Unit 1 Vector Calculus and Applications It is well known that accelerating electric charges produce time-varying fields which have an electric and magnetic component – thus the name electromagnetic fields. On the other hand, stationary charges produce static electric fields, while moving, but non-accelerating, charges give rise to static magnetic fields. Vector calculus is an extremely useful tool in characterizing and applying the properties of these fields. In as far as is feasible, in this section and others, we shall seek to illustrate the rudiments of vector calculus as they may be applied in the study of engineering electromagnetics. That is, rather than treating the subject from a purely mathematical viewpoint, we wish to investigate some of the more prominent features of the vector calculus in the context of the development of the fundamentals of electromagnetics. Some of the mathematics, in its most general application, proves to be formidable, but fortunately many of the important features may be developed in the study of basic electromagnetics problems. 1.1 Review - A Few Comments on the Properties of Vectors The following material has been encountered much earlier in the programme and, with the exception of the first one or two tutorials, will not receive any detailed treatment here: 1
Transcript
Unit 1
Vector Calculus and Applications
It is well known that accelerating electric charges produce time-varying fields which
have an electric and magnetic component – thus the name electromagnetic fields. On
the other hand, stationary charges produce static electric fields, while moving, but
non-accelerating, charges give rise to static magnetic fields. Vector calculus is an
extremely useful tool in characterizing and applying the properties of these fields. In
as far as is feasible, in this section and others, we shall seek to illustrate the rudiments
of vector calculus as they may be applied in the study of engineering electromagnetics.
That is, rather than treating the subject from a purely mathematical viewpoint,
we wish to investigate some of the more prominent features of the vector calculus
in the context of the development of the fundamentals of electromagnetics. Some
of the mathematics, in its most general application, proves to be formidable, but
fortunately many of the important features may be developed in the study of basic
electromagnetics problems.
1.1 Review - A Few Comments on the Properties
of Vectors
The following material has been encountered much earlier in the programme and, with
the exception of the first one or two tutorials, will not receive any detailed treatment
here:
1
1. Definition of a scalar: a quantity possessing magnitude only (eg., potential
difference);
2. Definition of a vector: a quantity possessing magnitude and direction (eg., elec-
tric field intensity); in these notes, non-time-varying vectors are indicate by an
arrow (eg., ~E) and time-varying vectors will be indicated by a tilde underscore
such as ~E˜. Ordinary phasors will be represented by a tilde overbar such as in
V and time-harmonic vector quantities will be represented in the same manner
(eg., E). The text uses boldface letters for vectors, but this is not convenient
for note-taking.
3. Definition of a unit vector: a vector whose magnitude is unity. For example,
r =~r
|~r|
is a unit vector in the direction of a vector, ~r, whose magnitude is |~r|.
4. We will also consider the following operations:
(a) addition (or subtraction) of vectors
(b) the scalar (or dot) product
(c) the vector (or cross) product
(d) triple products
A little consideration will be given to the specification of vectors in each of the
three most commonly used coordinate systems: Cartesian (or rectangular), cylindrical
and spherical.
1.1.1 Vectors and Vector Space – Discussion Based on theCartesian Coordinate System
Consider the right-handed rectangular or Cartesian coordinate system. A vector ~r
may be specified in this system by its components along the respective axes as shown:
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The x, y and z axes are mutually perpendicular. The position vector ~r is from the
origin, O, to some arbitrary point P (x, y, z) and is given by
~r = ~x + ~y + ~z .
Thus ~r may be given as an ordered triple (x, y, z). A collection of all such vectors
which has the following five properties constitute a vector space: Let ~r = (x, y, z),
~r1 = (x1, y1, z1), and ~r2 = (x2, y2, z2).
1. Vector equality: ~r1 = ~r2 means x1 = x2, y1 = y2, and z1 = z2.
2. Vector addition: ~r = ~r1 + ~r2 means x = x1 + x2, and so on.
3. Scalar multiplication: For a real number a, a~r = (ax, ay, az) and vice versa.
4. Negative of a vector: −~r = (−x,−y,−z) and vice versa.
5. Null vector: A vector ~0 implies x = 0, y = 0, and z = 0 and vice versa.
Since the vector components are real numbers, the following properties also hold:
Of course, there is nothing special about the notation. For any two vectors, ~A =
Axx + Ayy + Azz and ~B = Bxx + Byy + Bz z,
~A × ~B =
∣
∣
∣
∣
∣
∣
∣
x y zAx Ay Az
Bx By Bz
∣
∣
∣
∣
∣
∣
∣
~A and ~B may be any kind of vector quantities (eg., force ~F , electric field intensity,
~E, and so on).
Scalar and Vector Fields
A field may be defined as some function of a vector which is specified by a connection
from an arbitrary origin to a general point in space. (i.e., the field is a function of a
position vector, ~r).
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A vector field is a vector function – i.e., it has both magnitude and direction,
which will, in general, vary throughout the region.
A scalar field is a scalar function – i.e., it is a function of ~r but it has only
magnitude.
It is important in this course to realize that the value of a field may vary, in
general, with time as well as position.
Examples: (1) Electric Field Intensity ( ~E(~r))
The electric field intensity, ~E(~r), of a positive point charge, q, (at (0, 0, 0), say) is a
vector field. Recall that ~E(~r) is defined as the force per unit charge experienced by a
small positive test charge brought to position ~r.
Illustration:
~E(~r) =1
4πε0
q
|~r|2 r
The parameter, ε0, measured in farads per metre is the permittivity of free space, and
it has a value of 8.854 × 10−12 ≈ 10−9
36πF/m. The ~E field is radially away from the
positive point charge and the field unit is the volt per metre (V/m).
(2) Electric Potential V (~r)
We shall see later that the electric potential, V , which is defined as the work done
per unit charge in bringing that charge to a position, ~r, from some arbitraily chosen
zero reference, is strictly a scalar field. For example, for the above illustrated charge
with the zero reference at infinity, it will be observed that the potential, measured in
volts, is given by the scalar equation
V (~r) =1
4πε0
q
|~r| .
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Complex Representation and Time-Averaging of Time-Harmonic Scalarsand Vectors
1. Scalars
A time-harmonic scalar quantity, V (t), with amplitude V0, angular frequency ω = 2πf
(f ≡ frequency in hertz (Hz)), and phase φ may be written
V (t) = V0 cos(ωt + φ) .
Recalling that ejθ = cos θ + j sin θ – i.e. Euler’s identity–, V (t) may clearly be cast as
V (t) = Re{
V ejωt}
with
V = V0ejφ
which is called a phasor. Note that the phasor is not a function of time. Thus, we
may say
V (t) ↔ V
where ↔ may be read as “is equivalent in phasor form to” – in other words ↔ indicates
a kind of transformation.
Addition
It may be readily verified that if U(t) = Re{
Uejωt}
with U = U0ejφ, then
V (t) + U(t) ↔ V + U .
The same is NOT true of multiplication. (Check it out).
Time Derivative
Noting that∂V (t)
∂t= −ωV0 sin(ωt + φ) = Re
{
jωV0ejφejωt
}
,
∂V (t)
∂t↔ jωV .
That is, jω can replace the time derivative in phasor (complex) representation of
time-harmonic scalars, a useful feature when considering the time-varying form of
Maxwell’s equations (see Unit 2).
