) =?δ(r ? r
prime
).
In Green’s second identity let
φ(r) = G(r|r
a
), ψ(r) = G(r|r
b
),
where r
a
and r
b
are arbitrary points, and integrate over the unprimed coordinates. We
have
integraldisplay
V
[G(r|r
a
)?
2
G(r|r
b
)? G(r|r
b
)?
2
G(r|r
a
)] dV =
?
contintegraldisplay
S
bracketleftbigg
G(r|r
a
)
?G(r|r
b
)
?n
? G(r|r
b
)
?G(r|r
a
)
?n
bracketrightbigg
dS.
If G is the unbounded-space Green’s function, the surface integral must vanish since
S
B
→∞. It must also vanish under Dirichlet or Neumann boundary conditions. Since
?
2
G(r|r
a
) =?δ(r ? r
a
), ?
2
G(r|r
b
) =?δ(r ? r
b
),
we have
integraldisplay
V
[G(r|r
a
)δ(r ? r
b
)? G(r|r
b
)δ(r ? r
a
)] dV = 0,
hence
G(r
b
|r
a
) = G(r
a
|r
b
)
by the sifting property. By the arbitrariness of r
a
and r
b
, reciprocity is established.
Electrostatic shielding. The Dirichlet Green’s function can be used to explain elec-
trostatic shielding. We consider a closed, grounded, conducting shell with charge outside
butnotinside(Figure3.7).By(3.71)thepotentialatpointsinsidetheshellis
Phi1(r) =
contintegraldisplay
S
B
Phi1(r
prime
)
?G
D
(r|r
prime
)
?n
prime
dS
prime
,
Figure 3.7: Electrostatic shielding by a conducting shell.
where S
B
is tangential to the inner surface of the shell and we have used ρ = 0 within
the shell. Because Phi1(r
prime
) = 0 for all r
prime
on S
B
,wehave
Phi1(r) = 0
everywhere in the region enclosed by the shell. This result is independent of the charge
outside the shell, and the interior region is “shielded” from the e?ects of that charge.
Conversely, consider a grounded conducting shell with charge contained inside. If we
surround the outside of the shell by a surface S
1
and let S
B
recede to in?nity, then (3.71)
becomes
Phi1(r) = lim
S
B
→∞
contintegraldisplay
S
B
Phi1(r
prime
)
?G
D
(r|r
prime
)
?n
prime
dS
prime
+
contintegraldisplay
S
1
Phi1(r
prime
)
?G
D
(r|r
prime
)
?n
prime
dS
prime
.
Again there is no charge in V (since the charge lies completely inside the shell). The
contribution from S
B
vanishes. Since S
1
lies adjacent to the outer surface of the shell,
Phi1(r
prime
) ≡ 0 on S
1
.ThusPhi1(r) = 0 for all points outside the conducting shell.
Example solution to Poisson’s equation: planar layered media. For simple
geometries Poisson’s equation may be solved as part of a boundary value problem (§ A.4).
Occasionally such a solution has an appealing interpretation as the superposition of
potentials produced by the physical charge and its “images.” We shall consider here the
case of planar media and subsequently use the results to predict the potential produced
by charge near a conducting sphere.
Consider a layered dielectric medium where various regions of space are separated by
planes at constant values of z. Material region i occupies volume region V
i
and has
permittivity epsilon1
i
; it may or may not contain source charge. The solution to Poisson’s
equation is given by (3.56). The contribution
Phi1
p
(r) =
integraldisplay
V
G(r|r
prime
)
ρ(r
prime
)
epsilon1
dV
prime
produced by sources within V is known as the primary potential. The term
Phi1
s
(r) =
contintegraldisplay
S
bracketleftbigg
Phi1(r
prime
)
?G(r|r
prime
)
?n
prime
? G(r|r
prime
)
?Phi1(r
prime
)
?n
prime
bracketrightbigg
dS
prime
,
on the other hand, involves an integral over the surface ?elds and is known as the sec-
ondary potential. This term is linked to e?ects outside V. Since the “sources” of Phi1
s
(i.e., the surface ?elds)lie on the boundary of V, Phi1
s
satis?es Laplace’s equation within
V. We may therefore use other, more convenient, representations of Phi1
s
provided they
satisfy Laplace’s equation. However, as solutions to a homogeneous equation they are of
inde?nite form until linked to appropriate boundary values.
Since the geometry is invariant in the x and y directions, we represent each potential
function in terms of a 2-D Fourier transform over these variables. We leave the z depen-
dence intact so that we may apply boundary conditions directly in the spatial domain.
The transform representations of the Green’s functions for the primary and secondary
potentials are derived in Appendix A. From (A.55)we see that the primary potential
within region V
i
can be written as
Phi1
p
i
(r) =
integraldisplay
V
i
G
p
(r|r
prime
)
ρ(r
prime
)
epsilon1
i
dV
prime
(3.76)
where
G
p
(r|r
prime
) =
1
4π|r ? r
prime
|
=
1
(2π)
2
integraldisplay
∞
?∞
e
?k
ρ
|z?z
prime
|
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
(3.77)
is the primary Green’s function with k
ρ
= ?xk
x
+ ?yk
y
, k
ρ
=|k
ρ
|, and d
2
k
ρ
= dk
x
dk
y
.
We also ?nd in (A.56)that a solution of Laplace’s equation can be written as
Phi1
s
(r) =
1
(2π)
2
integraldisplay
∞
?∞
bracketleftbig
A(k
ρ
)e
k
ρ
z
+ B(k
ρ
)e
?k
ρ
z
bracketrightbig
e
jk
ρ
·r
d
2
k
ρ
(3.78)
where A(k
ρ
) and B(k
ρ
) must be found by the application of appropriate boundary con-
ditions.
As a simple example, consider a charge distribution ρ(r) in free space above a grounded
conducting plane located at z = 0. We wish to ?nd the potential in the region z > 0
using the Fourier transform representation of the potentials. The total potential is a sum
of primary and secondary terms:
Phi1(x, y, z) =
integraldisplay
V
bracketleftBigg
1
(2π)
2
integraldisplay
∞
?∞
e
?k
ρ
|z?z
prime
|
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
bracketrightBigg
ρ(r
prime
)
epsilon1
0
dV
prime
+
+
1
(2π)
2
integraldisplay
∞
?∞
bracketleftbig
B(k
ρ
)e
?k
ρ
z
bracketrightbig
e
jk
ρ
·r
d
2
k
ρ
,
where the integral is over the region z > 0. Here we have set A(k
ρ
) = 0 because e
k
ρ
z
grows with increasing z. Since the plane is grounded we must have Phi1(x, y,0) = 0.
Because z < z
prime
when we apply this condition, we have |z ? z
prime
|=z
prime
? z and thus
Phi1(x, y,0) =
1
(2π)
2
integraldisplay
∞
?∞
bracketleftBigg
integraldisplay
V
ρ(r
prime
)
epsilon1
0
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
+ B(k
ρ
)
bracketrightBigg
e
jk
ρ
·r
d
2
k
ρ
= 0.
Invoking the Fourier integral theorem we ?nd
B(k
ρ
) =?
integraldisplay
V
ρ(r
prime
)
epsilon1
0
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
,
Figure 3.8: Construction of electrostatic Green’s function for a ground plane.
hence the total potential is
Phi1(x, y, z) =
integraldisplay
V
bracketleftBigg
1
(2π)
2
integraldisplay
∞
?∞
e
?k
ρ
|z?z
prime
|
? e
?k
ρ
(z+z
prime
)
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
bracketrightBigg
ρ(r
prime
)
epsilon1
0
dV
prime
=
integraldisplay
V
G(r|r
prime
)
ρ(r
prime
)
epsilon1
0
dV
prime
where G(r|r
prime
) is the Green’s function for the region above a grounded planar conductor.
We can interpret this Green’s function as a sum of the primary Green’s function (3.77)
and a secondary Green’s function
G
s
(r|r
prime
) =?
1
(2π)
2
integraldisplay
∞
?∞
e
?k
ρ
(z+z
prime
)
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
. (3.79)
For z > 0 the term z+z
prime
can be replaced by |z+z
prime
|. Then, comparing (3.79)with (3.77),
we see that
G
s
(r|x
prime
, y
prime
, z
prime
) =?G
p
(r|x
prime
, y
prime
,?z
prime
) =?
1
4π|r ? r
prime
i
|
(3.80)
where r
prime
i
= ?xx
prime
+?yy
prime
??zz
prime
. Because the Green’s function is the potential of a point charge,
we may interpret the secondary Green’s function as produced by a negative unit charge
placedinaposition?z
prime
immediatelybeneaththepositiveunitchargethatproduces G
p
(Figure3.8).Thissecondarychargeisthe“image”oftheprimarycharge.Thattwosuch
charges would produce a null potential on the ground plane is easily veri?ed.
As a more involved example, consider a charge distribution ρ(r) above a planar in-
terface separating two homogeneous dielectric media. Region 1 occupies z > 0 and has
permittivity epsilon1
1
, while region 2 occupies z < 0 and has permittivity epsilon1
2
. In region 1 we
can write the total potential as a sum of primary and secondary components, discarding
the term that grows with z:
Phi1
1
(x, y, z) =
integraldisplay
V
bracketleftBigg
1
(2π)
2
integraldisplay
∞
?∞
e
?k
ρ
|z?z
prime
|
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
bracketrightBigg
ρ(r
prime
)
epsilon1
1
dV
prime
+
+
1
(2π)
2
integraldisplay
∞
?∞
bracketleftbig
B(k
ρ
)e
?k
ρ
z
bracketrightbig
e
jk
ρ
·r
d
2
k
ρ
. (3.81)
With no source in region 2, the potential there must obey Laplace’s equation and there-
fore consists of only a secondary component:
Phi1
2
(r) =
1
(2π)
2
integraldisplay
∞
?∞
bracketleftbig
A(k
ρ
)e
k
ρ
z
bracketrightbig
e
jk
ρ
·r
d
2
k
ρ
. (3.82)
To determine A and B we impose (3.36)and (3.37). By (3.36)we have
1
(2π)
2
integraldisplay
∞
?∞
bracketleftBigg
integraldisplay
V
ρ(r
prime
)
epsilon1
1
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
+ B(k
ρ
)? A(k
ρ
)
bracketrightBigg
e
jk
ρ
·r
d
2
k
ρ
= 0,
hence
integraldisplay
V
ρ(r
prime
)
epsilon1
1
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
+ B(k
ρ
)? A(k
ρ
) = 0
by the Fourier integral theorem. Applying (3.37)at z = 0 with ?n
12
= ?z, and noting that
there is no excess surface charge, we ?nd
integraldisplay
V
ρ(r
prime
)
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
?epsilon1
1
B(k
ρ
)?epsilon1
2
A(k
ρ
) = 0.
The solutions
A(k
ρ
) =
2epsilon1
1
epsilon1
1
+epsilon1
2
integraldisplay
V
ρ(r
prime
)
epsilon1
1
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
,
B(k
ρ
) =
epsilon1
1
?epsilon1
2
epsilon1
1
+epsilon1
2
integraldisplay
V
ρ(r
prime
)
epsilon1
1
e
?k
ρ
z
prime
2k
ρ
e
?jk
ρ
·r
prime
dV
prime
,
are then substituted into (3.81)and (3.82)to give
Phi1
1
(r) =
integraldisplay
V
bracketleftBigg
1
(2π)
2
integraldisplay
∞
?∞
e
?k
ρ
|z?z
prime
|
+
epsilon1
1
?epsilon1
2
epsilon1
1
+epsilon1
2
e
?k
ρ
(z+z
prime
)
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
bracketrightBigg
ρ(r
prime
)
epsilon1
1
dV
prime
=
integraldisplay
V
G
1
(r|r
prime
)
ρ(r
prime
)
epsilon1
1
dV
prime
,
Phi1
2
(r) =
integraldisplay
V
bracketleftBigg
1
(2π)
2
integraldisplay
∞
?∞
2epsilon1
2
epsilon1
1
+epsilon1
2
e
?k
ρ
(z
prime
?z)
2k
ρ
e
jk
ρ
·(r?r
prime
)
d
2
k
ρ
bracketrightBigg
ρ(r
prime
)
epsilon1
2
dV
prime
=
integraldisplay
V
G
2
(r|r
prime
)
ρ(r
prime
)
epsilon1
2
dV
prime
.
Since z
prime
> z for all points in region 2, we can replace z
prime
? z by |z ? z
prime
| in the formula for
Phi1
2
.
As with the previous example, let us compare the result to the form of the primary
Green’s function (3.77). We see that
G
1
(r|r
prime
) =
1
4π|r ? r
prime
|
+
epsilon1
1
?epsilon1
2
epsilon1
1
+epsilon1
2
1
4π|r ? r
prime
1
|
,
G
2
(r|r
prime
) =
2epsilon1
2
epsilon1
1
+epsilon1
2
1
4π|r ? r
prime
2
|
,
where r
prime
1
= ?xx
prime
+ ?yy
prime
? ?zz
prime
and r
prime
2
= ?xx
prime
+ ?yy
prime
+ ?zz
prime
. So we can also write
Phi1
1
(r) =
1
4π
integraldisplay
V
bracketleftbigg
1
|r ? r
prime
|
+
epsilon1
1
?epsilon1
2
epsilon1
1
+epsilon1
2
1
|r ? r
prime
1
|
bracketrightbigg
ρ(r
prime
)
epsilon1
1
dV
prime
,
Phi1
2
(r) =
1
4π
integraldisplay
V
bracketleftbigg
2epsilon1
2
epsilon1
1
+epsilon1
2
1
|r ? r
prime
2
|
bracketrightbigg
ρ(r
prime
)
epsilon1
2
dV
prime
.
Figure 3.9: Green’s function for a grounded conducting sphere.
Note that Phi1
2
→ Phi1
1
as epsilon1
2
→ epsilon1
1
.
There is an image interpretation for the secondary Green’s functions. The secondary
Green’s function for region 1 appears as a potential produced by an image of the primary
charge located at ?z
prime
in an in?nite medium of permittivity epsilon1
1
, and with an amplitude of
(epsilon1
1
?epsilon1
2
)/(epsilon1
1
+epsilon1
2
) times the primary charge. The Green’s function in region 2 is produced
by an image charge located at z
prime
(i.e., at the location of the primary charge)in an in?nite
medium of permittivity epsilon1
2
with an amplitude of 2epsilon1
2
/(epsilon1
1
+epsilon1
2
) times the primary charge.
Example solution to Poisson’s equation: conducting sphere. As an example
involving a nonplanar geometry, consider the potential produced by a source near a
groundedconductingsphereinfreespace(Figure3.9).Basedonourexperiencewith
planar layered media, we hypothesize that the secondary potential will be produced by
an image charge; hence we try the simple Green’s function
G
s
(r|r
prime
) =
A(r
prime
)
4π|r ? r
prime
i
|
where the amplitude A and location r
prime
i
of the image are to be determined. We further
assume, based on our experience with planar problems, that the image charge will reside
inside the sphere along a line joining the origin to the primary charge. Since r = a?r for
all points on the sphere, the total Green’s function must obey the Dirichlet condition
G(r|r
prime
)|
r=a
=
1
4π|r ? r
prime
|
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r=a
+
A(r
prime
)
4π|r ? r
prime
i
|
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r=a
=
1
4π|a?r ? r
prime
?r
prime
|
+
A(r
prime
)
4π|a?r ? r
prime
i
?r
prime
|
= 0
in order to have the potential, given by (3.56), vanish on the sphere surface. Factoring a
from the ?rst denominator and r
prime
i
from the second we obtain
1
4πa|?r ?
r
prime
a
?r
prime
|
+
A(r
prime
)
4πr
prime
i
|
a
r
prime
i
?r ? ?r
prime
|
= 0.
Now |k?r ? k
prime
?r
prime
|=k
2
+ k
prime2
? 2kk
prime
cosγ where γ is the angle between ?r and ?r
prime
and k,k
prime
are constants; this means that |k?r ? ?r
prime
|=|?r ? k?r
prime
|. Hence as long as we choose
r
prime
a
=
a
r
prime
i
,
A
r
prime
i
=?
1
a
,
the total Green’s function vanishes everywhere on the surface of the sphere. The image
charge is therefore located within the sphere at r
prime
i
= a
2
r
prime
/r
prime2
and has amplitude A =
?a/r
prime
. (Note that both the location and amplitude of the image depend on the location
of the primary charge.)With this Green’s function and (3.71), the potential of an
arbitrary source placed near a grounded conducting sphere is
Phi1(r) =
integraldisplay
V
ρ(r
prime
)
epsilon1
1
4π
bracketleftBigg
1
|r ? r
prime
|
?
a/r
prime
|r ?
a
2
r
prime2
r
prime
|
bracketrightBigg
dV
prime
.
The Green’s function may be used to compute the surface charge density induced on
the sphere by a unit point charge: it is merely necessary to ?nd the normal component of
electric ?eld from the gradient of Phi1(r). We leave this as an exercise for the reader, who
may then integrate the surface charge and thereby show that the total charge induced
on the sphere is equal to the image charge. So the total charge induced on a grounded
sphere by a point charge q at a point r = r
prime
is Q =?qa/r
prime
.
It is possible to ?nd the total charge induced on the sphere without ?nding the image
charge ?rst. This is an application of Green’s reciprocation theorem (§ 3.4.4). According
to (3.211), if we can ?nd the potential V
P
at a point r produced by the sphere when it is
isolated and carrying a total charge Q
0
, then the total charge Q induced on the grounded
sphere in the vicinity of a point charge q placed at r is given by
Q =?qV
P
/V
1
where V
1
is the potential of the isolated sphere. We can apply this formula by noting that
an isolated sphere carrying charge Q
0
produces a ?eld E(r) = ?rQ
0
/4πepsilon1r
2
. Integration
from a radius r to in?nity gives the potential referred to in?nity: Phi1(r) = Q
0
/4πepsilon1r. So
the potential of the isolated sphere is V
1
= Q
0
/4πepsilon1a, while the potential at radius r
prime
is
V
P
= Q
0
/4πepsilon1r
prime
. Substitution gives Q =?qa/r
prime
as before.