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2. Vectors
Consider the time-harmonic vector
~V (t) = [Vx cos(ωt + φx)] x + [Vy cos(ωt + φy)] y + [Vz cos(ωt + φz)] z
= Re{[(
Vxejφx
)
x +(
Vyejφy
)
y +(
Vzejφz
)
z]
ejωt}
.
The equivalent “phasor” notation notation (V ) for ~V (t) is
~V (t) ↔ V =(
Vxejφx
)
x +(
Vyejφy
)
y +(
Vzejφz
)
z
so that
~V (t) = Re{
V ejωt}
.
1. Time Averages
In considering such issues as power requirements in electronic communications sys-
tems, computation of time-average values of combinations of time-harmonic quantities
is important. It is trivial to show that the time-averaged value of a time-harmonic
quantity is identically zero. For example, using 〈〉 to indicate time averaging over a
period T (T = 1/f) of such a quantity, it is obvious that
〈V (t)〉 =1
T
∫ T
0
V0 cos(ωt + φ)dt = 0
The same is true for complex time-harmonic vector quantities, ~V (t). The importance
of non-zero time averaging shows up in the averaging of products of time-harmonic
quantities. Consider the following examples:
Example 1. Determine the time average of V 2(t) for V (t) = V0 cos(ωt + φ).
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Example 2. Consider two complex vectors E = ER + jEI and H = HR + jHI (we’ll
meet these later as complex electric and magnetic field intensity, respectively). We
wish to compute the time-averaged cross product 〈 ~E˜× ~H
˜〉. Remember ~E
˜≡ ~E(t)
etc..
It is easy to similarly show that for the dot product,
〈 ~E˜· ~H
˜〉 =
1
2Re
{
E · H∗
}
.
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1.2 Coordinate Systems
The geometry of many electromagnetics problems often dictates which coordinate
system is more appropriate in facilitating a solution. For example, in examining
problems associated with coaxial cables or line sources, (circular) cylindrical coordi-
nates are usually useful. In other instances, certain problem symmetries may suggest
the use of spherical coordinates and in some cases, the ordinary Carstesian or rect-
angular system may be all that is needed. The proper choice of coordinate system is
usually critical in ensuring the “quickest” solution. Of the several such systems, we’ll
consider only those mentioned above.
1.2.1 Cartesian (or Rectangular) Coordinates
We have already assumed knowledge of this system in the previous section. Here, we
will consider specification of differential elements used in various kinds of integrations
involving the (x, y, z) (i.e., Cartesian or rectangular) coordinates.
1. Specification of a Differential Length, d~L:
Consider a curve in space as shown:
x
y
z
dL
r r + dr
curve in space
~r = xx + yy + zz
The quantity, d~L is simply a differential vector of magnitude dL lying along the curve,
while ~r and ~r + d~r are position vectors specifying the beginning and ending of d~L.
Clearly, d~L = d~r. Therefore,
d~L = dxx + dyy + dzz
and the magnitude is given by
dL =√
d~L · d~L =√
dx2 + dy2 + dz2 .
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2. Specification of a Vector Differential Area, d~S:
Suppose dS is a differential element of a surface which encloses a volume. Then,
if n is the outward unit normal (positive) by definition
d~S = ndS Illustration:
If the surface is not closed, the positive unit normal depends on the direction in
which the perimeter is traversed. For a right-handed coordinate system this may be
illustrated as
Now, consider differential areas in each of the three coordinate planes as illustrated
below:
We have
d~Sx = dydz x is a differential surface vector along the x direction.
d~Sy = dxdz y is a differential surface vector along the y direction.
d~Sz = dxdy z is a differential surface vector along the z direction.
3. Specification of a Differential Volume, dV :
Consider the differential volume element, dV , in rectangular coordinates (Note
that the text uses dν):
dV = dxdydz .
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1.2.2 (Circular) Cylindrical Coordinates
We use the term “circular” cylindrical coordinates because, strictly speaking, there are
other kinds of cylindrical coordinate systems such as “elliptic” cylindrical coordinates.
Throughout this course, the term cylindical coordinates will be used in reference to
the former. It is one of the two curvilinear coordinate systems we will encounter, the
other being the spherical coordinate system of the next subsection.
1. Specification of the coordinate system:
Consider the following illustration:
Any point in the above illustrated space may be represented by three curvilinear
coordinates which are here labelled (ρ, φ, z). Just as in the Cartesian system where
the intersection of the planes
x = constant, y = constant, z = constant
describes any point (x, y, z), so the intersection of the surfaces
ρ = constant, φ = constant, z = constant
describes any point (ρ, φ, z). The three coordinate surfaces are
(i.) right circular cylinders having the z-axis as the common axis.
ρ =√
x2 + y2 = constant
Note that ρ is the perpendicular distance from the z-axis while r is the general
distance from the origin.
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(ii.) half planes through the z-axis:
φ = tan−1
(
y
x
)
= constant
(φ is measured counterclockwise (ccw) from the positive x-axis as shown.)
(iii.) planes parallel to the xy-plane as in the Cartesian system:
z = constant
The limits on ρ, φ and z are
0 ≤ ρ < ∞, 0 ≤ φ ≤ 2π, −∞ < z < ∞
Inverting the equations under (i), (ii) and (iii) above, or going directly to the
geometry of the above figure, gives the transformation from cylindrical to Cartesian
cordinates as:
x = ρ cos φy = ρ sin φz = z
2. Unit vectors in Cylindrical Coordinates:
Consider the following illustration:
ρ ≡ unit vector normal to the cylindricalsurface and pointing in the direction ofincreasing ρ.
φ ≡ unit vector (tangential to the cylindricalsurface), perpendicular to the half-planeφ = constant and pointing in thedirection of increasing azimuth angle, φ.
z ≡ unit vector pointing in the direction ofincreasing z.
Note thatρ · ρ = φ · φ = z · z = 1
whileρ × φ = z; φ × z = ρ; z × ρ = φ.
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3. Relationships Between Cartesian and Cylindrical Unit Vectors:
It is obvious that x and y, being perpendicular to z, have no component in the
z direction. However, both x and y may have projections along both the ρ and φ
directions (and vice versa).
Illustration: Because we are considering unit vectors, the following result:
x · ρ = cos φ x · φ = cos(90◦ + φ) = − sin φ
y · ρ = cos(90◦ − φ) = sin φ y · φ = cos(φ)
To summarize in table form:
x y zρ cos φ sin φ 0
φ − sin φ cos φ 0z 0 0 1
On this basis,
x = (x · ρ)ρ + (x · φ)φ = cos φρ − sin φφ
y = sin φρ + cos φφ
z = z
with
cos φ =x√
x2 + y2and sin φ =
y√x2 + y2
.
Similarly, it is clear from the table that
ρ = cos φx + sin φy
φ = − sin φx + cos φy
z = z
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4. Position Vector,~r, in Cylindrical Coordinates:
The above relationships between unit vectors may now be used to established the
form of the position vector, ~r, in cylindrical coordinates. We simply need to determine
the projections of the Cartesian components unto the new unit vectors:
~r =
Of course, this result could have easily established geometrically as depicted below:
It may be noted that any general vector given by ~A = Axx + Ayy + Az z may be
transformed to cylindrical coordinates as A = Aρρ + Aφφ + Az z by determining the
projections of Ax and Ay unto the ρ and φ directions. That is,
5. Vector Differential Length in Cylindrical Coordinates:
Recall, in general, if u is a function of v and w, then
du =∂u
∂vdv +
∂u
∂wdw
It is left as an exercise to verify, using partial derivatives and the relationships between
the unit vectors, that d~r = d~L = dxx + dyy + dzz transforms to
d~r = d~L = dρρ + ρdφφ + dzz
(Remember that both x and y are functions, in general, of ρ and φ.) Also, the above
result may be quickly established from simple geometry. Finally, the magnitude of
the differential length is given by
dL =√
d~L · d~L =√
(dρ)2 + ρ2(dφ)2 + (dz)2
6. Vector Differential Area and Differential Volume in Cylindrical Coordinates:
Consider the following diagram:
We note that
d~Sp = ρdφdzρ (vector differential area in ρ direction)
d~Sφ = dρdzφ (vector differential area in φ direction)
d~Sz = ρdρdφz (vector differential area in z direction)
It is easily deduced from the above diagram that the differential volume element
in cylindrical coordinates is given by
dV = ρdρdφdz
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1.2.3 Spherical Polar Coordinates
1. Specification of the coordinate system:
Consider the following illustration in which a general point, P , in space is located
on a spherical surface centred at (0, 0, 0) of the Cartesian coordinate system.
Point P may be specified by the three spherical polar coordinates (we’ll use, simply,
“spherical coordinates” in our terminology) (r, θ, φ). A point is uniquely specified by
the intersection of the following surfaces:
(i.) Concentric spheres centres at the origin and given by
r =√
x2 + y2 + z2 = constant
(ii.) Right circular cones centred on the z-axis (the polar axis) with vertices at the
origin and which are specified by the equation of their polar angle as
θ = cos−1
(
z√x2 + y2 + z2
)
= constant
(iii.) Half planes through the z-axis given by
φ = tan−1
(
y
x
)
= constant
(φ is the azimuth angle).
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The limits on r, θ and φ are
0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π .
The transformation from spherical to rectangular coordinates may be readily ver-
ified from the geometry as:
x = r sin θ cos φy = r sin θ sin φz = r cos θ
2. Unit vectors in Spherical Coordinates:
Consider the following illustration:
r ≡ unit vector normal to the sphericalsurface and pointing in the direction ofincreasing r.
θ ≡ unit vector tangential to the sphericalsurface, along a line of “longitude” as shownand pointing in the direction of increasing θ.
φ ≡ unit vector tangential to the sphericalsurface, perpendicular to the half-planeφ = constant and pointing in thedirection of increasing azimuth angle, φ.
It must be emphasized that the unit vectors r and θ vary in direction as the
angles θ and φ vary and φ varies with φ.
Note that
r · r = θ · θ = φ · φ = 1
while
r × θ = φ; θ × φ = r; φ × r = θ.
3. Relationships Between Cartesian and Spherical Unit Vectors:
We’ll consider the ideas involved for the specific case of the projection of r onto
x, y, and z and leave the other cases as an exercise.
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Illustration: Projections of r unto the Cartesian unit Vectors:
A consideration of the geometry shows that the projection of r unto x is given as
x · r = sin θ cos φ .
Similarly,
y · r = sin θ sin φ and z · r = cos θ .
We have seen from the cylindrical coordinate discussion that the φ projections unto
the Cartesian unit vectors are given by
x · φ = − sin φ ; y · φ = cos φ ; z · φ = 0
Furthermore, it is not too difficult to set up the geometry which shows that the θ
projections are given by
x · θ = cos θ cos φ
y · θ = cos θ sin φ
z · θ = − sin θ
To summarize in table form:
r θ φx sin θ cos φ cos θ cos φ − sin φy sin θ sin φ cos θ sin φ cos φz cos θ − sin θ 0
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On this basis,
r = (x · r)x + (y · r)y + (z · r)z = sin θ cos φx + sin θ sin φy + cos θz
θ = cos θ cos φx + cos θ sin φy − sin θz
φ = − sin θx + cos θy
∗∗It is left as an exercise to similarly determine x, y, and z in terms of the spherical
polar coordinate unit vectors.
Note that by definition ~r = rr. This may be arrived at algebraically by using ∗∗
and the fact that ~r = xx + yy + zz – i.e. this is a way of verifying the “correctness”
of the ∗∗ exercise.
As another exercise, ∗∗ may be used along with the appropriate partial derivatives
to show that the vector differential length is given by
d~r = drr + rdθθ + r sin θdφφ
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4. Vector Differential Area and Differential Volume in Spherical Coordinates:
Consider the following diagrams:
Differential Area
We note from the geometry that
d~Sr = r2 sin θdθdφr (vector differential area in r direction)
d~Sθ = r sin θdrdφθ (vector differential area in θ direction)
d~Sφ = rdrdθφ (vector differential area in φ direction)
Differential Volume
It is easily deduced from the above diagram that the differential volume element
in cylindrical coordinates is given by
dV = r2 sin θdrdθdφ
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1.3 Line Integrals – with Applications
In electromagnetics, it is common to encounter integrals whose integrands consist of
one or more operations involving vector quantities. We shall meet some of these in
this and subsequent sections of this unit.
For scalar fields, V (~r) say, and vector fields, ~A(~r) say, the following forms of line
integrals may be encountered:
In each case, one attempts to reduce the vector integral to scalar forms. The contour,
C, is the path over which the integral is evaluated. It may be an “open” contour in
which case the starting and end points are different or a “closed” contour which has
the same staring and end points. For closed contours it is common to use the symbol∮
Cor∮
.
Initially, let’s consider the second of the three types. Suppose, for example, that
~A(~r) = ~E(~r) which is the electric field intensity in V/m. Recall that d~r = d~L.
Illustration:
~E · d~L is “the projection of ~E onto d~L” multiplied
by |d~L| – i.e. ~E · d~L = | ~E||d~L| cos θ = ELdL.
Thus,∫
C. . . =
∫ B
A
~E · d~L is simply the sum of all such projections along the curve
AB.
Application 1. Energy, Potential Difference and Potential
Energy
Suppose that it is desired to move a charge, Q, over a displacement, d~L, in a
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region where the electric field intensity is ~E (which, for now, is a non-time-varying
field). The force, ~FE, which the field exerts on the charge is by definition of ~E given
by
~FE = Q~E (1.1)
By definition of work, the differential amount of work, dWE, done by the field is given
by
dWE = ~FE · d~L
and from (1.1),
dWE = Q~E · d~L (1.2)
Now, if an external source is to move Q in the field, the the force, Fext, required is
equal but opposite to FE (i.e. Fext = −~FE). The differential amount of external work,
dW , must therefore be given by
and using (1.2),
dW = −Q~E · d~L (1.3)
From our definition of the line integral, it is clear that the total work done by an
external force in moving a charge Q along a path, C, in a region where ~E exists is
therefore
or
W = −Q∫
C
~E · d~L (1.4)
Remember, equation (1.4) is the WORK DONE BY THE EXTERNAL SOURCE
NOT BY THE FIELD ITSELF – i.e. it is the ENERGY expended by an external
source. This definite integral is basic to the study of field theory. If∫
C
~E · d~L is the
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same for all paths C connecting two specific points (say A and B) in a field, then the
field is said to be conservative.