3.2.5 Force and energy
Maxwell’s stress tensor. The electrostatic version of Maxwell’s stress tensor can be
obtained from (2.288)by setting B = H = 0:
ˉ
T
e
=
1
2
(D · E)
ˉ
I ? DE. (3.83)
The total electric force on the charges in a region V bounded by the surface S is given
by the relation
F
e
=?
contintegraldisplay
S
ˉ
T
e
· dS =
integraldisplay
V
f
e
dV
where f
e
= ρE is the electric force volume density.
In particular, suppose that S is adjacent to a solid conducting body embedded in a
dielectric having permittivity epsilon1(r). Since all the charge is at the surface of the conductor,
the force within V acts directly on the surface. Thus, ?
ˉ
T
e
· ?n is the surface force density
(traction) t. Using D = epsilon1E, and remembering that the ?elds are normal to the conductor,
we ?nd that
ˉ
T
e
· ?n =
1
2
epsilon1E
2
n
?n ?epsilon1EE · ?n =?
1
2
epsilon1E
2
n
?n =?
1
2
ρ
s
E.
The surface force density is perpendicular to the surface.
As a simple but interesting example, consider the force acting on a rigid conducting
sphere of radius a carrying total charge Q in a homogeneous medium. At equilibrium
the charge is distributed uniformly with surface density ρ
s
= Q/4πa
2
, producing a ?eld
E = ?rQ/4πepsilon1r
2
external to the sphere. Hence a force density
t =
1
2
?r
Q
2
epsilon1(4πa
2
)
2
acts at each point on the surface. This would cause the sphere to expand outward if the
structural integrity of the material were to fail. Integration over the entire sphere yields
F =
1
2
Q
2
epsilon1(4πa
2
)
2
integraldisplay
S
?r dS = 0.
However, integration of t over the upper hemisphere yields
F =
1
2
Q
2
epsilon1(4πa
2
)
2
integraldisplay
2π
0
integraldisplay
π/2
0
?ra
2
sinθ dθ dφ.
Substitution of ?r = ?x sinθ cosφ+ ?y sinθ sinφ+ ?z cosθ leads immediately to F
x
= F
y
= 0,
but the z-component is
F
z
=
1
2
Q
2
epsilon1(4πa
2
)
2
integraldisplay
2π
0
integraldisplay
π/2
0
a
2
cosθ sinθ dθ dφ =
Q
2
32epsilon1πa
2
.
This result can also be obtained by integrating ?
ˉ
T
e
· ?n over the entire xy-plane with
?n =??z. Since ?
ˉ
T
e
·(??z) = ?z
epsilon1
2
E · E we have
F = ?z
1
2
Q
2
(4πepsilon1)
2
integraldisplay
2π
0
integraldisplay
∞
a
rdrdφ
r
4
= ?z
Q
2
32epsilon1πa
2
.
As a more challenging example, consider two identical line charges parallel to the z-
axis and located at x =±d/2, y = 0 in free space. We can ?nd the force on one line
charge due to the other by integrating Maxwell’s stress tensor over the yz-plane. From
(3.64)we ?nd that the total electric ?eld on the yz-plane is
E(y, z) =
y
y
2
+(d/2)
2
ρ
l
πepsilon1
0
?y
where ρ
l
is the line charge density. The force density for either line charge is ?
ˉ
T
e
· ?n,
where we use ?n =±?x to obtain the force on the charge at x =?d/2. The force density
for the charge at x =?d/2 is
ˉ
T
e
· ?n =
1
2
(D · E)
ˉ
I · ?x ? DE · ?x =
epsilon1
0
2
bracketleftbigg
y
y
2
+(d/2)
2
ρ
l
πepsilon1
0
bracketrightbigg
2
?x
and the total force is
F
?
=?
integraldisplay
∞
?∞
integraldisplay
∞
?∞
ρ
2
l
2π
2
epsilon1
0
y
2
bracketleftbig
y
2
+(d/2)
2
bracketrightbig
2
?x dydz.
On a per unit length basis the force is
F
?
l
=??x
ρ
2
l
2π
2
epsilon1
0
integraldisplay
∞
?∞
y
2
[y
2
+(d/2)
2
]
2
dy =??x
ρ
2
l
2πdepsilon1
0
.
Note that the force is repulsive as expected.
Figure 3.10: Computation of electrostatic stored energy via the assembly energy of a
charge distribution.
Electrostatic stored energy. In § 2.9.5 we considered the energy relations for the
electromagnetic ?eld. Those relations remain valid in the static case. Since our interpre-
tation of the dynamic relations was guided in part by our knowledge of the energy stored
in a static ?eld, we must, for completeness, carry out a study of that e?ect here.
The energy of a static con?guration is taken to be the work required to assemble the
con?guration from a chosen starting point. For a con?guration of static charges, the
stored electric energy is the energy required to assemble the con?guration, starting with
all charges removed to in?nite distance (the assumed zero potential reference). If the
assembled charges are not held in place by an external mechanical force they will move,
thereby converting stored electric energy into other forms of energy (e.g., kinetic energy
and radiation).
By (3.62), the work required to move a point charge q from a reservoir at in?nity to
a point P at r in a potential ?eld Phi1 is
W = qPhi1(r).
If instead we have a continuous charge density ρ present, and wish to increase this to
ρ +δρ by bringing in a small quantity of charge δρ, a total work
δW =
integraldisplay
V
∞
δρ(r)Phi1(r)dV (3.84)
is required, and the potential ?eld is increased to Phi1+δPhi1. Here V
∞
denotes all of space.
(We could restrict the integral to the region containing the charge, but we shall ?nd it
helpful to extend the domain of integration to all of space.)
NowconsiderthesituationshowninFigure3.10.Herewehavechargeintheformof
both volume densities and surface densities on conducting bodies. Also present may be
linear material bodies. We can think of assembling the charge in two distinctly di?erent
ways. We could, for instance, bring small portions of charge (or point charges)together
to form the distribution ρ. Or, we could slowly build up ρ by adding in?nitesimal, but
spatially identical, distributions. That is, we can create the distribution ρ from a zero
initial state by repeatedly adding a charge distribution
δρ(r) = ρ(r)/N,
where N is a large number. Whenever we add δρ we must perform the work given by
(3.84), but we also increase the potential proportionately (remembering that all materials
are assumed linear). At each step, more work is required. The total work is
W =
N
summationdisplay
n=1
integraldisplay
V
∞
δρ(r)[(n ? 1)δPhi1(r)] dV =
bracketleftBigg
N
summationdisplay
n=1
(n ? 1)
bracketrightBigg
integraldisplay
V
∞
ρ(r)
N
Phi1(r)
N
dV. (3.85)
We must use an in?nite number of steps so that no energy is lost to radiation at any step
(since the charge we add each time is in?nitesimally small). Using
N
summationdisplay
n=1
(n ? 1) = N(N ? 1)/2,
(3.85)becomes
W =
1
2
integraldisplay
V
∞
ρ(r)Phi1(r)dV (3.86)
as N →∞. Finally, since some assembled charge will be in the form of a volume density
and some in the form of the surface density on conductors, we can generalize (3.86)to
W =
1
2
integraldisplay
V
prime
ρ(r)Phi1(r)dV +
1
2
I
summationdisplay
i=1
Q
i
V
i
. (3.87)
Here V
prime
is the region outside the conductors, Q
i
is the total charge on the ith conductor
(i = 1,...,I), and V
i
is the absolute potential (referred to in?nity)of the ith conductor.
An intriguing property of electrostatic energy is that the charges on the conductors
will arrange themselves, while seeking static equilibrium, into a minimum-energy con?g-
uration (Thomson’s theorem).
In keeping with our ?eld-centered view of electromagnetics, we now wish to write the
energy (3.86)entirely in terms of the ?eld vectors E and D. Since ρ =?·D we have
W =
1
2
integraldisplay
V
∞
[?·D(r)]Phi1(r)dV.
Then, by (B.42),
W =
1
2
integraldisplay
V
∞
?·[Phi1(r)D(r)] dV ?
1
2
integraldisplay
V
∞
D(r)· [?Phi1(r)] dV.
Use of the divergence theorem and (3.30)leads to
W =
1
2
contintegraldisplay
S
∞
Phi1(r)D(r)· dS +
1
2
integraldisplay
V
∞
D(r)· E(r)dV
Figure 3.11: Multipole expansion.
where S
∞
is the bounding surface that recedes toward in?nity to encompass all of space.
Because Phi1 ～ 1/r and D ～ 1/r
2
as r →∞, the integral over S
∞
tends to zero and
W =
1
2
integraldisplay
V
∞
D(r)· E(r)dV. (3.88)
Hence we may compute the assembly energy in terms of the ?elds supported by the
charge ρ.
It is signi?cant that the assembly energy W is identical to the term within the time
derivative in Poynting’s theorem (2.299). Hence our earlier interpretation, that this term
represents the time-rate of change of energy “stored” in the electric ?eld, has a ?rm basis.
Of course, the assembly energy is a static concept, and our generalization to dynamic
?elds is purely intuitive. We also face similar questions regarding the meaning of energy
density, and whether energy can be “localized” in space. The discussions in § 2.9.5 still
apply.
3.2.6 Multipole expansion
Consider an arbitrary but spatially localized charge distribution of total charge Q
inanunboundedhomogeneousmedium(Figure3.11).Wehavealreadyobtainedthe
potential (3.61)of the source; as we move the observation point away, Phi1 should decrease
in a manner roughly proportional to 1/r. The actual variation depends on the nature
of the charge distribution and can be complicated. Often this dependence is dominated
by a speci?c inverse power of distance for observation points far from the source, and we
can investigate it by expanding the potential in powers of 1/r. Although such multipole
expansions of the potential are rarely used to perform actual computations, they can
provide insight into both the behavior of static ?elds and the physical meaning of the
polarization vector P.
Let us place our origin of coordinates somewhere within the charge distribution, as
showninFigure3.11,andexpandtheGreen’sfunctionspatialdependenceinathree-
dimensional Taylor series about the origin:
1
R
=
∞
summationdisplay
n=0
1
n!
(r
prime
·?
prime
)
n
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
=
1
r
+(r
prime
·?
prime
)
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+
1
2
(r
prime
·?
prime
)
2
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+···, (3.89)
where R =|r?r
prime
|. Convergence occurs if |r| > |r
prime
|. In the notation (r
prime
·?
prime
)
n
we interpret
a power on a derivative operator as the order of the derivative. Substituting (3.89)into
(3.61)and writing the derivatives in Cartesian coordinates we obtain
Phi1(r) =
1
4πepsilon1
integraldisplay
V
ρ(r
prime
)
bracketleftbigg
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+(r
prime
·?
prime
)
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+
1
2
(r
prime
·?
prime
)
2
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+···
bracketrightbigg
dV
prime
. (3.90)
For the second term we can use (3.57)to write
(r
prime
·?
prime
)
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
= r
prime
·
parenleftbigg
?
prime
1
R
parenrightbiggvextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
= r
prime
·
parenleftBigg
?
R
R
2
parenrightBigg
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
= r
prime
·
?r
r
2
. (3.91)
The third term is complicated. Let us denote (x, y, z) by (x
1
, x
2
, x
3
) and perform an
expansion in rectangular coordinates:
(r
prime
·?
prime
)
2
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
=
3
summationdisplay
i=1
3
summationdisplay
j=1
x
prime
i
x
prime
j
?
2
?x
prime
i
?x
prime
j
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
.
It turns out [172] that this can be written as
(r
prime
·?
prime
)
2
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
=
1
r
3
?r ·(3r
prime
r
prime
? r
prime2
ˉ
I)· ?r.
Substitution into (3.90)gives
Phi1(r) =
Q
4πepsilon1r
+
?r · p
4πepsilon1r
2
+
1
2
?r ·
ˉ
Q · ?r
4πepsilon1r
3
+···, (3.92)
which is the multipole expansion for Phi1(r). It converges for all r > r
m
where r
m
is the
radiusofthesmallestspherecompletelycontainingthechargecenteredat r
prime
= 0 (Figure
3.11).In(3.92)theterms Q, p,
ˉ
Q,andsoonarecalledthemultipolemomentsofρ(r).
The ?rst moment is merely the total charge
Q =
integraldisplay
V
ρ(r
prime
)dV
prime
.
The second moment is the electric dipole moment vector
p =
integraldisplay
V
r
prime
ρ(r
prime
)dV
prime
.
The third moment is the electric quadrupole moment dyadic
ˉ
Q =
integraldisplay
V
(3r
prime
r
prime
? r
prime2
ˉ
I)ρ(r
prime
)dV
prime
.
The expansion (3.92)allows us to identify the dominant power of r for r greatermuch r
m
.
The ?rst nonzero term in (3.92)dominates the potential at points far from the source.
Interestingly, the ?rst nonvanishing moment is independent of the location of the origin
of r
prime
, while all subsequent higher moments depend on the location of the origin [91]. We
can see this most easily through a few simple examples.
For a single point charge q located at r
0
we can write ρ(r) = qδ(r ? r
0
). The ?rst
moment of ρ is
Q =
integraldisplay
V
qδ(r
prime
? r
0
)dV
prime
= q.
Figure 3.12: A dipole distribution.
Note that this is independent of r
0
. The second moment
p =
integraldisplay
V
r
prime
qδ(r
prime
? r
0
)dV
prime
= qr
0
depends on r
0
, as does the third moment
ˉ
Q =
integraldisplay
V
(3r
prime
r
prime
? r
prime2
ˉ
I)qδ(r
prime
? r
0
)dV
prime
= q(3r
0
r
0
? r
2
0
ˉ
I).
If r
0
= 0 then only the ?rst moment is nonzero; that this must be the case is obvious
from (3.61).
ForthedipoleofFigure3.12wecanwrite
ρ(r) =?qδ(r ? r
0
+ d/2)+ qδ(r ? r
0
? d/2).
In this case
Q =?q + q = 0, p = qd,
ˉ
Q = q[3(r
0
d + dr
0
)? 2(r
0
· d)
ˉ
I].
Only the ?rst nonzero moment, in this case p, is independent of r
0
.Forr
0
= 0 the only
nonzero multipole moment would be the dipole moment p. If the dipole is aligned along
the z-axis with d = d ?z and r
0
= 0, then the exact potential is
Phi1(r) =
1
4πepsilon1
p cosθ
r
2
.
By (3.30)we have
E(r) =
1
4πepsilon1
p
r
3
(?r2 cosθ +
?
θ sinθ), (3.93)
which is the classic result for the electric ?eld of a dipole.
Finally,considerthequadrupoleshowninFigure3.13.Thechargedensityis
ρ(r) =?qδ(r ? r
0
)+ qδ(r ? r
0
? d
1
)+ qδ(r ? r
0
? d
2
)? qδ(r ? r
0
? d
1
? d
2
).
Figure 3.13: A quadrupole distribution.
Carrying through the details, we ?nd that the ?rst two moments of ρ vanish, while the
third is given by
ˉ
Q = q[?3(d
1
d
2
+ d
2
d
1
)+ 2(d
1
· d
2
)
ˉ
I].
As expected, it is independent of r
0
.
It is tedious to carry (3.92)beyond the quadrupole term using the Taylor expansion.
Another approach is to expand 1/R in spherical harmonics. Referring to Appendix E.3
we ?nd that
1
|r ? r
prime
|
= 4π
∞
summationdisplay
n=0
n
summationdisplay
m=?n
1
2n + 1
r
primen
r
n+1
Y
?
nm
(θ
prime
,φ
prime
)Y
nm
(θ,φ)
(see Jackson [91] or Arfken [5] for a detailed derivation). This expansion converges for
|r| > |r
m
|. Substitution into (3.61)gives
Phi1(r) =
1
epsilon1
∞
summationdisplay
n=0
1
r
n+1
bracketleftBigg
1
2n + 1
n
summationdisplay
m=?n
q
nm
Y
nm
(θ,φ)
bracketrightBigg
(3.94)
where
q
nm
=
integraldisplay
V
ρ(r
prime
)r
primen
Y
?
nm
(θ
prime
,φ
prime
)dV
prime
.
We can now identify any inverse power of r in the multipole expansion, but at the price
of dealing with a double summation. For a charge distribution with axial symmetry (no
φ-variation), only the coe?cient q
n0
is nonzero. The relation
Y
n0
(θ,φ) =
radicalbigg
2n + 1
4π
P
n
(cosθ)
allows us to simplify (3.94)and obtain
Phi1(r) =
1
4πepsilon1
∞
summationdisplay
n=0
1
r
n+1
q
n
P
n
(cosθ) (3.95)
where
q
n
= 2π
integraldisplay
r
prime
integraldisplay
θ
prime
ρ(r
prime
,θ
prime
)r
primen
P
n
(cosθ
prime
)r
prime2
sinθ
prime
dθ
prime
dr
prime
.
As a simple example consider a spherical distribution of charge given by
ρ(r) =
3Q
πa
3
cosθ, r ≤ a.
This can be viewed as two adjacent hemispheres carrying total charges ±Q. Since cosθ =
P
1
(cosθ), we compute
q
n
= 2π
integraldisplay
a
0
integraldisplay
π
0
3Q
πa
3
P
1
(cosθ
prime
)r
primen
P
n
(cosθ
prime
)r
prime2
sinθ
prime
dθ
prime
dr
prime
= 2π
3Q
πa
3
a
n+3
n + 3
integraldisplay
π
0
P
1
(cosθ)P
n
(cosθ
prime
)sinθ
prime
dθ
prime
.
Using the orthogonality relation (E.123)we ?nd
q
n
= 2π
3Q
πa
3
a
n+3
n + 3
δ
1n
2
2n + 1
.