Illustration:
If the field is non-time-varying (i.e. static), then this property of conservation will
exist. In general, for time-varying fields this property will NOT hold. It should be
intuitive that, if conservation holds, then for any CLOSED path
∮
C
~E · d~L = 0 (1.5)
Recall, Kirchhoff’s voltage law – here we have it without wires!!
Example: Determine the work done in carrying a 4.0-C charge from B(1, 0, 2) to
A(0.8, 0.6, 2) along the shorter arc of the circle given by C: x2 + y2 = 1; z = 2 if
~E = yx + xy + 2z.
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Potential Difference
By definition, the potential difference (VAB) between two points in an ~E-field is
the work per unit charge required to move a charge from point B to point A in that
field. From equation (1.4),
VAB =W
Q= −
∫ A
B
~E · d~L (1.6)
The unit is the joule per coulomb (J/C) or volt (V). The potential difference, VAB,
is positive if work is done (i.e., energy is expended by an external source) in moving
the charge from B to A. Notice that, with this definition for the volt, together with
equation (1.1) the ~E-field unit is indeed
as we have said.
Given equation (1.6), it is clear that the potential difference between points A and
B of the last example is VAB = −1.92 J/4.0 C = −0.48 V.
Potential (or Absolute Potential), V
The potential VA at a point A in an electric field is the work per unit charge
required to move a charge from some arbitrarily specified zero reference to point A.
Two of the most commonly used zero references are
1. the earth’s surface – i.e., “ground”
and
2. infinity.
For example, if infinity is chosen as the zero reference, using the same ideas as led to
equation (1.6), we have the potential at point A as
and that at point B as
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Thus,
VA − VB =
That is
VA − VB = VAB (1.7)
and hence the name “potential difference” for VAB.
Example: For a zero reference chosen at infinity, determine the potential at a distance
rA from the origin due to a point charge q at the origin. Assume free space. Finally,
extend the idea to the case of n separate point charges each of which is at position
~r ′
n – i.e., not necessarily at the origin.
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Application 2. The Biot-Savart Law and Ampere’s Law
The Biot-Savart Law proposed in 1820 by Ampere’s colleagues states that for a
differential length of filament, d~L′, located at a point, P1, and carrying a current I
(dc for now) produces at any field point P2 a (differential) magnetic field intensity
(d ~H) whose magnitude is proportional to the product of the current, the magnitude
of d~L′ and the sine of the angle lying between the filament and a line connecting the
filament to point P2. |d ~H| is inveresely proportional to the square of the distance,
|~R12|, from the position of the filament to P2.
Illustration:
Using SI units and a proportionality constant of 1/4π, this may be formalized as
d ~H =Id~L′ × R12
4πR212
; (The ~H unit is, clearly, A/m.) (1.8)
For the entire current loop, this may be written as
~H =∮ Id~L′ × R
4πR2(1.9)
This is the static ~H field at P2 due to the closed-loop current. R is the unit vector
from any d~L′ toward P2. (Notice that this is the third type of line integral initially
mentioned in this section and that the result is a vector quantity).
Example: Consider a square loop centred on the origin in the x-y plane and carrying
a dc current I as shown. Determine the magnetic field intensity at (0, 0, 0).
28
Ampere’ circuital law states that the line integral of the magnetic field intensity,
~H, around any closed path, C, is exactly equal to the direct current, Ie, enclosed by
the path. Symbolically,∮
C
~H · d~L = Ie (1.10)
where Ie is the current “cutting through the loop”. It is, in fact, possible to derive this
expression from the Biot-Savart law and some math that we have not yet encountered.
(For now, we shall accept the law as being capable of experimental verification). The
orientation of the integration contour, C, follows the right-hand rule with respect
to the direction of the current flow as illustrated below (thumb in direction of the
current and fingers curled in direction in which path is traversed positively).
Illustration 1. Current “Filament”:
Illustration 2. “Thick” Wire Carrying a DC Current (I):
In this illustration, ~J is referred to as the (volume) electric current density and it has
units of A/m2. If the cross section of the wire is A, then for the situation shown,
~J =I
Ar
For contour C1 enclosing area A, Ie = I.
29
For contour C2, Ie = | ~J |A′ =
(
A′
A
)
I.
For some special, yet often important, cases a contour may be chosen such that
(1) ~H is everywhere tangent to the contour
and
(2) | ~H| is constant along the contour.
As we shall see, this greatly simplifies the integration in equation (1.10) and Ampere’s
law may be used to find ~H if Ie is known.
Example: Consider an infinitely long thin wire along the z-axis which is carrying a
dc current, I. Determine the value of ~H at an arbitrary point, P , at a radial distance
ρ from the wire.
The contour C should be chosen with care. Here, let’s choose C parallel to the x-y
plane as shown. We make the following “arguments”:
(1) It can be easily seen from the fact that the wire stretches to infinity in either
direction along the z-axis that ~H has no z dependence.
(2) From the Biot-Savart law, since d~L ′× R (= d~z ′× R) always lies in the φ direction
for any point on the contour, ~H must also lie in the φ direction. Also, ~H will clearly
depend on ρ.
(3) From symmetry, there is clearly no reason why ~H should change in magnitude for
different values of φ.
30
Using these three “observations”, we may thus write
~H = Hφ(ρ)φ.
For the contour chosen, d~L = ρdφφ so that Ampere’s law∮
C
~H · d~L = Ie becomes
~H(ρ) =I
2πρφ (1.11)
Thus, for a dc current as specified (i.e. an “infinite line” current), the static magnetic
field intensity, ~H, is in the φ direction and is inversely proportional to the radial
distance, ρ, from the line. This example may be extended to the important case of
the coaxial line and we shall do this in a tutorial to follow shortly.
1.4 Surface and Volume Integrals – with Applica-
tions
We have already encountered the ideas of a “differential surface area” and “differential
volume” in three coordinate systems. We will now carry out applied integrations
involving these. Initially, because both surface and volume integrals may appear in a
single equation in electromagnetics, we briefly consider both types before proceeding
to applications.
1. Surface Integrals
Analogous to the line integral forms, surface integrals may appear (for scalars,
V (~r), and vectors, ~A(~r)) as
31
where d~S is a local normal to the surface S as discussed earlier. The dot product
is the most commonly encountered form. In fact, this form may be interpreted as a
“flow” or “flux” through the given surface (more later). If the surface is closed, it is
common to use the construct∮
Sto indicate this fact.
Illustration: Determine the surface area of a sphere of radius a.
2. Volume Integrals
Volume integrals tend to be somewhat simpler than surface integrals because the
volume element, dV , is a scalar quantity. For a vector ~A(~r) (in Cartesian coordinates
in this illustration)
so that the vector integral is essentially reduced to a vector sum of scalar integrals.
A similar situation exists for the other coordinate systems. Of course, we may also
encounter the usual scalar volume integrals, say for example,∫
vol
V (~r)dV where V (~r)
is a scalar field.
Example: Suppose a continuous charge distribution lies in a spherical shell defined
by 1 < r < 2. Let the volume charge density, ρv(~r), (i.e. the charge per volume in
32
C/m3) be given by
ρv(~r) =2
r3C/m3 .
Find the total charge Q within the shell.