Hence the only nonzero coe?cient is q
1
= Qa and
Phi1(r) =
1
4πepsilon1
1
r
2
QaP
1
(cosθ)=
Qa
4πepsilon1r
2
cosθ.
This is the potential of a dipole having moment p = ?zQa. Thus we could replace the
sphere with point charges ?Q at z =?a/2 without changing the ?eld for r > a.
Physical interpretation of the polarization vector in a dielectric. We have
used the Maxwell–Minkowski equations to determine the electrostatic potential of a
charge distribution in the presence of a dielectric medium. Alternatively, we can use
the Maxwell–Bo? equations
?×E = 0, (3.96)
?·E =
1
epsilon1
0
(ρ ??·P). (3.97)
Equation (3.96)allows us to de?ne a scalar potential through (3.30). Substitution into
(3.97)gives
?
2
Phi1(r) =?
1
epsilon1
0
[ρ(r)+ρ
P
(r)] (3.98)
where ρ
P
=??·P. This has the form of Poisson’s equation (3.50), but with charge
density term ρ(r)+ρ
P
(r). Hence the solution is
Phi1(r) =
1
4πepsilon1
0
integraldisplay
V
ρ(r
prime
)??
prime
· P(r
prime
)
|r ? r
prime
|
dV
prime
.
To this we must add any potential produced by surface sources such as ρ
s
. If there is a
discontinuity in the dielectric region, there is also a surface polarization source ρ
Ps
= ?n·P
according to (3.35). Separating the volume into regions with bounding surfaces S
i
across
which the permittivity is discontinuous, we may write
Phi1(r) =
1
4πepsilon1
0
integraldisplay
V
ρ(r
prime
)
|r ? r
prime
|
dV
prime
+
1
4πepsilon1
0
integraldisplay
S
ρ
s
(r
prime
)
|r ? r
prime
|
dS
prime
+
+
summationdisplay
i
bracketleftbigg
1
4πepsilon1
0
integraldisplay
V
i
??
prime
· P(r
prime
)
|r ? r
prime
|
dV
prime
+
1
4πepsilon1
0
contintegraldisplay
S
i
?n
prime
· P(r
prime
)
|r ? r
prime
|
dS
prime
bracketrightbigg
, (3.99)
where ?n points outward from region i. Using the divergence theorem on the fourth term
and employing (B.42), we obtain
Phi1(r) =
1
4πepsilon1
0
integraldisplay
V
ρ(r
prime
)
|r ? r
prime
|
dV
prime
+
1
4πepsilon1
0
integraldisplay
S
ρ
s
(r
prime
)
|r ? r
prime
|
dS
prime
+
+
summationdisplay
i
bracketleftbigg
1
4πepsilon1
0
integraldisplay
V
i
P(r
prime
)·?
prime
parenleftbigg
1
|r ? r
prime
|
parenrightbigg
dV
prime
bracketrightbigg
.
Since ?
prime
(1/R) =
?
R/R
2
, the third term is a sum of integrals of the form
1
4πepsilon1
integraldisplay
V
i
P(r
prime
)·
?
R
R
2
dV.
Comparing this to the second term of (3.92), we see that this integral represents a volume
superposition of dipole terms where P is a volume density of dipole moments.
Thus, a dielectric with permittivity epsilon1 is equivalent to a volume distribution of dipoles
in free space. No higher-order moments are required, and no zero-order moments are
needed since any net charge is included in ρ. Note that we have arrived at this conclusion
based only on Maxwell’s equations and the assumption of a linear, isotropic relationship
between D and E. Assuming our macroscopic theory is correct, we are tempted to make
assumptions about the behavior of matter on a microscopic level (e.g., atoms exposed to
?elds are polarized and their electron clouds are displaced from their positively charged
nuclei), but this area of science is better studied from the viewpoints of particle physics
and quantum mechanics.
Potential of an azimuthally-symmetric charged spherical surface. In several
of our example problems we shall be interested in evaluating the potential of a charged
spherical surface. When the charge is azimuthally-symmetric, the potential is particularly
simple.
We will need the value of the integral
F(r) =
1
4π
integraldisplay
S
f (θ
prime
)
|r ? r
prime
|
dS
prime
(3.100)
where r = r ?r describes an arbitrary observation point and r
prime
= a?r
prime
identi?es the source
point on the surface of the sphere of radius a. The integral is most easily done using the
expansion (E.200)for |r ? r
prime
|
?1
in spherical harmonics. We have
F(r) = a
2
∞
summationdisplay
n=0
n
summationdisplay
m=?n
Y
nm
(θ,φ)
2n + 1
r
n
<
r
n+1
>
integraldisplay
π
?π
integraldisplay
π
0
f (θ
prime
)Y
?
nm
(θ
prime
,φ
prime
)sinθ
prime
dθ
prime
dφ
prime
where r
<
= min{r,a} and r
>
= max{r,a}. Using orthogonality of the exponentials we
?nd that only the m = 0 terms contribute:
F(r) = 2πa
2
∞
summationdisplay
n=0
Y
n0
(θ,φ)
2n + 1
r
n
<
r
n+1
>
integraldisplay
π
0
f (θ
prime
)Y
?
n0
(θ
prime
,φ
prime
)sinθ
prime
dθ
prime
.
Finally, since
Y
n0
=
radicalbigg
2n + 1
4π
P
n
(cosθ)
we have
F(r) =
1
2
a
2
∞
summationdisplay
n=0
P
n
(cosθ)
r
n
<
r
n+1
>
integraldisplay
π
0
f (θ
prime
)P
n
(cosθ
prime
)sinθ
prime
dθ
prime
. (3.101)
As an example, suppose f (θ) = cosθ = P
1
(cosθ). Then
F(r) =
1
2
a
2
∞
summationdisplay
n=0
P
n
(cosθ)
r
n
<
r
n+1
>
integraldisplay
π
0
P
1
(cosθ
prime
)P
n
(cosθ
prime
)sinθ
prime
dθ
prime
.
The orthogonality of the Legendre polynomials can be used to show that
integraldisplay
π
0
P
1
(cosθ
prime
)P
n
(cosθ
prime
)sinθ
prime
dθ
prime
=
2
3
δ
1n
,
hence
F(r) =
a
2
3
cosθ
r
<
r
2
>
. (3.102)
3.2.7 Field produced by a permanently polarized body
Certain materials, called electrets, exhibit polarization in the absence of an external
electric ?eld. A permanently polarized material produces an electric ?eld both internal
and external to the material, hence there must be a charge distribution to support the
?elds. We can interpret this charge as being caused by the permanent separation of
atomic charge within the material, but if we are only interested in the macroscopic ?eld
then we need not worry about the microscopic implications of such materials. Instead, we
can use the Maxwell–Bo? equations and ?nd the potential produced by the material by
using (3.99). Thus, the ?eld of an electret with known polarization P occupying volume
region V in free space is dipolar in nature and is given by
Phi1(r) =
1
4πepsilon1
0
integraldisplay
V
??
prime
· P(r
prime
)
|r ? r
prime
|
dV
prime
+
1
4πepsilon1
0
contintegraldisplay
S
?n
prime
· P(r
prime
)
|r ? r
prime
|
dS
prime
where ?n points out of the volume region V.
As an example, consider a material sphere of radius a, permanently polarized along
its axis with uniform polarization P(r) = ?zP
0
. We have the equivalent source densities
ρ
p
=??·P = 0,ρ
Ps
= ?n · P = ?r · ?zP
0
= P
0
cosθ.
Then
Phi1(r) =
1
4πepsilon1
0
contintegraldisplay
S
ρ
Ps
(r
prime
)
|r ? r
prime
|
dS
prime
=
1
4πepsilon1
0
contintegraldisplay
S
P
0
cosθ
prime
|r ? r
prime
|
dS
prime
.
The integral takes the form (3.100), hence by (3.102) the solution is
Phi1(r) = P
0
a
2
3epsilon1
0
cosθ
r
<
r
2
>
. (3.103)
If we are interested only in the potential for r > a, we can use the multipole expansion
(3.95)to obtain
Phi1(r) =
1
4πepsilon1
0
∞
summationdisplay
n=0
1
r
n+1
q
n
P
n
(cosθ), r > a
where
q
n
= 2π
integraldisplay
π
0
ρ
Ps
(θ
prime
)a
n
P
n
(cosθ
prime
)a
2
sinθ
prime
dθ
prime
.
Substituting for ρ
Ps
and remembering that cosθ = P
1
(cosθ),wehave
q
n
= 2πa
n+2
P
0
integraldisplay
π
0
P
1
(cosθ
prime
)P
n
(cosθ
prime
)sinθ
prime
dθ
prime
.
Using the orthogonality relation (E.123)we ?nd
q
n
= 2πa
n+2
P
0
δ
1n
2
2n + 1
.
Therefore the only nonzero coe?cient is
q
1
=
4πa
3
P
0
3
and
Phi1(r) =
1
4πepsilon1
0
1
r
2
4πa
3
P
0
3
P
1
(cosθ)=
P
0
a
3
3epsilon1
0
r
2
cosθ, r > a.
This is a dipole ?eld, and matches (3.103)as expected.
3.2.8 Potential of a dipole layer
Surface charge layers sometimes occur in bipolar form, such as in the membrane sur-
rounding an animal cell. These can be modeled as a dipole layer consisting of parallel
surface charges of opposite sign.
Consider a surface S located in free space. Parallel to this surface, and a distance Delta1/2
below, is located a surface charge layer of density ρ
s
(r) = P
s
(r). Also parallel to S, but
a distance Delta1/2 above, is a surface charge layer of density ρ
s
(r) =?P
s
(r). We de?ne the
surface dipole moment density D
s
as
D
s
(r) = Delta1 P
s
(r). (3.104)
Letting the position vector r
prime
0
point to the surface S we can write the potential (3.61)
produced by the two charge layers as
Phi1(r) =
1
4πepsilon1
0
integraldisplay
S
+
P
s
(r
prime
)
1
|r ? r
prime
0
? ?n
prime
Delta1
2
|
dS
prime
?
1
4πepsilon1
0
integraldisplay
S
?
P
s
(r
prime
)
1
|r ? r
prime
0
+ ?n
prime
Delta1
2
|
dS
prime
.
Figure 3.14: A dipole layer.
We are interested in the case in which the two charge layers collapse onto the surface S,
and wish to compute the potential produced by a given dipole moment density. When
Delta1 → 0 we have r
prime
0
→ r
prime
and may write
Phi1(r) = lim
Delta1→0
1
4πepsilon1
0
integraldisplay
S
D
s
(r
prime
)
Delta1
bracketleftBigg
1
|R ? ?n
prime
Delta1
2
|
?
1
|R + ?n
prime
Delta1
2
|
bracketrightBigg
dS
prime
,
where R = r ? r
prime
. By the binomial theorem, the limit of the term in brackets can be
written as
lim
Delta1→0
?
?
bracketleftBigg
R
2
+
parenleftbigg
Delta1
2
parenrightbigg
2
? 2R · ?n
prime
Delta1
2
bracketrightBigg
?
1
2
?
bracketleftBigg
R
2
+
parenleftbigg
Delta1
2
parenrightbigg
2
+ 2R · ?n
prime
Delta1
2
bracketrightBigg
?
1
2
?
?
= lim
Delta1→0
parenleftBigg
R
?1
bracketleftBigg
1 +
?
R · ?n
prime
R
Delta1
2
bracketrightBigg
? R
?1
bracketleftBigg
1 ?
?
R · ?n
prime
R
Delta1
2
bracketrightBiggparenrightBigg
= Delta1?n
prime
·
R
R
3
.
Thus
Phi1(r) =
1
4πepsilon1
0
integraldisplay
S
D
s
(r
prime
)·
R
R
3
dS
prime
(3.105)
where D
s
= ?nD
s
is the surface vector dipole moment density. The potential of a dipole
layer decreases more rapidly (～ 1/r
2
)than that of a unipolar charge layer. We saw
similar behavior in the dipole term of the multipole expansion (3.92)for a general charge
distribution.
We can use (3.105)to study the behavior of the potential across a dipole layer. As
we approach the layer from above, the greatest contribution to Phi1 comes from the charge
region immediately beneath the observation point. Assuming that the surface dipole
moment density is continuous beneath the point, we can compute the di?erence in the
?elds across the layer at point r by replacing the arbitrary surface layer by a disk of
constant surface dipole moment density D
0
= D
s
(r). For simplicity we center the disk
at z = 0 inthe xy-planeasshowninFigure3.15andcomputethepotentialdi?erence
Delta1V across the layer; i.e., Delta1V = Phi1(h)?Phi1(?h) on the disk axis as h → 0. Using (3.105)
along with r
prime
=±h?z ?ρ
prime
?ρ
prime
, we obtain
Delta1V = lim
h→0
bracketleftBigg
1
4πepsilon1
0
integraldisplay
2π
0
integraldisplay
a
0
[?zD
0
] ·
?zh ? ?ρ
prime
ρ
prime
parenleftbig
h
2
+ρ
prime2
parenrightbig
3/2
ρ
prime
dρ
prime
dφ
prime
?
?
1
4πepsilon1
0
integraldisplay
2π
0
integraldisplay
a
0
[?zD
0
] ·
??zh ? ?ρ
prime
ρ
prime
parenleftbig
h
2
+ρ
prime2
parenrightbig
3/2
ρ
prime
dρ
prime
dφ
prime
bracketrightBigg
Figure 3.15: Auxiliary disk for studying the potential distribution across a dipole layer.
where a is the disk radius. Integration yields
Delta1V =
D
0
2epsilon1
0
lim
h→0
?
?
?2
radicalBig
1 +
parenleftbig
a
h
parenrightbig
2
+ 2
?
?
=
D
0
epsilon1
0
,
independent of a. Generalizing this to an arbitrary surface dipole moment density, we
?nd that the boundary condition on the potential is given by
Phi1
2
(r)?Phi1
1
(r) =
D
s
(r)
epsilon1
0
(3.106)
where “1” denotes the positive side of the dipole moments and “2” the negative side.
Physically, the potential di?erence in (3.106)is produced by the line integral of E “in-
ternal” to the dipole layer. Since there is no ?eld internal to a unipolar surface layer, V
is continuous across a surface containing charge ρ
s
but having D
s
= 0.
3.2.9 Behavior of electric charge density near a conducting edge
Sharp corners are often encountered in the application of electrostatics to practical ge-
ometries. The behavior of the charge distribution near these corners must be understood
in order to develop numerical techniques for solving more complicated problems. We can
use a simple model of a corner if we restrict our interest to the region near the edge.
ConsidertheintersectionoftwoplanesasshowninFigure3.16.Theregionnearthein-
tersection represents the corner we wish to study. We assume that the planes are held at
zero potential and that the charge on the surface is induced by a two-dimensional charge
distribution ρ(r), or by a potential di?erence between the edge and another conductor
far removed from the edge.
We can ?nd the potential in the region near the edge by solving Laplace’s equation in
cylindrical coordinates. This problem is studied in Appendix A where the separation of
variables solution is found to be either (A.127)or (A.128). Using (A.128)and enforcing
Phi1 = 0 at both φ = 0 and φ = β, we obtain the null solution. Hence the solution must
take the form (A.127):
Phi1(ρ,φ) = [A
φ
sin(k
φ
φ)+ B
φ
cos(k
φ
φ)][a
ρ
ρ
?k
φ
+ b
ρ
ρ
k
φ
]. (3.107)
Figure 3.16: A conducting edge.
Since the origin is included we cannot have negative powers of ρ and must put a
ρ
= 0.
The boundary condition Phi1(ρ,0) = 0 requires B
φ
= 0. The condition Phi1(ρ,β) = 0 then
requires sin(k
φ
β)= 0, which holds only if k
φ
= nπ/β, n = 1,2,.... The general solution
for the potential near the edge is therefore
Phi1(ρ,φ) =
N
summationdisplay
n=1
A
n
sin
parenleftbigg
nπ
β
φ
parenrightbigg
ρ
nπ/β
(3.108)
where the constants A
n
depend on the excitation source or system of conductors. (Note
that if the corner is held at potential V
0
negationslash= 0, we must merely add V
0
to the solution.)
The charge on the conducting surfaces can be computed from the boundary condition
on normal D. Using (3.30)we have
E
φ
=?
1
ρ
?
?φ
N
summationdisplay
n=1
A
n
sin
parenleftbigg
nπ
β
φ
parenrightbigg
ρ
nπ/β
=?
N
summationdisplay
n=1
A
n
nπ
β
cos
parenleftbigg
nπ
β
φ
parenrightbigg
ρ
(nπ/β)?1
,
hence
ρ
s
(x) =?epsilon1
N
summationdisplay
n=1
A
n
nπ
β
x
(nπ/β)?1
on the surface at φ = 0. Near the edge, at small values of x, the variation of ρ
s
is dom-
inated by the lowest power of x. (Here we ignore those special excitation arrangements
that produce A
1
= 0.)Thus
ρ
s
(x) ～ x
(π/β)?1
.
The behavior of the charge clearly depends on the wedge angle β. For a sharp edge
(half plane)we put β = 2π and ?nd that the ?eld varies as x
?1/2
. This square-root edge
singularity is very common on thin plates, ?ns, etc., and means that charge tends to
accumulate near the edge of a ?at conducting surface. For a right-angle corner where
β = 3π/2, there is the somewhat weaker singularity x
?1/3
. When β = π, the two
surfaces fold out into an in?nite plane and the charge, not surprisingly, is invariant with
x to lowest order near the folding line. When β<πthe corner becomes interior and we
?nd that the charge density varies with a positive power of distance from the edge. For
very sharp interior angles the power is large, meaning that little charge accumulates on
the inner surfaces near an interior corner.
3.2.10 Solution to Laplace’s equation for bodies immersed in an im-
pressed ?eld
An important class of problems is based on the idea of placing a body into an existing
electric ?eld, assuming that the ?eld arises from sources so remote that the introduction
of the body does not alter the original ?eld. The pre-existing ?eld is often referred to as
the applied or impressed ?eld, and the solution external to the body is usually formulated
as the sum of the applied ?eld and a secondary or scattered ?eld that satis?es Laplace’s
equation. This total ?eld di?ers from the applied ?eld, and must satisfy the appropriate
boundary condition on the body. If the body is a conductor then the total potential must
be constant everywhere on the boundary surface. If the body is a solid homogeneous
dielectric then the total potential ?eld must be continuous across the boundary.
As an example, consider a dielectric sphere of permittivity epsilon1 and radius a, centered at
the origin and immersed in a constant electric ?eld E
0
(r) = E
0
?z. By (3.30)the applied
potential ?eld is Phi1
0
(r) =?E
0
z =?E
0
r cosθ (to within a constant). Outside the sphere
(r > a)we write the total potential ?eld as
Phi1
2
(r) = Phi1
0
(r)+Phi1
s
(r)
where Phi1
s
(r) is the secondary or scattered potential. Since Phi1
s
must satisfy Laplace’s
equation, we can write it as a separation of variables solution (§ A.4). By azimuthal
symmetry the potential has an r-dependence as in (A.146), and a θ-dependence as in
(A.142)with B
θ
= 0 and m = 0.ThusPhi1
s
has a representation identical to (A.147),
except that we cannot use terms that are unbounded as r →∞. We therefore use
Phi1
s
(r,θ)=
∞
summationdisplay
n=0
B
n
r
?(n+1)
P
n
(cosθ). (3.109)
The potential inside the sphere also obeys Laplace’s equation, so we can use the same
form (A.147)while discarding terms unbounded at the origin. Thus
Phi1
1
(r,θ)=
∞
summationdisplay
n=0
A
n
r
n
P
n
(cosθ) (3.110)
for r < a. To ?nd the constants A
n
and B
n
we apply (3.36)and (3.37)to the total ?eld.
Application of (3.36)at r = a gives
?E
0
a cosθ +
∞
summationdisplay
n=0
B
n
a
?(n+1)
P
n
(cosθ)=
∞
summationdisplay
n=0
A
n
a
n
P
n
(cosθ).
Multiplying through by P
m
(cosθ)sinθ, integrating from θ = 0 to θ = π, and using the
orthogonality relationship (E.123), we obtain
?E
0
a + a
?2
B
1
= A
1
a, (3.111)
B
n
a
?(n+1)
= A
n
a
n
, n negationslash= 1, (3.112)
where we have used P
1
(cosθ)= cosθ. Next, since ρ
s
= 0, equation (3.37)requires that
epsilon1
1
?Phi1
1
(r)
?r
= epsilon1
2
?Phi1
2
(r)
?r
at r = a. This gives
?epsilon1
0
E
0
cosθ +epsilon1
0
∞
summationdisplay
n=0
[?(n + 1)B
n
]a
?n?2
P
n
(cosθ)= epsilon1
∞
summationdisplay
n=0
[nA
n
]a
n?1
P
n
(cosθ).
By orthogonality of the Legendre functions we have
?epsilon1
0
E
0
? 2epsilon1
0
B
1
a
?3
= epsilon1A
1
, (3.113)
?epsilon1
0
(n + 1)B
n
a
?n?2
= epsilon1nA
n
a
n?1
, n negationslash= 1. (3.114)
Equations (3.112)and (3.114)cannot hold simultaneously unless A
n
= B
n
= 0 for n negationslash= 1.
Solving (3.111)and (3.113)we have
A
1
=?E
0
3epsilon1
0
epsilon1 + 2epsilon1
0
, B
1
= E
0
a
3
epsilon1 ?epsilon1
0
epsilon1 + 2epsilon1
0
.
Hence
Phi1
1
(r) =?E
0
3epsilon1
0
epsilon1 + 2epsilon1
0
r cosθ =?E
0
z
3epsilon1
0
epsilon1 + 2epsilon1
0
, (3.115)
Phi1
2
(r) =?E
0
r cosθ + E
0
a
3
r
2
epsilon1 ?epsilon1
0
epsilon1 + 2epsilon1
0
cosθ. (3.116)
Interestingly, the electric ?eld
E
1
(r) =??Phi1
1
(r) = ?zE
0
3epsilon1
0
epsilon1 + 2epsilon1
0
inside the sphere is constant with position and is aligned with the applied external ?eld.
However, it is weaker than the applied ?eld since epsilon1>epsilon1
0
. To explain this, we compute
the polarization charge within and on the sphere. Using D = epsilon1E = epsilon1
0
E + P we have
P
1
= ?z(epsilon1 ?epsilon1
0
)E
0
3epsilon1
0
epsilon1 + 2epsilon1
0
. (3.117)
The volume polarization charge density ?? · P is zero, while the polarization surface
charge density is
ρ
Ps
= ?r · P = (epsilon1 ?epsilon1
0
)E
0
3epsilon1
0
epsilon1 + 2epsilon1
0
cosθ.
Hence the secondary electric ?eld can be attributed to an induced surface polarization
charge, and is in a direction opposing the applied ?eld. According to the Maxwell–Bo?
viewpoint we should be able to replace the sphere by the surface polarization charge
immersed in free space, and use the formula (3.61)to reproduce (3.115)and (3.116).
This is left as an exercise for the reader.
3.3 Magnetostatics
The large-scale forms of the magnetostatic ?eld equations are
contintegraldisplay
Gamma1
H · dl =
integraldisplay
S
J · dS, (3.118)
contintegraldisplay
S
B · dS = 0, (3.119)
while the point forms are
?×H(r) = J(r), (3.120)
?·B(r) = 0. (3.121)
Note the interesting dichotomy between the electrostatic ?eld equations and the magne-
tostatic ?eld equations. Whereas the electrostatic ?eld exhibits zero curl and a divergence
proportional to the source (charge), the magnetostatic ?eld has zero divergence and a
curl proportional to the source (current). Because the vector relationship between the
magnetostatic ?eld and its source is of a more complicated nature than the scalar rela-
tionship between the electrostatic ?eld and its source, more e?ort is required to develop a
strong understanding of magnetic phenomena. Also, it must always be remembered that
although the equations describing the electrostatic and magnetostatic ?eld sets decou-
ple, the phenomena themselves remain linked. Since current is moving charge, electrical
phenomena are associated with the establishment of the current that supports a magne-
tostatic ?eld. We know, for example, that in order to have current in a wire an electric
?eld must be present to drive electrons through the wire.
The magnetic scalar potential. Under certain conditions the equations of magne-
tostatics have the same form as those of electrostatics. If J = 0 in a region V, the
magnetostatic equations are
?×H(r) = 0, (3.122)
?·B(r) = 0; (3.123)
compare with (3.5)–(3.6) when ρ = 0. Using (3.122)we can de?ne a magnetic scalar
potential Phi1
m
:
H =??Phi1
m
. (3.124)
The negative sign is chosen for consistency with (3.30). We can then de?ne a magnetic
potential di?erence between two points as
V
m21
=?
integraldisplay
P
2
P
1
H · dl =?
integraldisplay
P
2
P
1
??Phi1
m
(r)· dl =
integraldisplay
P
2
P
1
dPhi1
m
(r) = Phi1
m
(r
2
)?Phi1
m
(r
1
).
Unlike the electrostatic potential di?erence, V
m21
isnotunique.ConsiderFigure3.17,
which shows a plane passing through the cross-section of a wire carrying total current I.
Although there is no current within the region V (external to the wire), equation (3.118)
still gives
integraldisplay
Gamma1
2
H · dl ?
integraldisplay
Gamma1
3
H · dl = I.
Thus
integraldisplay
Gamma1
2
H · dl =
integraldisplay
Gamma1
3
H · dl + I,
and the integral
integraltext
Gamma1
H · dl is not path-independent. However,
integraldisplay
Gamma1
1
H · dl =
integraldisplay
Gamma1
2
H · dl
since no current passes through the surface bounded by Gamma1
1
? Gamma1
2
. So we can arti?cially
impose uniqueness by demanding that no path cross a cut such as that indicated by the
line L in the ?gure.
Figure 3.17: Magnetic potential.
Because V
m21
is not unique, the ?eld H is nonconservative. In point form this is
shown by the fact that ?×H is not identically zero. We are not too concerned about
energy-related implications of the nonconservative nature of H; the electric point charge
has no magnetic analogue that might fail to conserve potential energy if moved around
in a magnetic ?eld.
Assuming a linear, isotropic region where B(r) = μ(r)H(r), we can substitute (3.124)
into (3.123)and expand to obtain
?μ(r)·?Phi1
m
(r)+μ(r)?
2
Phi1
m
(r) = 0.
For a homogeneous medium this reduces to Laplace’s equation
?
2
Phi1
m
= 0.
We can also obtain an analogue to Poisson’s equation of electrostatics if we use
B = μ
0
(H + M) =?μ
0
?Phi1
m
+μ
0
M
in (3.123); we have
?
2
Phi1
m
=?ρ
M
(3.125)
where
ρ
M
=??·M
is called the equivalent magnetization charge density. This form can be used to describe
?elds of permanent magnets in the absence of J. Comparison with (3.98)shows that ρ
M
is analogous to the polarization charge ρ
P
.
Since Phi1
m
obeys Poisson’s equation, the details regarding uniqueness and the construc-
tion of solutions follow from those of the electrostatic case. If we include the possibility of
a surface density of magnetization charge, then the integral solution for Phi1
m
in unbounded
space is
Phi1
m
(r) =
1
4π
integraldisplay
V
ρ
M
(r
prime
)
|r ? r
prime
|
dV
prime
+
1
4π
integraldisplay
S
ρ
Ms
(r
prime
)
|r ? r
prime
|
dS
prime
. (3.126)
Here ρ
Ms
, the surface density of magnetization charge, is identi?ed as ?n · M in the
boundary condition (3.152).
3.3.1 The magnetic vector potential
Although the magnetic scalar potential is useful for describing ?elds of permanent
magnets and for solving certain boundary value problems, it does not include the e?ects of
source current. A second type of potential function, called the magnetic vector potential,
can be used with complete generality to describe the magnetostatic ?eld. Because ?·B =
0, we can write by (B.49)
B(r) =?×A(r) (3.127)
where A is the vector potential. Now A is not determined by (3.127)alone, since the
gradient of any scalar ?eld can be added to A without changing the value of ?×A.
Such “gauge transformations” are discussed in Chapter 5, where we ?nd that ?·A must
also be speci?ed for uniqueness of A.
The vector potential can be used to develop a simple formula for the magnetic ?ux
passing through an open surface S:
Psi1
m
=
integraldisplay
S
B · dS =
integraldisplay
S
(?×A)· dS =
contintegraldisplay
Gamma1
A · dl, (3.128)
where Gamma1 is the contour bounding S.
In the linear isotropic case where B = μH we can ?nd a partial di?erential equation
for A by substituting (3.127)into (3.120). Using (B.43)we have
?×
bracketleftbigg
1
μ(r)
?×A(r)
bracketrightbigg
= J(r),
hence
1
μ(r)
?×[?×A(r)] ? [?×A(r)] ×?
parenleftbigg
1
μ(r)
parenrightbigg
= J(r).
In a homogeneous region we have
?×(?×A) = μJ (3.129)
or
?(?·A)??
2
A = μJ (3.130)
by (B.47). As mentioned above we must eventually specify ?·A. Although the choice is
arbitrary, certain selections make the computation of A both mathematically tractable
and physically meaningful. The “Coulomb gauge condition” ?·A = 0 reduces (3.130)
to
?
2
A =?μJ. (3.131)
The vector potential concept can also be applied to the Maxwell–Bo? magnetostatic
equations
?×B = μ
0
(J +?×M), (3.132)
?·B = 0. (3.133)
By (3.133)we may still de?ne A through (3.127). Substituting this into (3.132) we have,
under the Coulomb gauge,
?
2
A =?μ
0
[J + J
M
] (3.134)
where J
M
=?×M is the magnetization current density.
Figure 3.18: Circular loop of wire.
The di?erential equations (3.131)and (3.134)are vector versions of Poisson’s equation,
and may be solved quite easily for unbounded space by decomposing the vector source
into rectangular components. For instance, dotting (3.131)with ?x we ?nd that
?
2
A
x
=?μJ
x
.
This scalar version of Poisson’s equation has solution
A
x
(r) =
μ
4π
integraldisplay
V
J
x
(r
prime
)
|r ? r
prime
|
dV
prime
in unbounded space. Repeating this for each component and assembling the results, we
obtain the solution for the vector potential in an unbounded homogeneous medium:
A(r) =
μ
4π
integraldisplay
V
J(r
prime
)
|r ? r
prime
|
dV
prime
. (3.135)
Any surface sources can be easily included through a surface integral:
A(r) =
μ
4π
integraldisplay
V
J(r
prime
)
|r ? r
prime
|
dV
prime
+
μ
4π
integraldisplay
S
J
s
(r
prime
)
|r ? r
prime
|
dS
prime
. (3.136)
In unbounded free space containing materials represented by M,wehave
A(r) =
μ
0
4π
integraldisplay
V
J(r
prime
)+ J
M
(r
prime
)
|r ? r
prime
|
dV
prime
+
μ
0
4π
integraldisplay
S
J
s
(r
prime
)+ J
Ms
(r
prime
)
|r ? r
prime
|
dV
prime
(3.137)
where J
Ms
=??n × M is the surface density of magnetization current as described in
(3.153). It may be veri?ed directly from (3.137) that ?·A = 0.
Field of a circular loop. Consider a circular loop of line current of radius a in
unboundedspace(Figure3.18).Using J(r
prime
)= I
?
φ
prime
δ(z
prime
)δ(ρ
prime
? a )andnotingthat r =
ρ ?ρ+ z?z and r
prime
= a ?ρ
prime
, we can write (3.136)as
A(r) =
μI
4π
integraldisplay
2π
0
?
φ
prime
adφ
prime
bracketleftbig
ρ
2
+ a
2
+ z
2
? 2aρ cos(φ ?φ
prime
)
bracketrightbig
1/2
.
Because
?
φ
prime
=??x cosφ
prime
+ ?y sinφ
prime
we ?nd that
A(r) =
μIa
4π
?
φ
integraldisplay
2π
0
cosφ
prime
bracketleftbig
ρ
2
+ a
2
+ z
2
? 2aρ cosφ
prime
bracketrightbig
1/2
dφ
prime
.
We put the integral into standard form by setting φ
prime
= π ? 2x:
A(r) =?
μIa
4π
?
φ
integraldisplay
π/2
?π/2
1 ? 2 sin
2
x
bracketleftbig
ρ
2
+ a
2
+ z
2
+ 2aρ(1 ? 2 sin
2
x)
bracketrightbig
1/2
2 dx.
Letting
k
2
=
4aρ
(a +ρ)
2
+ z
2
, F
2
= (a +ρ)
2
+ z
2
,
we have
A(r) =?
μIa
4π
?
φ
4
F
integraldisplay
π/2
0
1 ? 2 sin
2
x
[1 ? k
2
sin
2
x]
1/2
dx.
Then, since
1 ? 2 sin
2
x
[1 ? k
2
sin
2
x]
1/2
=
k
2
? 2
k
2
[1 ? k
2
sin
2
x]
?1/2
+
2
k
2
[1 ? k
2
sin
2
x]
1/2
,
we have
A(r) =
?
φ
μI
πk
radicalbigg
a
ρ
bracketleftbiggparenleftbigg
1 ?
1
2
k
2
parenrightbigg
K(k
2
)? E(k
2
)
bracketrightbigg
. (3.138)
Here
K(k
2
) =
integraldisplay
π/2
0
du
[1 ? k
2
sin
2
u]
1/2
, E(k
2
) =
integraldisplay
π/2
0
[1 ? k
2
sin
2
u]
1/2
du,
are complete elliptic integrals of the ?rst and second kinds, respectively.
We have k
2
lessmuch 1 when the observation point is far from the loop (r
2
= ρ
2
+ z
2
greatermuch a
2
).
Using the expansions [47]
K(k
2
) =
π
2
bracketleftbigg
1 +
1
4
k
2
+
9
64
k
4
+···
bracketrightbigg
, E(k
2
) =
π
2
bracketleftbigg
1 ?
1
4
k
2
?
3
64
k
4
?···
bracketrightbigg
,
in (3.138)and keeping the ?rst nonzero term, we ?nd that
A(r) ≈
?
φ
μI
4πr
2
(πa
2
)sinθ. (3.139)
De?ning the magnetic dipole moment of the loop as
m = ?zIπa
2
,
we can write (3.139)as
A(r) =
μ
4π
m × ?r
r
2
. (3.140)
Generalization to an arbitrarily-oriented circular loop with center located at r
0
is accom-
plished by writing m = ?nIAwhere A is the loop area and ?n is normal to the loop in the
right-hand sense. Then
A(r) =
μ
4π
m ×
r ? r
0
|r ? r
0
|
3
.
We shall ?nd, upon investigating the general multipole expansion of A below, that this
holds for any planar loop.
The magnetic ?eld of the loop can be found by direct application of (3.127). For the
case r
2
greatermuch a
2
we take the curl of (3.139)and ?nd that
B(r) =
μ
4π
m
r
3
(?r 2 cosθ +
?
θ sinθ). (3.141)
Comparison with (3.93)shows why we often refer to a small loop as a magnetic dipole.
But (3.141)is approximate, and since there are no magnetic monopoles we cannot con-
struct an exact magnetic analogue to the electric dipole. On the other hand, we shall
?nd below that the multipole expansion of a ?nite-extent steady current begins with the
dipole term (since the current must form closed loops). We may regard small loops as
the elemental units of steady current from which all other currents may be constructed.