Application 1. Electric Flux, Flux Density and Gauss’s Law (Electric)
Before using the surface integrals in an electrostatics application, we need to
consider a few related details on electric flux.
In (approximately) 1837, Michael Faraday, being interested in static electric fields
and the effects which various insulating materials (or dielectrics) had on these fields,
devised the following experiment:
Faraday had two concentric spheres constructed in such a way that the outer one
could be dismantled into two hemispheres. With the equipment taken apart, the inner
sphere was given a known positive charge. Then, using about 2 cm of “perfect” (ideal)
dielectric material in the intervening space, the outer shell was clamped around the
inner. Next, the outer shell was discharged by connecting it momentarily to ground.
The outer shell was then carefully separated and the negative charge induced on each
hemisphere was measured.
Faraday found that the magnitude of the charge induced on the outer sphere was
equal to the that of the charge on the inner sphere, irrespective of the dielectric used.
He concluded that there was some kind of “displacement” from the inner to the outer
33
sphere which was independent of the medium. This is now referred to severally as
displacement, displacement flux, or, as we shall use, electric flux. (Of course, the idea
of “electric flux lines” as entities streaming away from electric charge is simply an
invention to aid our conceptualization of the presence of an electric field). This flux,
denoted by Ψ (unfortunately, the text uses F ), is in SI units related to the charge,
Q, producing it via a dimensionless proportionality constant of unity; i.e. the electric
flux in coulombs is given by
Ψ = Q .
The path of the flux lines is radially away from the inner sphere as shown:
An important entity in electromagnetics is the idea of electric flux density, ~D. For
example, in the above illustration, at the surface of the inner sphere,
~D =Ψ
area of surfacer =
Q
4πa2r
while at the outer surface
~D =Q
4πb2r .
Clearly, ~D is measured in coulombs/metre2. If we think of the inner sphere as shrink-
ing to a point charge, Q, then at a distance r from the charge
~D =Q
4πr2r . (1.12)
From this last expression, it may be seen that, for free space,
~D = ε0~E (1.13)
Equation (1.13) is one of the important so-called constitutive relations which are
essential in solving electromagnetics problems.
34
If the charge is distributed within a volume, such that the charge density is ρv,
then the ideas used in developing equation (1.13) lead to a volume integral
~D =∫
V ′
ρvdV ′
4πR2R . (1.14)
Here R is the distance from the differential volume, dV ′, under consideration to the
point of observation, and R is the unit vector in that direction.
If the region of interest is NOT effectively free space, then the permittivity, ε0,
must be replaced with the permittivity, ε, of the region. If the region is not electrically
isotropic – i.e. not “same” electrical properties in all directions – then ε could be a
matrix rather than a single number. In passing, we note that it is common to tabulate
relative permittivities, εR, where
εR =ε
ε0
. (1.15)
[Aside 1. Without discussing the mathematical details, it is important to realize
that when a material (not free space) is placed in an electric field, there is generallly
motion of the bound charge within the material as it “reacts” to the field. A measure
of the tendency of bound charge to do this is called the susceptibility, χe (a dimen-
sionless constant). If the material is isotropic and if this “reaction effect” is linearly
related to the applied field, this effect adds directly to the applied electric field and is
proportional to that field. The overall electric flux density, ~D, is therefore increased
by an amount, ~P , called the polarization. ~D becomes
~D = ε0~E + ~P
= ε0~E + χeε0
~E
= (1 + χe)ε0~E .
On comparing the last expression with equation (1.15) and using the fact that ~D = ε ~E,
we see that
εR = (1 + χe) .
35
It should be pointed out that εR is dependent on frequency (in time-varying fields).
For example, at HF (3–30 MHz), sea water has a relative permittivity of εR ≈ 80
whereas at optical frequencies, the value is about 2. End of Aside 1.]
[Aside 2. The idea of charge density is obviously very important in e-m.
For line charges, Q =∫
L′
ρLdL′ where ρL is the linear charge density in C/m.
For surface charges, Q =∫
S′
ρSdS ′ where ρs is the surface charge density in C/m2.
Volume charge density was encountered in an example given earlier in this section.
End of Aside 2.]
The generalization of Faraday’s experiments led to the following formalization
known as Gauss’s Law:
The electric flux passing through any closed surface is equal to the total
free charge enclosed by that surface.
In general, the closed surface may take any form we wish to visualize – some will
obviously be more convenient to use in calculations than others.
Illustration: (in terms of the vector differential area, d~S).
Since the differential flux, dΨ, crossing the differential area must be the product of
the normal component of ~D and the differential surface d~S:
dΨ = ~D · d~S = | ~D| |d~S| cos θ
Therefore,
Ψ =∮
dΨ =∮
S
~D · d~S = Q Gauss’s Law (1.16)
36
since Ψ = Q where Q is the charge enclosed by S. Thus, too,
∮
S
~D · d~S =∫
V ′
ρvdV ′ = Q (1.17)
Of course, line or surface integrals may replace the volume integral as circumstances
warrant.
Facilitating appplication of Gauss’s law is dependent on a suitable choice of the
closed surface for integration. In fact, if the charge distribution is known, equation
(1.17) can be used to obtain ~D in an easy manner if it is possible to choose a closed
surface, S, which satisfies the following two properties:
(1) ~D is everywhere either normal or tangential to S so that ~D · d~S = DdS or 0,
respectively.
AND
(2) On that portion of S where ~D · d~S 6= 0, the magnitude, | ~D| = D is constant.
Example: An infinite uniform line charge with linear density ρL lies along the z-axis.
Determine ~D and ~E at a distance ρ from the z-axis (see p. 37a).
37
Application 2. Electric Flux Density and Gauss’s Law (Magnetic)
Before using the surface integrals in a magnetostatics application, we need to
consider a few related details on magnetic flux density.
In addition to the constitutive relationship, ~D = ε ~E, relating electric flux density,
~D, to the electric field intensity, ~E, there is another such relationship which relates
the magnetic flux density, ~B, to the magnetic field intensity, ~H. ~B is defined by
~B = µ ~H
where µ in henrys per metre (H/m) is called the permeability of the material and, for
free space, we write µ0 = 4π × 10−7 H/m. ~B is measured in webers/metre2 (Wb/m2
or tesla, T) and since ~H is measured in A/m, the weber is the product of henrys and
amperes.
A discussion of the value of µ for non-free-space parallels that which we briefly
considered for ε in the previous class handout. Again without discussing the theoret-
ical detail, we “observe” that orbiting and spinning electrons within a material may
be thought of as current loops which produce B-fields without external excitation.
When an external field is applied to the material, the interactions between it and the
internal structure may cause the magnetic flux density, ~B, to increase or decrease
from its free-space value. This is reflected in a value of µ which is either greater or
smaller, respectively, than µ0. The B-field is changed from what it would be in free
space according to the expression
~B = µ0( ~H + ~M)
where ~M , called the magnetization or magnetic-moment density, is a measure of
how the material reacts internally in a “magnetic” sense when an external H-field is
applied. A dimensionless constant, χm, called the magnetic susceptibility is defined
such that
~M = χm~H
38
and
~B = µ0(1 + χm) ~H ;
i.e. µ = µ0(1 + χm) and a relative permeability, µR, may be defined as
µR =µ
µ0
= (1 + χm) .