3.3.2 Multipole expansion
It is possible to derive a general multipole expansion for A analogous to (3.94). But
the vector nature of A requires that we use vector spherical harmonics, hence the result
is far more complicated than (3.94). A simpler approach yields the ?rst few terms and
requires only the Taylor expansion of 1/R. Consider a steady current localized near the
origin and contained within a sphere of radius r
m
. We substitute the expansion (3.89)
into (3.135)to obtain
A(r) =
μ
4π
integraldisplay
V
J(r
prime
)
bracketleftbigg
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+(r
prime
·?
prime
)
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+
1
2
(r
prime
·?
prime
)
2
1
R
vextendsingle
vextendsingle
vextendsingle
vextendsingle
r
prime
=0
+···
bracketrightbigg
dV
prime
, (3.142)
which we view as
A(r) = A
(0)
(r)+ A
(1)
(r)+ A
(2)
(r)+···.
The ?rst term is merely
A
(0)
(r) =
μ
4πr
integraldisplay
V
J(r
prime
)dV
prime
=
μ
4πr
3
summationdisplay
i=1
?x
i
integraldisplay
V
J
i
(r
prime
)dV
prime
where (x, y, z) = (x
1
, x
2
, x
3
). However, by (3.26)each of the integrals is zero and we have
A
(0)
(r) = 0;
the leading term in the multipole expansion of A for a general steady current distribution
vanishes.
Using (3.91)we can write the second term as
A
(1)
(r) =
μ
4πr
3
integraldisplay
V
J(r
prime
)
3
summationdisplay
i=1
x
i
x
prime
i
dV
prime
=
μ
4πr
3
3
summationdisplay
j=1
?x
j
3
summationdisplay
i=1
x
i
integraldisplay
V
x
prime
i
J
j
(r
prime
)dV
prime
. (3.143)
By adding the null relation (3.28)we can write
integraldisplay
V
x
prime
i
J
j
dV
prime
=
integraldisplay
V
x
prime
i
J
j
dV
prime
+
integraldisplay
V
[x
prime
i
J
j
+ x
prime
j
J
i
] dV
prime
= 2
integraldisplay
V
x
prime
i
J
j
dV
prime
+
integraldisplay
V
x
prime
j
J
i
dV
prime
or
integraldisplay
V
x
prime
i
J
j
dV
prime
=
1
2
integraldisplay
V
[x
prime
i
J
j
? x
prime
j
J
i
] dV
prime
. (3.144)
Figure 3.19: A planar wire loop.
By this and (3.143)the second term in the multipole expansion is
A
(1)
(r) =
μ
4πr
3
1
2
integraldisplay
V
3
summationdisplay
j=1
?x
j
3
summationdisplay
i=1
x
i
[x
prime
i
J
j
? x
prime
j
J
i
] dV
prime
=?
μ
4πr
3
1
2
integraldisplay
V
r × [r
prime
× J(r
prime
)] dV
prime
.
De?ning the dipole moment vector
m =
1
2
integraldisplay
V
r × J(r)dV (3.145)
we have
A
(1)
(r) =
μ
4π
m ×
parenleftbigg
?r
r
2
parenrightbigg
=?
μ
4π
m ×?
1
r
. (3.146)
This is the dipole moment potential for the steady current J. Since steady currents of
?nite extent consist of loops, the dipole component is generally the ?rst nonzero term
in the expansion of A. Higher-order components may be calculated, but extension of
(3.142)beyond the dipole term is quite tedious and will not be attempted.
As an example let us compute the dipole moment of the planar but otherwise arbitrary
loopshowninFigure3.19.Specializing(3.145)fo ralinecurrentwehave
m =
I
2
contintegraldisplay
Gamma1
r × dl.
ExaminingFigure3.19,weseethat
1
2
r × dl = ?n dS
where dS is the area of the sector swept out by r as it moves along dl, and ?n is the
normal to the loop in the right-hand sense. Thus
m = ?nIA (3.147)
where A is the area of the loop.
Physical interpretation of M in a magnetic material. In (3.137)we presented
an expression for the vector potential produced by a magnetized material in terms of
equivalent magnetization surface and volume currents. Suppose a magnetized medium
is separated into volume regions with bounding surfaces across which the permeability
is discontinuous. With J
M
=?×M and J
Ms
=??n × M we obtain
A(r) =
μ
0
4π
integraldisplay
V
J(r
prime
)
|r ? r
prime
|
dV
prime
+
μ
0
4π
integraldisplay
S
J
s
(r
prime
)
|r ? r
prime
|
dS
prime
+
+
summationdisplay
i
μ
0
4π
bracketleftbiggintegraldisplay
V
i
?
prime
× M(r
prime
)
|r ? r
prime
|
dV
prime
+
integraldisplay
S
i
??n
prime
× M(r
prime
)
|r ? r
prime
|
dS
prime
bracketrightbigg
. (3.148)
Here ?n points outward from region V
i
. Using the curl theorem on the fourth term and
employing the vector identity (B.43), we have
A(r) =
μ
0
4π
integraldisplay
V
J(r
prime
)
|r ? r
prime
|
dV
prime
+
μ
0
4π
integraldisplay
S
J
s
(r
prime
)
|r ? r
prime
|
dS
prime
+
+
summationdisplay
i
bracketleftbigg
μ
0
4π
integraldisplay
V
i
M(r
prime
)×?
prime
parenleftbigg
1
|r ? r
prime
|
parenrightbigg
dV
prime
bracketrightbigg
. (3.149)
But ?
prime
(1/R) =
?
R/R
2
, hence the third term is a sum of integrals of the form
μ
0
4π
integraldisplay
V
i
M(r
prime
)×
?
R
R
2
dV
prime
.
Comparison with (3.146)shows that this integral represents a volume superposition of
dipole moments where M is a volume density of magnetic dipole moments. Hence a
magnetic material with permeability μ is equivalent to a volume distribution of magnetic
dipoles in free space. As with our interpretation of the polarization vector in a dielectric,
we base this conclusion only on Maxwell’s equations and the assumption of a linear,
isotropic relationship between B and H.
3.3.3 Boundary conditions for the magnetostatic ?eld
The boundary conditions found for the dynamic magnetic ?eld remain valid in the
magnetostatic case. Hence
?n
12
×(H
1
? H
2
) = J
s
(3.150)
and
?n
12
·(B
1
? B
2
) = 0, (3.151)
where ?n
12
points into region 1 from region 2. Since the magnetostatic curl and divergence
equations are independent, so are the boundary conditions (3.150)and (3.151). We can
also write (3.151)in terms of equivalent sources by (3.118):
?n
12
·(H
1
? H
2
) = ρ
Ms1
+ρ
Ms2
, (3.152)
where ρ
Ms
= ?n · M is called the equivalent magnetization surface charge density. Here ?n
points outward from the material body.
For a linear, isotropic material described by B = μH, equation (3.150)becomes
?n
12
×
parenleftbigg
B
1
μ
1
?
B
2
μ
2
parenrightbigg
= J
s
.
With (3.118)we can also write (3.150)as
?n
12
×(B
1
? B
2
) = μ
0
(J
s
+ J
Ms1
+ J
Ms2
) (3.153)
where J
Ms
=??n × M is the equivalent magnetization surface current density.
We may also write the boundary conditions in terms of the scalar or vector potential.
Using H =??Phi1
m
, we can write (3.150)as
Phi1
m1
(r) = Phi1
m2
(r) (3.154)
provided that the surface current J
s
= 0. As was the case with (3.36), the possibility of
an additive constant here is generally ignored. To write (3.151)in terms of Phi1
m
we ?rst
note that B/μ
0
? M =??Phi1
m
; substitution into (3.151)gives
?Phi1
m1
?n
?
?Phi1
m2
?n
=?ρ
Ms1
?ρ
Ms2
(3.155)
where the normal derivative is taken in the direction of ?n
12
. For a linear isotropic material
where B = μH we have
μ
1
?Phi1
m1
?n
= μ
2
?Phi1
m2
?n
. (3.156)
Note that (3.154)and (3.156)are independent.
Boundary conditions on A may be derived using the approach of § 2.8.2. Consider
Figure2.6.Herethesurfacemaycarryeitheranelectricsurfacecurrent J
s
or an equiv-
alent magnetization current J
Ms
, and thus may be a surface of discontinuity between
di?ering magnetic media. If we integrate ?×A over the volume regions V
1
and V
2
and
add the results we ?nd that
integraldisplay
V
1
?×A dV +
integraldisplay
V
2
?×A dV =
integraldisplay
V
1
+V
2
B dV.
By the curl theorem
integraldisplay
S
1
+S
2
?n × A dS+
integraldisplay
S
10
??n
10
× A
1
dS+
integraldisplay
S
20
??n
20
× A
2
dS =
integraldisplay
V
1
+V
2
B dV
where A
1
is the ?eld on the surface S
10
and A
2
is the ?eld on S
20
.Asδ → 0 the surfaces S
1
and S
2
combine to give S. Also S
10
and S
20
coincide, as do the normals ?n
10
=??n
20
= ?n
12
.
Thus
integraldisplay
S
(?n × A)dS?
integraldisplay
V
B dV =
integraldisplay
S
10
?n
12
×(A
1
? A
2
)dS. (3.157)
Now let us integrate over the entire volume region V including the surface of discontinuity.
This gives
integraldisplay
S
(?n × A)dS?
integraldisplay
V
B dV = 0,
and for agreement with (3.157)we must have
?n
12
×(A
1
? A
2
) = 0. (3.158)
A similar development shows that
?n
12
·(A
1
? A
2
) = 0. (3.159)
Therefore A is continuous across a surface carrying electric or magnetization current.
3.3.4 Uniqueness of the magnetostatic ?eld
Because the uniqueness conditions established for the dynamic ?eld do not apply to
magnetostatics, we begin with the magnetostatic ?eld equations. Consider a region of
space V bounded by a surface S. There may be source currents and magnetic materials
both inside and outside V. Assume (B
1
,H
1
) and (B
2
,H
2
) are solutions to the magne-
tostatic ?eld equations with source J. We seek conditions under which B
1
= B
2
and
H
1
= H
2
.
The di?erence ?eld H
0
= H
2
? H
1
obeys ?×H
0
= 0. Using (B.44)we examine the
quantity
?·(A
0
× H
0
) = H
0
·(?×A
0
)? A
0
·(?×H
0
) = H
0
·(?×A
0
)
where A
0
is de?ned by B
0
= B
2
? B
1
=?×A
0
=?×(A
2
? A
1
). Integrating over V
we obtain
contintegraldisplay
S
(A
0
× H
0
)· dS =
integraldisplay
V
H
0
·(?×A
0
)dV =
integraldisplay
V
H
0
· B
0
dV.
Then, since (A
0
× H
0
)· ?n =?A
0
·(?n × H
0
), we have
?
contintegraldisplay
S
A
0
·(?n × H
0
)dS =
integraldisplay
V
H
0
· B
0
dV. (3.160)
If A
0
= 0 or ?n × H
0
= 0 everywhere on S,orA
0
= 0 on part of S and ?n × H
0
= 0 on the
remainder, then
integraldisplay
V
H
0
· B
0
dS = 0. (3.161)
So H
0
= 0 or B
0
= 0 by arbitrariness of V. Assuming H and B are linked by the
constitutive relations, we have H
1
= H
2
and B
1
= B
2
. The ?elds within V are unique
provided that A, the tangential component of H, or some combination of the two, is
speci?ed over the bounding surface S.
One other condition will cause the left-hand side of (3.160)to vanish. If S recedes to
in?nity then, provided that the potential functions vanish su?ciently fast, the condition
(3.161)still holds and uniqueness is guaranteed. Equation (3.135)shows that A ～ 1/r
as r →∞, hence B,H ～ 1/r
2
. So uniqueness is ensured by the speci?cation of J in
unbounded space.
3.3.5 Integral solution for the vector potential
We have used the scalar Green’s theorem to ?nd a solution for the electrostatic poten-
tial within a region V in terms of the source charge in V and the values of the potential
and its normal derivative on the boundary surface S. Analogously, we may ?nd A within
V in terms of the source current in V and the values of A and its derivatives on S. The
vector relationship between B and A complicates the derivation somewhat, requiring
Green’s second identity for vector ?elds.
Let P and Q be continuous with continuous ?rst and second derivatives throughout
V and on S. The divergence theorem shows that
integraldisplay
V
?·[P ×(?×Q)] dV =
integraldisplay
S
[P ×(?×Q)] · dS.
By virtue of (B.44)we have
integraldisplay
V
[(?×Q)·(?×P)? P ·(? ×{? × Q})] dV =
integraldisplay
S
[P ×(?×Q)] · dS.
We now interchange P and Q and subtract the result from the above, obtaining
integraldisplay
V
[Q ·(? ×{? × P})? P ·(? ×{? × Q})] dV =
integraldisplay
S
[P ×(?×Q)? Q ×(?×P)] · dS. (3.162)
Note that ?n points outward from V. This is Green’s second identity for vector ?elds.
Now assume that V contains a magnetic material of uniform permeability μ and set
P = A(r
prime
), Q =
c
R
,
in (3.162)written in terms of primed coordinates. Here c is a constant vector, nonzero
but otherwise arbitrary. We ?rst examine the volume integral terms. Note that
?
prime
×(?
prime
× Q) =?
prime
×
parenleftBig
?
prime
×
c
R
parenrightBig
=??
prime
2
parenleftBig
c
R
parenrightBig
+?
prime
bracketleftBig
?
prime
·
parenleftBig
c
R
parenrightBigbracketrightBig
.
By (B.162)and (3.58)we have
?
prime
2
parenleftBig
c
R
parenrightBig
=
1
R
?
prime
2
c + c?
prime
2
parenleftbigg
1
R
parenrightbigg
+ 2
parenleftbigg
?
prime
1
R
·?
prime
parenrightbigg
c = c?
prime
2
parenleftbigg
1
R
parenrightbigg
=?c4πδ(r ? r
prime
),
hence
P · [?
prime
×(?
prime
× Q)] = 4πc · Aδ(r ? r
prime
)+ A ·?
prime
bracketleftBig
?
prime
·
parenleftBig
c
R
parenrightBigbracketrightBig
.
Since ?·A = 0 the second term on the right-hand side can be rewritten using (B.42):
?
prime
·(ψA) = A ·(?
prime
ψ)+ψ?
prime
· A = A ·(?
prime
ψ).
Thus
P · [?
prime
×(?
prime
× Q)] = 4πc · Aδ(r ? r
prime
)+?
prime
·
bracketleftbigg
A
braceleftbigg
c ·?
prime
parenleftbigg
1
R
parenrightbiggbracerightbiggbracketrightbigg
,
where we have again used (B.42). The other volume integral term can be found by
substituting from (3.129):
Q · [?
prime
×(?
prime
× P)] = μ
1
R
c · J(r
prime
).
Next we investigate the surface integral terms. Consider
?n
prime
·
bracketleftbig
P ×(?
prime
× Q)
bracketrightbig
= ?n
prime
·
braceleftBig
A ×
bracketleftBig
?
prime
×
parenleftBig
c
R
parenrightBigbracketrightBigbracerightBig
= ?n
prime
·
braceleftbigg
A ×
bracketleftbigg
1
R
?
prime
× c ? c ×?
prime
parenleftbigg
1
R
parenrightbiggbracketrightbiggbracerightbigg
=??n
prime
·
braceleftbigg
A ×
bracketleftbigg
c ×?
prime
parenleftbigg
1
R
parenrightbiggbracketrightbiggbracerightbigg
.
This can be put in slightly di?erent form by the use of (B.8). Note that
(A × B)·(C × D) = A · [B ×(C × D)]
= (C × D)·(A × B)
= C · [D ×(A × B)],
hence
?n
prime
·
bracketleftbig
P ×(?
prime
× Q)
bracketrightbig
=?c ·
bracketleftbigg
?
prime
parenleftbigg
1
R
parenrightbigg
×(?n
prime
× A)
bracketrightbigg
.
The other surface term is given by
?n
prime
· [Q ×(?
prime
× P)] = ?n
prime
·
bracketleftBig
c
R
×(?
prime
× A)
bracketrightBig
= ?n
prime
·
parenleftBig
c
R
× B
parenrightBig
=?
c
R
·(?n
prime
× B).
We can now substitute each of the terms into (3.162)and obtain
μc ·
integraldisplay
V
J(r
prime
)
R
dV
prime
? 4πc ·
integraldisplay
V
A(r
prime
)δ(r ? r
prime
)dV
prime
? c ·
contintegraldisplay
S
[?n
prime
· A(r
prime
)]?
prime
parenleftbigg
1
R
parenrightbigg
dS
prime
=?c ·
contintegraldisplay
S
?
prime
parenleftbigg
1
R
parenrightbigg
× [?n
prime
× A(r
prime
)] dS
prime
+ c ·
contintegraldisplay
S
1
R
?n
prime
× B(r
prime
)dS
prime
.
Since c is arbitrary we can remove the dot products to obtain a vector equation. Then
A(r) =
μ
4π
integraldisplay
V
J(r
prime
)
R
dV
prime
?
1
4π
contintegraldisplay
S
braceleftbigg
[?n
prime
× A(r
prime
)] ×?
prime
parenleftbigg
1
R
parenrightbigg
+
+
1
R
?n
prime
× B(r
prime
)+ [?n
prime
· A(r
prime
)]?
prime
parenleftbigg
1
R
parenrightbiggbracerightbigg
dS
prime
. (3.163)
We have expressed A in a closed region in terms of the sources within the region and
the values of A and B on the surface. While uniqueness requires speci?cation of either
A or ?n×B on S, the expression (3.163)includes both quantities. This is similar to (3.56)
for electrostatic ?elds, which required both the scalar potential and its normal derivative.
The reader may be troubled by the fact that we require P and Q to be somewhat well
behaved, then proceed to involve the singular function c/R and integrate over the singu-
larity. We choose this approach to simplify the presentation; a more rigorous approach
which excludes the singular point with a small sphere also gives (3.163). This approach
was used in § 3.2.4 to establish (3.58). The interested reader should see Stratton [187]
for details on the application of this technique to obtain (3.163).
It is interesting to note that as S →∞the surface integral vanishes since A ～ 1/r
and B ～ 1/r
2
, and we recover (3.135). Moreover, (3.163) returns the null result when
evaluated at points outside S (see Stratton [187]). We shall see this again when studying
the integral solutions for electrodynamic ?elds in § 6.1.3.
Finally, with
Q =?
prime
parenleftbigg
1
R
parenrightbigg
× c
we can ?nd an integral expression for B within an enclosed region, representing a gen-
eralization of the Biot–Savart law (Problem 3.20). However, this case will be covered in
the more general development of § 6.1.1.
The Biot–Savart law. We can obtain an expression for B in unbounded space by
performing the curl operation directly on the vector potential:
B(r) =?×
μ
4π
integraldisplay
V
J(r
prime
)
|r ? r
prime
|
dV
prime
=
μ
4π
integraldisplay
V
?×
J(r
prime
)
|r ? r
prime
|
dV
prime
.
Using (B.43)and ?×J(r
prime
) = 0, we have
B(r) =?
μ
4π
integraldisplay
V
J ×?
1
|r ? r
prime
|
dV
prime
.
The Biot–Savart law
B(r) =
μ
4π
integraldisplay
V
J(r
prime
)×
?
R
R
2
dV
prime
(3.164)
follows from (3.57).
For the case of a line current we can replace J dV
prime
by I dl
prime
and obtain
B(r) = I
μ
4π
integraldisplay
Gamma1
dl
prime
×
?
R
R
2
. (3.165)
For an in?nitely long line current on the z-axis we have
B(r) = I
μ
4π
∞
integraldisplay
?∞
?z ×
?z(z ? z
prime
)+ ?ρρ
[(z ? z
prime
)
2
+ρ
2
]
3/2
dz
prime
=
?
φ
μI
2πρ
. (3.166)
This same result follows from taking ?×A after direct computation of A, or from direct
application of the large-scale form of Ampere’s law.
3.3.6 Force and energy
Ampere force on a system of currents. If a steady current J(r) occupying a region
V is exposed to a magnetic ?eld, the force on the moving charge is given by the Lorentz
force law
dF(r) = J(r)× B(r). (3.167)
This can be integrated to give the total force on the current distribution:
F =
integraldisplay
V
J(r)× B(r)dV. (3.168)
It is apparent that the charge ?ow comprising a steady current must be constrained in
some way, or the Lorentz force will accelerate the charge and destroy the steady nature
of the current. This constraint is often provided by a conducting wire.
As an example, consider an in?nitely long wire of circular cross-section centered on
the z-axis in free space. If the wire carries a total current I uniformly distributed over
the cross-section, then within the wire J = ?zI/(πa
2
) where a is the wire radius. The
resulting ?eld can be found through direct integration using (3.164), or by the use of
symmetry and either (3.118)or (3.120). Since B(r) =
?
φB
φ
(ρ), equation (3.118)shows
that
integraldisplay
2π
0
B
φ
(ρ)ρ dφ =
braceleftBigg
μ
0
I
a
2
ρ
2
,ρ≤ a
μ
0
I,ρ≥ a.
Thus
B(r) =
braceleftBigg
?
φμ
0
Iρ/2πa
2
,ρ≤ a,
?
φμ
0
I/2πρ, ρ ≥ a.
(3.169)
The force density within the wire,
dF = J × B =??ρ
μ
0
I
2
ρ
2π
2
a
4
,
is directed inward and tends to compress the wire. Integration over the wire volume gives
F = 0 because
integraldisplay
2π
0
?ρdφ = 0;
however, a section of the wire may experience a net force. For instance, we can compute
the force on one half of the wire split down its axis by using ?ρ = ?x cosφ + ?y sinφ to
obtain F
x
= 0 and
F
y
=?
μ
0
I
2
2π
2
a
4
integraldisplay
dz
integraldisplay
a
0
ρ
2
dρ
integraldisplay
π
0
sinφ dφ =?
μ
0
I
2
3π
2
a
integraldisplay
dz.
The force per unit length
F
l
=??y
μ
0
I
2
3π
2
a
(3.170)
is directed toward the other half as expected.
If the wire takes the form of a loop carrying current I, then (3.167)becomes
dF(r) = I dl(r)× B(r) (3.171)
and the total force acting is
F = I
contintegraldisplay
Gamma1
dl(r)× B(r).
We can write the force on J in terms of the current producing B. Assuming this latter
current J
prime
occupies region V
prime
, the Biot–Savart law (3.164)yields
F =
μ
4π
integraldisplay
V
J(r)×
integraldisplay
V
prime
J(r
prime
)×
r ? r
prime
|r ? r
prime
|
3
dV
prime
dV. (3.172)
This can be specialized to describe the force between line currents. Assume current 1,
following a path Gamma1
1
along the direction dl, carries current I
1
, while current 2, following
path Gamma1
2
along the direction dl
prime
, carries current I
2
. Then the force on current 1 is
F
1
= I
1
I
2
μ
4π
contintegraldisplay
Gamma1
1
contintegraldisplay
Gamma1
2
dl ×
parenleftbigg
dl
prime
×
r ? r
prime
|r ? r
prime
|
3
parenrightbigg
.
This equation, known as Ampere’s force law, can be written in a better form for compu-
tational purposes. We use (B.7)and ?(1/R) from (3.57):
F
1
= I
1
I
2
μ
4π
contintegraldisplay
Gamma1
2
dl
prime
contintegraldisplay
Gamma1
1
dl ·?
prime
parenleftbigg
1
|r ? r
prime
|
parenrightbigg
? I
1
I
2
μ
4π
contintegraldisplay
Gamma1
1
contintegraldisplay
Gamma1
2
(dl · dl
prime
)
r ? r
prime
|r ? r
prime
|
3
. (3.173)
The ?rst term involves an integral of a perfect di?erential about a closed path, producing
a null result. Thus
F
1
=?I
1
I
2
μ
4π
contintegraldisplay
Gamma1
1
contintegraldisplay
Gamma1
2
(dl · dl
prime
)
r ? r
prime
|r ? r
prime
|
3
. (3.174)
Figure 3.20: Parallel, current carrying wires.
Asasimpleexample,considerparallelwiresseparatedbyadistance d (Figure3.20).
In this case
F
1
=?I
1
I
2
μ
4π
integraldisplay bracketleftbiggintegraldisplay
∞
?∞
?d ?x +(z ? z
prime
)?z
[d
2
+(z ? z
prime
)
2
]
3/2
dz
prime
bracketrightbigg
dz = I
1
I
2
μ
2πd
?x
integraldisplay
dz
so the force per unit length is
F
1
l
= ?xI
1
I
2
μ
2πd
. (3.175)
The force is attractive if I
1
I
2
≥ 0 (i.e., if the currents ?ow in the same direction).
Maxwell’s stress tensor. The magnetostatic version of the stress tensor can be ob-
tained from (2.288)by setting E = D = 0:
ˉ
T
m
=
1
2
(B · H)
ˉ
I ? BH. (3.176)
The total magnetic force on the current in a region V surrounded by surface S is given
by
F
m
=?
contintegraldisplay
S
ˉ
T
m
· dS =
integraldisplay
V
f
m
dV
where f
m
= J × B is the magnetic force volume density.
Let us compute the force between two parallel wires carrying identical currents in free
space (let I
1
= I
2
= I inFigure3.20)an dcomparetheresultwith(3.175) .Theforce
on the wire at x =?d/2 can be computed by integrating
ˉ
T
m
· ?n over the yz-plane with
?n = ?x. Using (3.166)we see that in this plane the total magnetic ?eld is
B =??xμ
0
I
π
y
y
2
+ d
2
/4
.
Therefore
ˉ
T
m
· ?n =
1
2
B
x
B
x
μ
0
?x ? ?xB
x
B
x
μ
0
=?μ
0
I
2
2π
2
y
2
[y
2
+ d
2
/4]
2
?x
and by integration
F
1
= μ
0
I
2
2π
2
?x
integraldisplay
dz
integraldisplay
∞
?∞
y
2
[y
2
+ d
2
/4]
2
dy = I
2
μ
0
2πd
?x
integraldisplay
dz.
The resulting force per unit length agrees with (3.175)when I
1
= I
2
= I.
Torque in a magnetostatic ?eld. The torque exerted on a current-carrying conduc-
tor immersed in a magnetic ?eld plays an important role in many engineering applica-
tions. If a rigid body is exposed to a force ?eld of volume density dF(r), the torque on
that body about a certain origin is given by
T =
integraldisplay
V
r × dF dV (3.177)
where integration is performed over the body and r extends from the origin of torque.
If the force arises from the interaction of a current with a magnetostatic ?eld, then
dF = J × B and
T =
integraldisplay
V
r ×(J × B)dV. (3.178)
For a line current we can replace J dV with I dl to obtain
T = I
integraldisplay
Gamma1
r ×(dl × B).
If B is uniform then by (B.7)we have
T =
integraldisplay
V
[J(r · B)? B(r · J)] dV.
The second term can be written as
integraldisplay
V
B(r · J)dV = B
3
summationdisplay
i=1
integraldisplay
V
x
i
J
i
dV = 0
where (x
1
, x
2
, x
3
) = (x, y, z), and where we have employed (3.27). Thus
T =
integraldisplay
V
J(r · B)dV =
3
summationdisplay
j=1
?x
j
integraldisplay
V
J
j
3
summationdisplay
i=1
x
i
B
i
dV =
3
summationdisplay
i=1
B
i
3
summationdisplay
j=1
?x
j
integraldisplay
V
J
j
x
i
dV.
We can replace the integral using (3.144)to get
T =
1
2
integraldisplay
V
3
summationdisplay
j=1
?x
j
3
summationdisplay
i=1
B
i
[x
i
J
j
? x
j
J
i
] dV =?
1
2
integraldisplay
V
B ×(r × J)dV.
Since B is uniform we have, by (3.145),
T = m × B (3.179)
where m is the dipole moment. For a planar loop we can use (3.147)to obtain
T = IA?n × B.
Joule’s law. In § 2.9.5 we showed that when a moving charge interacts with an electric
?eld in a volume region V, energy is transferred between the ?eld and the charge. If the
source of that energy is outside V, the energy is carried into V as an energy ?ux over the
boundary surface S. The energy balance described by Poynting’s theorem (3.299)also
holds for static ?elds supported by steady currents: we must simply recognize that we
have no time-rate of change of stored energy. Thus
?
integraldisplay
V
J · E dV =
contintegraldisplay
S
(E × H)· dS. (3.180)
The term
P =?
integraldisplay
V
J · E dV (3.181)
describes the rate at which energy is supplied to the ?elds by the current within V;we
have P > 0 if there are sources within V that result in energy transferred to the ?elds,
and P < 0 if there is energy transferred to the currents. The latter case occurs when
there are conducting materials in V. Within these conductors
P =?
integraldisplay
V
σE · E dV. (3.182)
Here P < 0; energy is transferred from the ?elds to the currents, and from the currents
into heat (i.e., into lattice vibrations via collisions). Equation (3.182) is called Joule’s
law, and the transfer of energy from the ?elds into heat is Joule heating. Joule’s law is
the power relationship for a conducting material.
An important example involves a straight section of conducting wire having circular
cross-section. Assume a total current I is uniformly distributed over the cross-section
of the wire, and that the wire is centered on the z-axis and extends between the planes
z = 0, L. Let the potential di?erence between the ends be V. Using (3.169)we see that
at the surface of the wire
H =
?
φ
I
2πa
, E = ?z
V
L
.
The corresponding Poynting ?ux E × H is ??ρ-directed, implying that energy ?ows into
wire volume through the curved side surface. We can verify (3.180):
?
integraldisplay
V
J · E dV =
integraldisplay
L
0
integraldisplay
2π
0
integraldisplay
a
0
?z
I
πa
2
· ?z
V
L
ρ dρ dφ dz =?IV,
contintegraldisplay
S
(E × H)· dS =
integraldisplay
2π
0
integraldisplay
L
0
parenleftbigg
??ρ
IV
2πaL
parenrightbigg
· ?ρadφ dz =?IV.
Stored magnetic energy. We have shown that the energy stored in a static charge
distribution may be regarded as the “assembly energy” required to bring charges from
in?nity against the Coulomb force. By proceeding very slowly with this assembly, we are
able to avoid any complications resulting from the motion of the charges.
Similarly, we may equate the energy stored in a steady current distribution to the en-
ergy required for its assembly from current ?laments
6
brought in from in?nity. However,
the calculation of assembly energy is more complicated in this case: moving a current
6
Recall that a ?ux tube of a vector ?eld is bounded by streamlines of the ?eld. A current ?lament is a
?ux tube of current having vanishingly small, but nonzero, cross-section.
Figure 3.21: Calculation of work to move a ?lamentary loop in an applied magnetic ?eld.
?lament into the vicinity of existing ?laments changes the total magnetic ?ux passing
through the existing loops, regardless of how slowly we assemble the ?laments. As de-
scribed by Faraday’s law, this change in ?ux must be associated with an induced emf,
which will tend to change the current ?owing in the ?lament (and any existing ?laments)
unless energy is expended to keep the current constant (by the application of a battery
emf in the opposite direction). We therefore regard the assembly energy as consisting
of two parts: (1)the energy required to bring a ?lament with constant current from
in?nity against the Ampere force, and (2)the energy required to keep the current in this
?lament, and any existing ?laments, constant. We ignore the energy required to keep
the steady current ?owing through an isolated loop (i.e., the energy needed to overcome
Joule losses).
We begin by computing the amount of energy required to bring a ?lament with current
I from in?nity to a given position within an applied magnetostatic ?eld B(r). In this
?rst step we assume that the ?eld is supported by localized sources, hence vanishes at
in?nity, and that it will not be altered by the motion of the ?lament. The force on each
small segment of the ?lament is given by Ampere’s force law (3.171), and the total force
is found by integration. Suppose an external agent displaces the ?lament incrementally
from a starting position 1 to an ending position 2 along a vector δr as shown in Figure
3.21.Theworkrequiredis
δW =?(I dl × B)·δr = (I dl ×δr)· B
foreachsegmentofthewire.Figure3.21showsthat dl ×δr describesasmallpatchof
surface area between the starting and ending positions of the ?lament, hence ?(dl×δr)·B
is the outward ?ux of B through the patch. Integrating over all segments comprising the
?lament, we obtain
Delta1W = I
contintegraldisplay
Gamma1
(dl ×δr)· B =?I
integraldisplay
S
0
B · dS
for the total work required to displace the entire ?lament through δr; here the surface S
0
is described by the superposition of all patches. If S
1
and S
2
are the surfaces bounded
by the ?lament in its initial and ?nal positions, respectively, then S
1
, S
2
, and S
0
taken
together form a closed surface. The outward ?ux of B through this surface is
contintegraldisplay
S
0
+S
1
+S
2
B · dS = 0
so that
Delta1W =?I
integraldisplay
S
0
B · dS = I
integraldisplay
S
1
+S
2
B · dS
where ?n is outward from the closed surface. Finally, let Psi1
1,2
be the ?ux of B through S
1,2
in the direction determined by dl and the right-hand rule. Then
Delta1W =?I(Psi1
2
?Psi1
1
) =?IDelta1Psi1. (3.183)
Now suppose that the initial position of the ?lament is at in?nity. We bring the ?lament
into a ?nal position within the ?eld B through a succession of small displacements,
each requiring work (3.183). By superposition over all displacements, the total work is
W =?I(Psi1?Psi1
∞
) where Psi1
∞
and Psi1 are the ?uxes through the ?lament in its initial and
?nal positions, respectively. However, since the source of the ?eld is localized, we know
that B is zero at in?nity. Therefore Psi1
∞
= 0 and
W =?IPsi1 =?I
integraldisplay
S
B · ?n dS (3.184)
where ?n is determined from dl in the right-hand sense.
Now let us ?nd the work required to position two current ?laments in a ?eld-free region
of space, starting with both ?laments at in?nity. Assume ?lament 1 carries current I
1
and ?lament 2 carries current I
2
, and that we hold these currents constant as we move
the ?laments into position. We can think of assembling these ?laments in two ways: by
placing ?lament 1 ?rst, or by placing ?lament 2 ?rst. In either case, placing the ?rst
?lament requires no work since (3.184)is zero. The work required to place the second
?lament is W
1
=?I
1
Psi1
1
if ?lament 2 is placed ?rst, where Psi1
1
is the ?ux passing through
?lament 1 in its ?nal position, caused by the presence of ?lament 2. If ?lament 1 is
placed ?rst, the work required is W
2
=?I
2
Psi1
2
. Since the work cannot depend on which
loop is placed ?rst, we have W
1
= W
2
= W where we can use either W =?I
1
Psi1
1
or
W =?I
2
Psi1
2
. It is even more convenient, as we shall see, to average these values and use
W =?
1
2
(I
1
Psi1
1
+ I
2
Psi1
2
). (3.185)
We must determine the energy required to keep the currents constant as we move the
?laments into position. When moving the ?rst ?lament into place there is no induced
emf, since no applied ?eld is yet present. However, when moving the second ?lament
into place we will change the ?ux linked by both the ?rst and second loops. This change
of ?ux will induce an emf in each of the loops, and this will change the current. To keep
the current constant we must supply an opposing emf. Let dW
emf
/dt be the rate of work
required to keep the current constant. Then by (3.153)and (3.181)we have
dW
emf
dt
=?
integraldisplay
V
J · E dV =?I
integraldisplay
E · dl =?I
dPsi1
dt
.
Integrating, we ?nd the total work Delta1W required to keep the current constant in either
loop as the ?ux through the loop is changed by an amount Delta1Psi1:
Delta1W
em f
= IDelta1Psi1.