Finally, we note that for magnetically anisotropic materials, the constitutive relation-
ship
~B = µ ~H
will be a matrix equation. (In this course, we will consider only magnetically isotropic
materials so that µ will be an ordinary scalar).
We are now in a position to discuss Gauss’s law as it applies to magnetic fields.
Unlike electric flux lines which terminate on negative charges and begin on positive
charges, no such sources or “sinks” have been discovered for lines of magnetic flux.
– that is, no single magnetic “charges” or “poles” have ever been isolated. For this
reason, Gauss’s law for magnetic flux density ~B, which is analogous to equation (1.16)
or equation (1.17) becomes∮
S
~B · d~S = 0 (1.18)
Equation (1.18) indicates that the magnetic “charge” contained within a closed surface
is zero. If the integral is not closed
∫
S
~B · d~S = Φ (1.19)
where Φ is the magnetic flux, measured in webers, “linking” the surface.
Example: Find the magnetic flux, Φ, in a region of length d between the conductors
of a coaxial cable if each conductor carries a current I. Take the radii of the inner
and outer conductor to be a and b, respectively.
39
Summary
Electric and Magnetic Field Results (Static) From Sections 1-3 and 1-4
From our application of line, surface and volume integrals, we have developed the
following equations which govern static electric and (steady) magnetic fields. These
are the non-time-varying integral forms of the famous, and essential, Maxwell Equa-
tions. We shall shortly find other forms of these equations using the so-called differ-
ential operators.
Conservative ~E field:∮
C
~E · d~L = 0 (1.20)
Ampere’s Law:∮
C
~H · d~L = Ie =∫
S
~J · d~S (1.21)
Gauss’s Law (Electric):∮
S
~D · d~S =∫
v′ρvdV ′ = Q (1.22)
Gauss’s Law (Magnetic):∮
S
~B · d~S = 0 (1.23)
The right hand side of equation (1.21) is a generalization of | ~J|A′ when ~J varies over
the area, S, carrying the current and is not necessarily perpendicular to that area as
illustrated.
In addition to MAxwell’s equations, we have two constitutive relationships:
~D = ε ~E (1.24)
~B = µ ~H (1.25)
40
1.5 Vector Differential Operators
In the study of electromagnetics, as in many other physical sciences, the concept
of vector operators provides a convenient means of writing in concise form what
could otherwise be tedious and lengthy formulations of certain important physical
relationships. In this section, we wish to (1) define, (2) interpret and (3) apply such
operators in a variety of contexts.
1.5.1 The Gradient of a Scalar
Definition: Consider a scalar point function ϕ(x, y, z). The “del” operator, symbolized
~∇, applied to this function is defined in cartesian coordinates as
~∇ϕ = x∂ϕ
∂x+ y
∂ϕ
∂y+ z
∂ϕ
∂z; (1.26)
i.e.
~∇ ≡ x∂
∂x+ y
∂
∂y+ z
∂
∂z. (1.27)
The expression ~∇ϕ is referred to as the gradient of the scalar function ϕ. Thus, the
application of (1.27) to a scalar function results in a vector. At this point, we issue a
caution that the del operator may be treated as a vector in the gradient (and other
forms to follow) ONLY in cartesian coordinates. In the other coordinate systems we
have studied, this idea will require modification.
Illustration: If f(x, y, z) = x2yz, ~∇f = (2xyz)x + (x2z)y + (x2y)z .
Interpretation
Initially, let us consider the dot product
~∇ϕ · d~r
where d~r = dx x + dy y + dz z. Clearly,
~∇ϕ · d~r =∂ϕ
∂xdx +
∂ϕ
∂ydy +
∂ϕ
∂zdz = dϕ (1.28)
41
by definition (i.e. the dot product in (1.28) furnishes the change in the scalar function,
ϕ, corresponding to a change in position d~r).
Now the expression, ϕ(x, y, z) = C where C is a constant describes some (arbi-
trary) surface in space. Consider two points, P and Q, separated by a displacement
d~r on such a surface. Then, moving from P to Q, the change in ϕ(x, y, z) = C is
given, using (1.28), by
dϕ = (~∇ϕ) · d~r = 0 , (1.29)
where the zero results because we stay on the “constant” surface. Equation (1.29)
shows that ~∇ϕ ⊥ d~r. This is true for any direction of d~r from P to Q as long as the
two points are restricted to the surface. Thus, ~∇ϕ is seen to be normal to the surface
ϕ = constant (A VERY IMPORTANT RESULT).
Next, as illustrated, let us permit d~r to extend from one surface ϕ = C1 to an
adjacent surface ϕ = C2 so that invoking equation (1.28)
dϕ = C2 − C1 = ∆C = (~∇ϕ) · d~r . (1.30)
Now, for a given dϕ, |d~r| is a minimum when it is chosen parallel to ~∇ϕ (since then
cos θ = 1); or, for a given |d~r|, the change in the scalar function ϕ is maximized by
choosing d~r parallel to ~∇ϕ. This identifies ~∇ϕ as a vector having the direction of the
maximum space rate of change of ϕ.
42
Application:
We have seen that in a conservative ~E field, the potential V at position P (x, y, z)
is unique – i.e. it is independent of the path of integration. We had that
VP = −∫ P
∞
~E · d~L .
Because VP is a single-valued function, V (x, y, z), we may differentiate the latter
equation to give
dV
dL= −E cos θ
or, equivalently,
dV = − ~E · d~L = − ~E · d~r . (1.31)
Comparing equations (1.30) and (1.31) immediately allows us to write that
~E = −~∇V (1.32)
That is, the magnitude of ~E is given by the maximum space rate of change of V , and,
based on the previous discussion, the direction of ~E is normal to the equipotential
surface (a surface where V is constant) in the direction of decreasing potential (as
indicated by the the minus sign). Equation (1.32) reads “The electric field intensity,
~E, is the negative of the gradient of the potential, V ”. We now have a powerful way
of determining the electric field once the potential field is known.
GRADIENT IN CYLINDRICAL COORDINATES
For a scalar function ϕ(ρ, φ, z) in cylindrical coordinates,
dϕ =∂ϕ
∂ρdρ +
∂ϕ
∂φdφ +
∂ϕ
∂zdz
and d~r = dρρ + ρdφφ + dzz. With a view to equation (1.28), we write
dϕ =
(
∂ϕ
∂ρρ +
1
ρ
∂ϕ
∂φφ +
∂ϕ
∂zz
)
· dρρ + ρdφφ + dzz
=
(
∂ϕ
∂ρρ +
1
ρ
∂ϕ
∂φφ +
∂ϕ
∂zz
)
· d~r .
43
Comparing this last expression with (1.28), we identify the gradient in cylindrical
coordinates as
~∇ϕ =
(
∂ϕ
∂ρρ +
1
ρ
∂ϕ
∂φφ +
∂ϕ
∂zz
)
(1.33)
GRADIENT IN SPHERICAL COORDINATES
A procedure completely analogous to that which led to equation (1.33) may be
used when the scalar function is given in spherical coordinates as ϕ(r, θ, φ). The result
for the gradient is
~∇ϕ =
(
∂ϕ
∂rr +
1
r
∂ϕ
∂θθ +
1
r sin θ
∂ϕ
∂φφ
)
(1.34)
1.5.2 The Divergence of a Vector
The del operator, ~∇, is also used to operate on vector functions, say ~A. We shall first
consider the dot product,~∇ · ~A.