So the total work required to keep I
1
constant as the loops are moved from in?nity (where
the ?ux is zero)to their ?nal positions is I
1
Psi1
1
. Similarly, a total work I
2
Psi1
2
is required
to keep I
2
constant during the same process. Adding these to (3.185), the work required
to position the loops, we obtain the complete assembly energy
W =
1
2
(I
1
Psi1
1
+ I
2
Psi1
2
)
for two ?laments. The extension to N ?laments is
W
m
=
1
2
N
summationdisplay
n=1
I
n
Psi1
n
. (3.186)
Consequently, the energy of a single current ?lament is
W
m
=
1
2
IPsi1. (3.187)
We may interpret this as the “assembly energy” required to bring the single loop into
existence by bringing vanishingly small loops (magnetic dipoles)in from in?nity. We
may also interpret it as the energy required to establish the current in this single ?lament
against the back emf. That is, if we establish I by slowly increasing the current from
zero in N small steps Delta1I = I/N, an energy Psi1
n
Delta1I will be required at each step. Since
Psi1
n
increases proportionally to I,wehave
W
m
=
N
summationdisplay
n=1
I
N
bracketleftbigg
(n ? 1)
Psi1
N
bracketrightbigg
where Psi1 is the ?ux when the current is fully established. Since
summationtext
N
n=1
(n?1) = N(N?1)/2
we obtain
W
m
=
1
2
IPsi1 (3.188)
as N →∞.
A volume current J can be treated as though it were composed of N current ?laments.
Equations (3.128)and (3.186)give
W
m
=
1
2
N
summationdisplay
n=1
I
n
contintegraldisplay
Gamma1
n
A · dl.
Since the total current is
I =
integraldisplay
CS
J · dS =
N
summationdisplay
n=1
I
n
where CS denotes the cross-section of the steady current, we have as N →∞
W
m
=
1
2
integraldisplay
V
A · J dV. (3.189)
Alternatively, using (3.135), we may write
W
m
=
1
2
integraldisplay
V
integraldisplay
V
J(r)· J(r
prime
)
|r ? r
prime
|
dV dV
prime
.
Note the similarity between (3.189)and (3.86). We now manipulate (3.189)into a
form involving only the electromagnetic ?elds. By Ampere’s law
W
m
=
1
2
integraldisplay
V
A ·(?×H)dV.
Using (B.44)and the divergence theorem we can write
W
m
=
1
2
contintegraldisplay
S
(H × A)· dS +
1
2
integraldisplay
V
H ·(?×A)dV.
We now let S expand to in?nity. This does not change the value of W
m
since we do not
enclose any more current; however, since A ～ 1/r and H ～ 1/r
2
, the surface integral
vanishes. Thus, remembering that ?×A = B,wehave
W
m
=
1
2
integraldisplay
V
∞
H · B dV (3.190)
where V
∞
denotes all of space.
Although we do not provide a derivation, (3.190)is also valid within linear materials.
For nonlinear materials, the total energy required to build up a magnetic ?eld from B
1
to B
2
is
W
m
=
1
2
integraldisplay
V
∞
bracketleftbiggintegraldisplay
B
2
B
1
H · dB
bracketrightbigg
dV. (3.191)
This accounts for the work required to drive a ferromagnetic material through its hystere-
sis loop. Readers interested in a complete derivation of (3.191)should consult Stratton
[187].
As an example, consider two thin-walled, coaxial, current-carrying cylinders having
radii a,b (b > a). The intervening region is a linear magnetic material having perme-
ability μ. Assume that the inner and outer conductors carry total currents I in the ±z
directions, respectively. From the large-scale form of Ampere’s law we ?nd that
H =
?
?
?
?
?
0,ρ≤ a,
?
φ I/2πρ, a ≤ ρ ≤ b,
0,ρ>b,
(3.192)
hence by (3.190)
W
m
=
1
2
integraldisplay
dz
integraldisplay
2π
0
integraldisplay
b
a
μI
2
(2πρ)
2
ρ dρ dφ,
and the stored energy is
W
m
l
= μ
I
2
4π
ln
parenleftbigg
b
a
parenrightbigg
(3.193)
per unit length.
Suppose instead that the inner cylinder is solid and that current is spread uniformly
throughout. Then the ?eld between the cylinders is still given by (3.192)but within the
inner conductor we have
H =
?
φ
Iρ
2πa
2
by (3.169). Thus, to (3.193) we must add the energy
W
m,inside
l
=
1
2
integraldisplay
2π
0
integraldisplay
a
0
μ
0
I
2
ρ
2
(2πa
2
)
2
ρ dρ dφ =
μ
0
I
2
16π
stored within the solid wire. The result is
W
m
l
=
μ
0
I
2
4π
bracketleftbigg
μ
r
ln
parenleftbigg
b
a
parenrightbigg
+
1
4
bracketrightbigg
.
3.3.7 Magnetic ?eld of a permanently magnetized body
We now have the tools necessary to compute the magnetic ?eld produced by a perma-
nent magnet (a body with permanent magnetization M). As an example, we shall ?nd
the ?eld due to a uniformly magnetized sphere in three di?erent ways: by computing the
vector potential integral and taking the curl, by computing the scalar potential integral
and taking the gradient, and by ?nding the scalar potential using separation of variables
and applying the boundary condition across the surface of the sphere.
Consider a magnetized sphere of radius a, residing in free space and having permanent
magnetization
M(r) = M
0
?z.
The equivalent magnetization current and charge densities are given by
J
M
=?×M = 0, (3.194)
J
Ms
=??n × M =??r × M
0
?z = M
0
?
φsinθ, (3.195)
and
ρ
M
=??·M = 0, (3.196)
ρ
Ms
= ?n · M = ?r · M
0
?z = M
0
cosθ. (3.197)
The vector potential is produced by the equivalent magnetization surface current.
Using (3.137)we ?nd that
A(r) =
μ
0
4π
integraldisplay
S
J
Ms
|r ? r
prime
|
dS
prime
=
μ
0
4π
integraldisplay
π
?π
integraldisplay
π
0
M
0
?
φ
prime
sinθ
prime
|r ? r
prime
|
sinθ
prime
dθ
prime
dφ
prime
.
Since
?
φ
prime
=??x sinφ
prime
+ ?y cosφ
prime
, the rectangular components of A are
braceleftbigg
?A
x
A
y
bracerightbigg
=
μ
0
4π
integraldisplay
π
?π
integraldisplay
π
0
M
0
sinφ
prime
cosφ
prime
sinθ
prime
|r ? r
prime
|
a
2
sinθ
prime
dθ
prime
dφ
prime
. (3.198)
The integrals are most easily computed via the spherical harmonic expansion (E.200)for
the inverse distance |r ? r
prime
|
?1
:
braceleftbigg
?A
x
A
y
bracerightbigg
= μ
0
M
0
a
2
∞
summationdisplay
n=0
n
summationdisplay
m=?n
Y
nm
(θ,φ)
2n + 1
r
n
<
r
n+1
>
integraldisplay
π
?π
integraldisplay
π
0
sinφ
prime
cosφ
prime
sin
2
θ
prime
Y
?
nm
(θ
prime
,φ
prime
)dθ
prime
dφ
prime
.
Because the φ
prime
variation is sinφ
prime
or cosφ
prime
, all terms in the sum vanish except n = 1,
m =±1. Since
Y
1,?1
(θ,φ) =
radicalbigg
3
8π
sinθe
?jφ
, Y
1,1
(θ,φ) =?
radicalbigg
3
8π
sinθe
jφ
,
we have
braceleftbigg
?A
x
A
y
bracerightbigg
= μ
0
M
0
a
2
3
r
<
r
2
>
3
8π
sinθ
integraldisplay
π
0
sin
3
θ
prime
dθ
prime
·
·
bracketleftbigg
e
?jφ
integraldisplay
π
?π
sinφ
prime
cosφ
prime
e
jφ
prime
dφ
prime
+ e
jφ
integraldisplay
π
?π
sinφ
prime
cosφ
prime
e
?jφ
prime
dφ
prime
bracketrightbigg
.
Carrying out the integrals we ?nd that
braceleftbigg
?A
x
A
y
bracerightbigg
= μ
0
M
0
a
2
3
r
<
r
2
>
sinθ
braceleftbigg
sinφ
cosφ
bracerightbigg
or
A = μ
0
M
0
a
2
3
r
<
r
2
>
sinθ
?
φ.
Finally, B =?×A gives
B =
braceleftBigg
2μ
0
M
0
3
?z, r < a,
μ
0
M
0
a
3
3r
3
parenleftBig
?r 2 cosθ +
?
θ sinθ
parenrightBig
, r > a.
(3.199)
Hence B within the sphere is uniform and in the same direction as M, while B outside
the sphere has the form of the magnetic dipole ?eld with moment
m =
parenleftbigg
4
3
πa
3
parenrightbigg
M
0
.
We can also compute B by ?rst ?nding the scalar potential through direct computation
of the integral (3.126). Substituting for ρ
Ms
from (3.197), we have
Phi1
m
(r) =
1
4π
integraldisplay
S
ρ
Ms
(r
prime
)
|r ? r
prime
|
dS
prime
=
1
4π
integraldisplay
π
?π
integraldisplay
π
0
M
0
cosθ
prime
|r ? r
prime
|
sinθ
prime
dθ
prime
dφ
prime
.
This integral has the form of (3.100)with f (θ) = M
0
cosθ. Thus, from (3.102),
Phi1
m
(r) = M
0
a
2
3
cosθ
r
<
r
2
>
. (3.200)
The magnetic ?eld H is then
H =??Phi1
m
=
braceleftBigg
?
M
0
3
?z, r < a,
M
0
a
3
3r
3
parenleftBig
?r 2 cosθ +
?
θ sinθ
parenrightBig
, r > a.
.
Inside the sphere B is given by B = μ
0
(H + M), while outside the sphere it is merely
B = μ
0
H. These observations lead us again to (3.199).
Since the scalar potential obeys Laplace’s equation both inside and outside the sphere,
as a last approach to the problem we shall write Phi1
m
in terms of the separation of variables
solution discussed in § A.4. We can repeat our earlier arguments for the dielectric sphere
in an impressed electric ?eld (§ 3.2.10). Copying equations (3.109) and (3.110), we can
write for r ≤ a
Phi1
m1
(r,θ)=
∞
summationdisplay
n=0
A
n
r
n
P
n
(cosθ), (3.201)
and for r ≥ a
Phi1
m2
(r,θ)=
∞
summationdisplay
n=0
B
n
r
?(n+1)
P
n
(cosθ). (3.202)
The boundary condition (3.154)at r = a requires that
∞
summationdisplay
n=0
A
n
a
n
P
n
(cosθ)=
∞
summationdisplay
n=0
B
n
a
?(n+1)
P
n
(cosθ);
upon application of the orthogonality of the Legendre functions, this becomes
A
n
a
n
= B
n
a
?(n+1)
. (3.203)
We can write (3.155)as
?
?Phi1
m1
?r
+
?Phi1
m2
?r
=?ρ
Ms
so that at r = a
?
∞
summationdisplay
n=0
A
n
na
n?1
P
n
(cosθ)?
∞
summationdisplay
n=0
B
n
(n + 1)a
?(n+2)
P
n
(cosθ)=?M
0
cosθ.
After application of orthogonality this becomes
A
1
+ 2B
1
a
?3
= M
0
, (3.204)
na
n?1
A
n
=?(n + 1)B
n
a
?(n+2)
, n negationslash= 1. (3.205)
Solving (3.203)and (3.204)simultaneously for n = 1 we ?nd that
A
1
=
M
0
3
, B
1
=
M
0
3
a
3
.
We also see that (3.203)and (3.205)are inconsistent unless A
n
= B
n
= 0, n negationslash= 1.
Substituting these results into (3.201)and (3.202), we have
Phi1
m
=
braceleftBigg
M
0
3
r cosθ, r ≤ a,
M
0
3
a
3
r
2
cosθ, r ≥ a,
which is (3.200).
3.3.8 Bodiesimmersedinanimpressedmagnetic?eld: magnetostatic
shielding
A highly permeable enclosure can provide partial shielding from external magnetostatic
?elds.Considerasphericalshellofhighlypermeablematerial(Figure3.22);assumeit
is immersed in a uniform impressed ?eld H
0
= H
0
?z. We wish to determine the internal
?eld and the factor by which it is reduced from the external applied ?eld. Because there
are no sources (the applied ?eld is assumed to be created by sources far removed), we
may use magnetic scalar potentials to represent the ?elds everywhere. We may represent
the scalar potentials using a separation of variables solution to Laplace’s equation, with
a contribution only from the n = 1 term in the series. In region 1 we have both scattered
Figure 3.22: Spherical shell of magnetic material.
and applied potentials, where the applied potential is just Phi1
0
=?H
0
z =?H
0
r cosθ,
since H
0
=??Phi1
0
= H
0
?z. We have
Phi1
1
(r) = A
1
r
?2
cosθ ? H
0
r cosθ, (3.206)
Phi1
2
(r) = (B
1
r
?2
+ C
1
r)cosθ, (3.207)
Phi1
3
(r) = D
1
r cosθ. (3.208)
We choose (3.109)for the scattered potential in region 1 so that it decays as r →∞,
and (3.110)for the scattered potential in region 3 so that it remains ?nite at r = 0.In
region 2 we have no restrictions and therefore include both contributions. The coe?cients
A
1
, B
1
,C
1
, D
1
are found by applying the appropriate boundary conditions at r = a and
r = b. By continuity of the scalar potential across each boundary we have
A
1
b
?2
? H
0
b = B
1
b
?2
+ C
1
b,
B
1
a
?2
+ C
1
a = D
1
a.
By (3.156), the quantity μ?Phi1/?r is also continuous at r = a and r = b; this gives two
more equations:
μ
0
(?2A
1
b
?3
? H
0
) = μ(?2B
1
b
?3
+ C
1
),
μ(?2B
1
a
?3
+ C
1
) = μ
0
D
1
.
Simultaneous solution yields
D
1
=?
9μ
r
K
H
0
where
K = (2 +μ
r
)(1 + 2μ
r
)? 2(a/b)
3
(μ
r
? 1)
2
.
Substituting this into (3.208)and using H =??Phi1
m
, we ?nd that
H = κH
0
?z
within the enclosure, where κ = 9μ
r
/K. This ?eld is uniform and, since κ<1 for μ
r
> 1,
it is weaker than the applied ?eld. For μ
r
greatermuch 1 we have K ≈ 2μ
2
r
[1 ?(a/b)
3
]. Denoting
the shell thickness by Delta1 = b ? a, we ?nd that K ≈ 6μ
2
r
Delta1/a when Delta1/a lessmuch 1.Thus
κ =
3
2
1
μ
r
Delta1
a
describes the coe?cient of shielding for a highly permeable spherical enclosure, valid
when μ
r
greatermuch 1 and Delta1/a lessmuch 1. A shell for which μ
r
= 10,000 and a/b = 0.99 can reduce
the enclosure ?eld to 0.15% of the applied ?eld.
3.4 Static ?eld theorems
3.4.1 Mean value theorem of electrostatics
The average value of the electrostatic potential over a sphere is equal to the potential
at the center of the sphere, provided that the sphere encloses no electric charge. To see
this, write
Phi1(r) =
1
4πepsilon1
integraldisplay
V
ρ(r
prime
)
R
dV
prime
+
1
4π
contintegraldisplay
S
bracketleftBigg
?Phi1(r
prime
)
?
R
R
2
+
?
prime
Phi1(r
prime
)
R
bracketrightBigg
· dS
prime
;
put ρ ≡ 0 in V, and use the obvious facts that if S is a sphere centered at point r then
(1) R is constant on S and (2) ?n
prime
=?
?
R:
Phi1(r) =
1
4π R
2
contintegraldisplay
S
Phi1(r
prime
)dS
prime
?
1
4π R
contintegraldisplay
S
E(r
prime
)· dS
prime
.
The last term vanishes by Gauss’s law, giving the desired result.
3.4.2 Earnshaw’s theorem
It is impossible for a charge to rest in stable equilibrium under the in?uence of elec-
trostatic forces alone. This is an easy consequence of the mean value theorem of electro-
statics, which precludes the existence of a point where Phi1 can assume a maximum or a
minimum.
3.4.3 Thomson’s theorem
Static charge on a system of perfect conductors distributes itself so that the electric
storedenergyisaminimum.Figure3.23showsasystemof n conductingbodiesheldat
potentials Phi1
1
,...,Phi1
n
. Suppose the potential ?eld associated with the actual distribution
of charge on these bodies is Phi1, giving
W
e
=
epsilon1
2
integraldisplay
V
E · E dV =
epsilon1
2
integraldisplay
V
?Phi1·?Phi1dV
for the actual stored energy. Now assume a slightly di?erent charge distribution, resulting
in a new potential Phi1
prime
= Phi1+δPhi1 that satis?es the same boundary conditions (i.e., assume
δPhi1 = 0 on each conducting body). The stored energy associated with this hypothetical
situation is
W
prime
e
= W
e
+δW
e
=
epsilon1
2
integraldisplay
V
?(Phi1+δPhi1)·?(Phi1+δPhi1)dV
Figure 3.23: System of conductors used to derive Thomson’s theorem.
so that
δW
e
= epsilon1
integraldisplay
V
?Phi1·?(δPhi1)dV +
epsilon1
2
integraldisplay
V
|?(δPhi1)|
2
dV;
Thomson’s theorem will be proved if we can show that
integraldisplay
V
?Phi1·?(δPhi1)dV = 0, (3.209)
because then we shall have
δW
e
=
epsilon1
2
integraldisplay
V
|?(δPhi1)|
2
dV ≥ 0.
To establish (3.209), we use Green’s ?rst identity
integraldisplay
V
(?u ·?v + u?
2
v)dV =
contintegraldisplay
S
u?v · dS
with u = δPhi1 and v = Phi1:
integraldisplay
V
?Phi1·?(δPhi1)dV =
contintegraldisplay
S
δPhi1?Phi1· dS.
Here S is composed of (1)the exterior surfaces S
k
(k = 1,...,n)of the n bodies, (2)
the surfaces S
c
of the “cuts” that are introduced in order to keep V a simply-connected
region (a condition for the validity of Green’s identity), and (3) the sphere S
∞
of very
large radius r.Thus
integraldisplay
V
?Phi1·?(δPhi1)dV =
n
summationdisplay
k=1
integraldisplay
S
k
δPhi1?Phi1· dS +
integraldisplay
S
c
δPhi1?Phi1· dS +
integraldisplay
S
∞
δPhi1?Phi1· dS.
The ?rst term on the right vanishes because δPhi1 = 0 on each S
k
. The second term
vanishes because the contributions from opposite sides of each cut cancel (note that ?n
occurs in pairs that are oppositely directed). The third term vanishes because Phi1 ～ 1/r,
?Phi1 ～ 1/r
2
, and dS ～ r
2
where r →∞for points on S
∞
.
Figure 3.24: System of conductors used to derive Green’s reciprocation theorem.
3.4.4 Green’s reciprocation theorem
Considerasystemof n conductingbodiesasinFigure3.24.Anassociatedmathemat-
ical surface S
t
consists of the exterior surfaces S
1
,...,S
n
of the n bodies, taken together
with a surface S that enclosed all of the bodies. Suppose Phi1 and Phi1
prime
are electrostatic
potentials produced by two distinct distributions of stationary charge over the set of
conductors. Then ?
2
Phi1 = 0 =?
2
Phi1
prime
and Green’s second identity gives
contintegraldisplay
S
t
parenleftbigg
Phi1
?Phi1
prime
?n
?Phi1
prime
?Phi1
?n
parenrightbigg
dS = 0
or
n
summationdisplay
k=1
integraldisplay
S
k
Phi1
?Phi1
prime
?n
dS+
integraldisplay
S
Phi1
?Phi1
prime
?n
dS =
n
summationdisplay
k=1
integraldisplay
S
k
Phi1
prime
?Phi1
?n
dS+
integraldisplay
S
Phi1
prime
?Phi1
?n
dS.
Now let S be a sphere of very large radius R so that at points on S we have
Phi1,Phi1
prime
～
1
R
,
?Phi1
?n
,
?Phi1
prime
?n
～
1
R
2
, dS ～ R
2
;
as R →∞then,
n
summationdisplay
k=1
integraldisplay
S
k
Phi1
?Phi1
prime
?n
dS =
n
summationdisplay
k=1
integraldisplay
S
k
Phi1
prime
?Phi1
?n
dS.
Furthermore, the conductors are equipotentials so that
n
summationdisplay
k=1
Phi1
k
integraldisplay
S
k
?Phi1
prime
?n
dS =
n
summationdisplay
k=1
Phi1
prime
k
integraldisplay
S
k
?Phi1
?n
dS
and we therefore have
n
summationdisplay
k=1
q
prime
k
Phi1
k
=
n
summationdisplay
k=1
q
k
Phi1
prime
k
(3.210)
where the kth conductor (k = 1,...,n)has potential Phi1
k
when it carries charge q
k
,
and has potential Phi1
prime
k
when it carries charge q
prime
k
. This is Green’s reciprocation theorem.
A classic application is to determine the charge induced on a grounded conductor by
Figure 3.25: Application of Green’s reciprocation theorem. (a)The “unprimed situation”
permits us to determine the potential V
P
at point P produced by a charge q placed on
body 1. Here V
1
is the potential of body 1. (b)In the “primed situation” we ground
body 1 and induce a charge q
prime
by bringing a point charge q
prime
P
into proximity.
a nearby point charge. This is accomplished as follows. Let the conducting body of
interest be designated as body 1, and model the nearby point charge q
P
as a very small
conducting body designated as body 2 and located at point P in space. Take
q
1
= q, q
2
= 0,Phi1
1
= V
1
,Phi1
2
= V
P
,
and
q
prime
1
= q
prime
, q
prime
2
= q
prime
P
,Phi1
prime
1
= 0,Phi1
prime
2
= V
prime
P
,
givingthetwosituationsshowninFigure3.25.SubstitutionintoGreen’sreciprocation
theorem
q
prime
1
Phi1
1
+ q
prime
2
Phi1
2
= q
1
Phi1
prime
1
+ q
2
Phi1
prime
2
gives q
prime
V
1
+ q
prime
P
V
P
= 0 so that
q
prime
=?q
prime
P
V
P
/V
1
. (3.211)
3.5 Problems
3.1 The z-axis carries a line charge of nonuniform density ρ
l
(z). Show that the electric
?eld in the plane z = 0 is given by
E(ρ,φ) =
1
4πepsilon1
bracketleftbigg
?ρρ
integraldisplay
∞
?∞
ρ
l
(z
prime
)dz
prime
(ρ
2
+ z
prime
2
)
3/2
? ?z
integraldisplay
∞
?∞
ρ
l
(z
prime
)z
prime
dz
prime
(ρ
2
+ z
prime
2
)
3/2
bracketrightbigg
.
Compute E when ρ
l
= ρ
0
sgn(z), where sgn(z) is the signum function (A.6).
3.2 The ring ρ = a, z = 0, carries a line charge of nonuniform density ρ
l
(φ). Show that
the electric ?eld at an arbitrary point on the z-axis is given by
E(z) =
?a
2
4πepsilon1(a
2
+ z
2
)
3/2
bracketleftbigg
?x
integraldisplay
2π
0
ρ
l
(φ
prime
)cosφ
prime
dφ
prime
+ ?y
integraldisplay
2π
0
ρ
l
(φ
prime
)sinφ
prime
dφ
prime
bracketrightbigg
+
+ ?z
az
4πepsilon1(a
2
+ z
2
)
3/2
integraldisplay
2π
0
ρ
l
(φ
prime
)dφ
prime
.
Figure 3.26: Geometry for computing Green’s function for parallel plates.
Compute E when ρ
l
(φ) = ρ
0
sinφ. Repeat for ρ
l
(φ) = ρ
0
cos
2
φ.
3.3 The plane z = 0 carries a surface charge of nonuniform density ρ
s
(ρ,φ). Show that
at an arbitrary point on the z-axis the rectangular components of E are given by
E
x
(z) =?
1
4πepsilon1
integraldisplay
∞
0
integraldisplay
2π
0
ρ
s
(ρ
prime
,φ
prime
)ρ
prime
2
cosφ
prime
dφ
prime
dρ
prime
(ρ
prime
2
+ z
2
)
3/2
,
E
y
(z) =?
1
4πepsilon1
integraldisplay
∞
0
integraldisplay
2π
0
ρ
s
(ρ
prime
,φ
prime
)ρ
prime
2
sinφ
prime
dφ
prime
dρ
prime
(ρ
prime
2
+ z
2
)
3/2
,
E
z
(z) =
z
4πepsilon1
integraldisplay
∞
0
integraldisplay
2π
0
ρ
s
(ρ
prime
,φ
prime
)ρ
prime
dφ
prime
dρ
prime
(ρ
prime
2
+ z
2
)
3/2
.
Compute E when ρ
s
(ρ,φ) = ρ
0
U(ρ ? a) where U(ρ) is the unit step function (A.5).
Repeat for ρ
s
(ρ,φ) = ρ
0
[1 ? U(ρ ? a)].
3.4 The sphere r = a carries a surface charge of nonuniform density ρ
s
(θ). Show that
the electric intensity at an arbitrary point on the z-axis is given by
E(z) = ?z
a
2
2epsilon1
integraldisplay
π
0
ρ
s
(θ
prime
)(z ? a cosθ
prime
)sinθ
prime
dθ
prime
(a
2
+ z
2
? 2az cosθ
prime
)
3/2
.
Compute E(z) when ρ
s
(θ) = ρ
0
, a constant. Repeat for ρ
s
(θ) = ρ
0
cos
2
θ.
3.5 Beginning with the postulates for the electrostatic ?eld
?×E = 0, ?·D = ρ,
use the technique of § 2.8.2 to derive the boundary conditions (3.32)–(3.33).
3.6 A material half space of permittivity epsilon1
1
occupies the region z > 0, while a second
material half space of permittivity epsilon1
2
occupies z < 0. Find the polarization surface charge
densities and compute the total induced polarization charge for a point charge Q located
at z = h.
3.7 Consider a point charge between two grounded conducting plates as shown in
Figure3.26.WritetheGreen’sfunctionasthesumofprimaryandsecondarytermsand
apply the boundary conditions to show that the secondary Green’s function is
G
s
(r|r
prime
) =
1
(2π)
2
integraldisplay
∞
?∞
integraldisplay
∞
?∞
bracketleftbigg
?e
?k
ρ
(d?z)
sinh k
ρ
z
prime
sinh k
ρ
d
? e
?k
ρ
z
sinh k
ρ
(d ? z
prime
)
sinh k
ρ
d
bracketrightbigg
e
?jk
ρ
·r
prime
2k
ρ
d
2
k
ρ
.
(3.212)
3.8 Use the expansion
1
sinh k
ρ
d
= csch k
ρ
d = 2
∞
summationdisplay
n=0
e
?(2n+1)k
ρ
d
to show that the secondary Green’s function for parallel conducting plates (3.212)may
be written as an in?nite sequence of images of the primary point charge. Identify the
geometrical meaning of each image term.
3.9 Find the Green’s functions for a dielectric slab of thickness d placed over a perfectly
conducting ground plane located at z = 0.
3.10 Find the Green’s functions for a dielectric slab of thickness 2d immersed in free
space and centered on the z = 0 plane. Compare to the Green’s function found in
Problem 3.9.
3.11ReferringtothesystemofFigure3.9,?ndthechargedensityonthesurfaceof
the sphere and integrate to show that the total charge is equal to the image charge.
3.12 Use the method of Green’s functions to ?nd the potential inside a conducting
sphere for ρ inside the sphere.
3.13 Solve for the total potential and electric ?eld of a grounded conducting sphere
centered at the origin within a uniform impressed electric ?eld E = E
0
?z. Find total
charge induced on the sphere.
3.14 Consider a spherical cavity of radius a centered at the origin within a homogeneous
dielectric material of permittivity epsilon1 = epsilon1
0
epsilon1
r
. Solve for total potential and electric ?eld
inside the cavity in the presence of an impressed ?eld E = E
0
?z. Show that the ?eld in
the cavity is stronger than the applied ?eld, and explain this using polarization surface
charge.
3.15 Find the ?eld of a point charge Q located at z = d above a perfectly conducting
ground plane at z = 0. Use the boundary condition to ?nd the charge density on the
plane and integrate to show that the total charge is ?Q. Integrate Maxwell’s stress
tensor over the surface of the ground plane and show that the force on the ground plane
is the same as the force on the image charge found from Coulomb’s law.
3.16 Consider in free space a point charge ?q at r = r
0
+ d, a point charge ?q at
r = r
0
? d, and a point charge 2q at r
0
. Find the ?rst three multipole moments and the
resulting potential produced by this charge distribution.
3.17 A spherical charge distribution of radius a in free space has the density
ρ(r) =
Q
πa
3
cos 2θ.
Compute the multipole moments for the charge distribution and ?nd the resulting poten-
tial. Find a suitable arrangement of point charges that will produce the same potential
?eld for r > a as produced by the spherical charge.
3.18Computethemagnetic?uxdensity B forthecircularwireloopofFigure3.18by
(a) using the Biot–Savart law (3.165), and (b) computing the curl of (3.138).
Figure 3.27: Parallel plate capacitor.
3.19 Two circular current-carrying wires are arranged coaxially along the z-axis. Loop
1 has radius a
1
, carries current I
1
, and is centered in the z = 0 plane. Loop 2 has radius
a
2
, carries current I
2
, and is centered in the z = d plane. Find the force between the
loops.
3.20 Choose Q =?
prime
parenleftbig
1
R
parenrightbig
× c in (3.162)and derive the following expression for B:
B(r) =
μ
4π
integraldisplay
V
J(r
prime
)×?
prime
parenleftbigg
1
R
parenrightbigg
dV
prime
?
?
1
4π
contintegraldisplay
S
bracketleftbigg
[?n
prime
× B(r
prime
)] ×?
prime
parenleftbigg
1
R
parenrightbigg
+ [?n
prime
· B(r
prime
)]?
prime
parenleftbigg
1
R
parenrightbiggbracketrightbigg
dS
prime
,
where ?n is the normal vector outward from V. Compare to the Stratton–Chu formula
(6.8).
3.21 Compute the curl of (3.163)to obtain the integral expression for B given in Prob-
lem 3.20. Compare to the Stratton–Chu formula (6.8).
3.22 Obtain (3.170)by integration of Maxwell’s stress tensor over the xz-plane.
3.23 Consider two thin conducting parallel plates embedded in a region of permittivity
epsilon1(Figure3.27).Thebottomplateisconnectedtoground,andweapplyanexcesscharge
+Q to the top plate (and thus ?Q is drawn onto the bottom plate.)Neglecting fringing,
(a)solve Laplace’s equation to show that
Phi1(z) =
Q
Aepsilon1
z.
Use (3.87)to show that
W =
Q
2
d
2Aepsilon1
.
(b)Verify W using (3.88). (c) Use F =??zdW/dz to show that the force on the top plate
is
F =??z
Q
2
2Aepsilon1
.
(d)Verify F by integrating Maxwell’s stress tensor over a closed surface surrounding the
top plate.
3.24 Consider two thin conducting parallel plates embedded in a region of permittivity
epsilon1(Figure3.27).Thebottomplateisconnectedtoground,andweapplyapotential V
0
to
the top plate using a battery. Neglecting fringing, (a)solve Laplace’s equation to show
that
Phi1(z) =
V
0
d
z.
Use (3.87)to show that
W =
V
2
0
Aepsilon1
2d
.
(b)Verify W using (3.88). (c) Use F =??zdW/dz to show that the force on the top plate
is
F =??z
V
2
0
Aepsilon1
2d
2
.
(d)Verify F by integrating Maxwell’s stress tensor over a closed surface surrounding the
top plate.
3.25 A group of N perfectly conducting bodies is arranged in free space. Body n is
held at potential V
n
with respect to ground, and charge Q
n
is induced upon its surface.
By linearity we may write
Q
m
=
N
summationdisplay
n=1
c
mn
V
n
where the c
mn
are called the capacitance coe?cients. Using Green’s reciprocation the-
orem, demonstrate that c
mn
= c
nm
. Hint: Use (3.210). Choose one set of voltages so
that V
k
= 0, k negationslash= n, and place V
n
at some potential, say V
n
= V
0
, producing the set of
charges {Q
k
}. For the second set choose V
prime
k
= 0, k negationslash= m, and V
m
= V
0
, producing {Q
prime
k
}.
3.26 For the set of conductors of Problem 3.25, show that we may write
Q
m
= C
mm
V
m
+
summationdisplay
knegationslash=m
C
mk
(V
m
? V
k
)
where
C
mn
=?c
mn
, m negationslash= n, C
mm
=
N
summationdisplay
k=1
c
mk
.
Here C
mm
, called the self capacitance, describes the interaction between the mth con-
ductor and ground, while C
mn
, called the mutual capacitance, describes the interaction
between the mth and nth conductors.
3.27 For the set of conductors of Problem 3.25, show that the stored electric energy is
given by
W =
1
2
N
summationdisplay
m=1
N
summationdisplay
n=1
c
mn
V
n
V
m
.
3.28Agroupof N wiresisarrangedinfreespaceasshowninFigure3.28.Wire n
carries a steady current I
n
, and a ?ux Psi1
n
passes through the surface de?ned by its
contour Gamma1
n
. By linearity we may write
Psi1
m
=
N
summationdisplay
n=1
L
mn
I
n
Figure 3.28: A system of current-carrying wires.
where the L
mn
are called the coe?cients of inductance. Derive Neumann’s formula
L
mn
=
μ
0
4π
contintegraldisplay
Gamma1
n
contintegraldisplay
Gamma1
m
dl · dl
prime
|r ? r
prime
|
,
and thereby demonstrate the reciprocity relation L
mn
= L
nm
.
3.29ForthegroupofwiresshowninFigure3.28,showthatthestoredmagneticenergy
is given by
W =
1
2
N
summationdisplay
m=1
N
summationdisplay
n=1
L
mn
I
n
I
m
.
3.30 Prove the minimum heat generation theorem: steady electric currents distribute
themselves in a conductor in such a way that the dissipated power is a minimum. Hint:
Let J be the actual distribution of current in a conducting body, and let the power it
dissipates be P. Let J
prime
= J +δJ be any other current distribution, and let the power it
dissipates be P
prime
= P +δP. Show that
δP =
1
2
integraldisplay
V
1
σ
|δJ|
2
dV ≥ 0.
| 课件名称： | 《电磁学》电子书（英文版） |
| 课件分类： | 物理 |
| 课件类型： | 电子图书 |
| 文件大小： | 4.09MB |
| 下载次数： | 5 |
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- 1. 《电磁学》电子书（英文版）：Appendix-A
- 2. 《电磁学》电子书（英文版）：Appendix-B
- 3. 《电磁学》电子书（英文版）：Appendix-C
- 4. 《电磁学》电子书（英文版）：Appendix-D
- 5. 《电磁学》电子书（英文版）：Appendix-E
- 6. 《电磁学》电子书（英文版）：chap01
- 7. 《电磁学》电子书（英文版）：chap02
- 8. 《电磁学》电子书（英文版）：chap03
- 9. 《电磁学》电子书（英文版）：chap04
- 10. 《电磁学》电子书（英文版）：chap05
- 11. 《电磁学》电子书（英文版）：chap06
- 12. 《电磁学》电子书（英文版）：reference
- 13. 《电磁学》电子书（英文版）：toc