Definition: The divergence of a vector function ~A (i.e. ~A(x, y, z) in cartesian coordi-
nates) is defined as
~∇ · ~A =
(
x∂
∂x+ y
∂
∂y+ z
∂
∂z
)
· (Axx + Ayy + Azz) .
Therefore,
~∇ · ~A =∂Ax
∂x+
∂Ay
∂y+
∂Az
∂z(1.35)
Notice that the divergence of a vector is a scalar! As with the gradient (1.35) will not
have such a simple form in cylindrical and spherical coordinates.
Illustration: ~E = (x2yz)x + (xyz2)y + xz .
The divergence of ~E ( sometimes written, div ~E), is
~∇ · ~E = 2xyz + xz2 .
44
Interpretation
For the sake of simplicity, consider a vector function, ~A, that has only an x
component. From (1.35),
~∇ · ~A =∂Ax
∂x.
With reference to the figure below and noting that the rectangular block is of dif-
ferential volume,(
dV = lim∆x,∆y,∆z→0
∆x∆y∆z = lim∆v→0
∆v)
, the definition of partial
differentiation gives
∂Ax
∂x= lim
∆x→0
Axright face − A
xleft face
∆x
or
∂Ax
∂x= lim
∆v→0
Axright face∆y∆z − A
xleft face∆y∆z
∆v. (1.36)
We next observe that lim∆y,∆z→0
∆y∆z = |d~Sx|.
In (1.36), the first term in the numerator is the flow out while the second is the flow
into the block through dSx. Therefore, (1.36) may be written as
∂Ax
∂x= lim
∆v→0
∫
Sx
~A · d~Sx
∆v(1.37)
while recalling that d~Sx points away from the block at both the right and left faces.
Making identical arguments for ∂Ay
∂yand ∂Az
∂z, (1.37) may be generalized to include all
45
surfaces by writing
∂Ax
∂x+
∂Ay
∂y+
∂Az
∂z= lim
∆v→0
∮
S
~A · d~S
∆v(1.38)
where S is now the whole surface of the block. Thus,
~∇ · ~A = lim∆v→0
∮
S
~A · d~S
∆v(1.39)
and the divergence of a vector field ~A represents “the net outflow of the vector ~A per
unit volume”. For example, ~A could be the electric flux density, ~D, so that ~∇ · ~D
would be the “net flux per unit volume leaving” a particular region. Hence the term
divergence.
~∇ · ~A > 0 if there is a net outflow (i.e. a source region) and
~∇ · ~A < 0 if there is a net inflow (i.e. a sink region)
DIVERGENCE IN CYLINDRICAL COORDINATES
Starting with equation (1.39) and working backwards in cylindrical coordinates, it is
not too difficult to “show” that the expression for divergence in cylindrical coordinates
is:
~∇ · ~A =1
ρ
∂(ρAρ)
∂ρ+
1
ρ
∂Aφ
∂φ+
∂Az
∂z(1.40)
DIVERGENCE IN SPHERICAL COORDINATES
A similar procedure in spherical coordinates leads to
~∇ · ~A =1
r2
∂(r2Ar)
∂r+
1
r sin θ
∂(Aθ sin θ)
∂θ+
1
r sin θ
∂Aφ
∂φ(1.41)
46
APPLICATIONS
(1) Gauss’ Law
We have seen the integral form of Gauss’ Law as
∮
S
~D · d~S =∫
volρvdV = Q (1.42)
where Q is the charged enclosed by the surface, S. Now from equation (1.39)
~∇ · ~D = lim∆v→0
∮
S~D · d~S
∆v= lim
∆v→0
Q
∆v(1.43)
where we have indicated the volume, ∆v, to be shrinking to infinitesimal proportions.
In this case,
lim∆v→0
Q
∆v= ρv
with ρv being the volume charge density. Therefore, equation (1.43) becomes
~∇ · ~D = ρv (1.44)
Equation (1.44) is referred to as the “point form” of Gauss’ Law, so called because
the volume is becoming vanishingly small – i.e. shrinking to a point.
The magnetic Gauss’ Law (see equation (1.23)) in point form clearly becomes
~∇ · ~B = 0 (1.45)
Example: Given ~D = ρz2 sin2 φ ρ+ρz2 sin φ cos φ φ+ρ2z sin2 φ z C/m2, determine the
volume charge density, ρv, in the region where ~D exists.
47
(2.) Gauss’ Theorem or the Divergence Theorem
Now, from equation (1.44), we may carry out the volume integration
∫
vol
~∇ · ~D dV =∫
volρv dV ,
which, on comparison with equation (1.42), gives the DIVERGENCE THEOREM
∮
S
~D · d~S =∫
vol
~∇ · ~D dV (1.46)
Equation (1.46) is true of any vector field, ~A, i.e.∮
S
~A · d~S =∫
vol
~∇ · ~A dV , and we
see that the divergence theorem “changes” a surface integral to a volume integral. In
words, Gauss’ theorem or the divergence theorem says that
“the integral of the normal component of a vector field over a closed surface is
equal to the integral of the divergence of this vector field throughout the volume
enclosed by that surface.”
(3.) Conservation of Charge
The principle of conservation of charge states that charges can neither be created
nor destroyed. Now, considering a region bounded by a closed surface, the current
through that surface is
I =∮
S
~J · d~S .
Recall that ~J is the current density while I is the current flowing outward through
the surface, S. The rate of change of the charge, Q, inside the surface is −dQ
dt; i.e.
I = −dQ
dt
or∮
S
~J · d~S = −dQ
dt, (1.47)
which is the equation of conservation of charge (or continuity of current) in integral
form. It implies that an outward flow of positive charge through the closed surface
48
must be balanced by a decrease of positive charge (or an increase in negative charge)
within the region enclosed by that surface.
Using the divergence theorem and the fact that Q =∫
volρv dV we get, from equa-
tion (1.47),∫
vol
~∇ · ~J dV = − d
dt
∫
volρv dV .
Then, if the surface around the volume is constant,d
dtmay be replaced with a partial
derivative to give∫
vol
~∇ · ~J dV =∫
vol−∂ρv
∂tdV .
Since the equation is true for any volume, a “point” form of the conservation of charge
may be written as
~∇ · ~J = −∂ρv
∂t.
or
~∇ · ~J +∂ρv
∂t= 0 (1.48)
In summary, equation (1.48) implies that “the current, or charge per unit time, di-
verging from a small volume per unit volume is equal to the time rate of decrease of
charge per unit volume at every point”.
(4.) Poisson’s and Laplace’s Equations
Gauss’ law, in point form, says that
~∇ · ~D = ρv
which implies
~∇ · ~E =ρv
ε
since ~D = ε ~E. However, as an example of the gradient of a scalar quantity we saw
that
~E = −~∇V (V ≡ potential) .
49
Therefore,
~∇ · ~∇V = −ρv
ε(1.49)
This is the first important example we have seen which consists of a combination of
del operators. Equation (1.49) is known as Poisson’s equation. We must now interpret
~∇ · ~∇V :
~∇ · ~∇V =
(
x∂
∂x+ y
∂
∂y+ z
∂
∂z
)
·(
x∂V
∂x+ y
∂V
∂y+ z
∂V
∂z
)
or
~∇ · ~∇V =
(
∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2
)
= ~∇2V
where ~∇2V is read “ del-squared V ”. This expression is known as the Laplacian. It
is important to realize that any scalar field may replace V here.
If the volume charge density is zero (i.e. ρv = 0, even though there may still be
point, line, or surface charge at singular locations), equation (1.49) becomes
~∇2V = 0 (1.50)
This is known as Laplace’s equation and equations of this form appear in many fields
other than electromagnetics.
We finally note that the Laplacian of any scalar, V , in cylindrical coordinates is
given by
~∇2V =1
ρ
∂
∂ρ
(
ρ∂V
∂ρ
)
+1
ρ2
(
∂2V
∂φ2
)
+∂2V
∂z2Laplacian (Cylindrical)
In spherical coordinates,
~∇2V =1
r2
∂
∂r
(
r2∂V
∂r
)
+1
r2 sin θ
∂
∂θ
(
sin θ∂V
∂θ
)
+1
r2 sin2 θ
∂2V
∂φ2Laplacian (Spherical)
(5.) Green’s Theorem
As a final example of multiple application of the del operator, we consider an
important corollary to Gauss’ theorem (or the divergence theorem) which is know as
50
Green’s theorem. We state without proof (the proof appears in a tutorial session) that
for two continuous scalar functions, ε and ϕ, whose derivatives are also continuous,
the following identities hold:
~∇ ·(
ε~∇ϕ)
= ε~∇ · ~∇ϕ +(
~∇ε)
·(
~∇ϕ)
A.
and
~∇ ·(
ϕ~∇ε)
= ϕ~∇ · ~∇ε +(
~∇ϕ)
·(
~∇ε)
B.
Subtracting B from A,
~∇ ·(
ε~∇ϕ)
− ~∇ ·(
ϕ~∇ε)
= ε~∇ · ~∇ϕ − ϕ~∇ · ~∇ε
which implies
~∇ ·(
ε~∇ϕ − ϕ~∇ε)
= ε~∇ · ~∇ϕ − ϕ~∇ · ~∇ε .
Integrating over a volume, we have
∫
vol
~∇ ·(
ε~∇ϕ − ϕ~∇ε)
dV =∫
vol
(
ε~∇ · ~∇ϕ − ϕ~∇ · ~∇ε)
dV .
From the divergence theorem (equation (1.46))
∫
vol
(
ε~∇ · ~∇ϕ − ϕ~∇ · ~∇ε)
dV =∮
S
(
ε~∇ϕ − ϕ~∇ε)
· d~S (1.51)
This important result is known as Green’s theorem and it is commonly encountered
in advanced topics in electromagnetics. Another form of this theorem which follows
directly from equation A and the divergence theorem is
∮
S
(
ε~∇ϕ)
· d~S =∫
vol
(
ε~∇ · ~∇ϕ)
dV +∫
vol
(
~∇ε)
·(
~∇ϕ)
dV (1.52)
51
1.5.3 The Curl of a Vector
The del operator, ~∇, may also be “crossed” into a general vector, ~A. The symbolism
is ~∇× ~A and the operation is termed “the curl of ~A”.
Definition: The curl of a vector function ~A (i.e. ~A(x, y, z) in cartesian coordinates) is
defined as
~∇× ~A =
(
x∂
∂x+ y
∂
∂y+ z
∂
∂z
)
× (Axx + Ayy + Az z) .
Therefore, the vector which results is given by
~∇× ~A = x
(
∂Az
∂y− ∂Ay
∂z
)
+ y
(
∂Ax
∂z− ∂Az
∂x
)
+ z
(
∂Ay
∂x− ∂Ax
∂y
)
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
x y z∂
∂x
∂
∂y
∂
∂zAx Ay Az
∣
∣
∣
∣
∣
∣
∣
∣
∣
(1.53)
In expanding this determinant, the derivative nature of ~∇ must be considered. The
determinant in equation (1.53) must be expanded from the top down so that we get
the derivatives shown in the second row. It might be noted that expressions such as
~A× ~∇ are defined only as differential operators and the normal rules which apply to
the cross products of ordinary vectors do NOT apply here. For example, in general,
~A × ~∇ 6= −~∇× ~A.
Notice that the curl of a vector is a vector! As with the previous vector operators
encountered, (1.53) will not have such a simple form in cylindrical and spherical
coordinates.
Illustration: ~r = xx + yy + zz .
The ~∇× ~r, sometimes denoted as curl~r, is
~∇× ~r =
52
Interpretation
Some of the finer mathematical details are not included in the following discussion.
However, the basic ideas required for interpretation of the curl are discussed. For
the sake of simplicity, consider the x component of ~∇ × ~A, whose magnitude from
equation (1.53) is clearly given by
(
~∇× ~A)
· x =∂Az
∂y− ∂Ay
∂z.
With reference to the diagram below, which is a small (i.e. differential) rectangle on
the y-z plane, the definition of partial differentiation leads us to
∂Az
∂y= lim
∆y→0
(Az on b) − (Az on d)
∆y
and
∂Ay
∂z= lim
∆z→0
(Ay on c) − (Ay on a)
∆z
Thus, getting a common denominator gives
(
~∇× ~A)
·x = lim∆y,∆z→0
(Ay on a) ∆y + (Az on b) ∆z + (−Ay on c)∆y − (Az on d)∆z
∆y∆z
Using the limits we may legitimately write, on recalling from earlier discussions the
idea of the differential surface area, that
(
~∇× ~A)
· x = lim∆Sx→0
∮
abcd
~A · d~L
∆Sx
(1.54)
53
From this last expression, we see that the curl of ~A in the x direction may be
gotten by traversing the differential rectangle in the y-z plane as shown, following
the right-hand rule (i.e. fingers curled in direction of travel with thumb, in this case,
pointing in the x direction). Again, it must be emphasized that the rectangle is of
differential proportions and it is assumed that Ay and Az do not vary along a given di-
rection of travel – a more rigorous treatment of these ideas using higher order terms
in a Taylor series expansion would lead properly to the same interpretation. The
above discussion may be summarized as follows:
The “circulation” of the vector field per unit area around a differential area in the
y-z plane is given by the x component of ~∇× ~A. A similar statement could be made
about the other components.
If we allow for all components, then equation (1.54) may be generalized as
(
~∇× ~A)
· n = lim∆S→0
∮
C∆S
~A · d~L
∆S(1.55)
where n is the unit normal to the incremental surface, ∆S, and C∆S is the perimeter
of that surface.
To get an intuitive feeling for the curl, consider the following: If we have a paddle
wheel which we dip into some “force” field, the paddle wheel will spin differently
depending on its orientation in the field. If the wheel does not rotate at all, it means
that the curl is zero in the direction of the wheel’s axis. Larger angular “velocities”
of the wheel indicate larger values of the the curl. In order to find the direction of the
vector curl, we hunt around for the orientation that provides the greatest “torque”.
The direction of the curl is then along the axis of the wheel, as given by the right-hand
rule.
54
CURL IN CYLINDRICAL COORDINATES
We state without proof that the curl in cylindrical coordinates is given by
~∇× ~A =
(
1
ρ
∂Az
∂φ− ∂Aφ
∂z
)
ρ +
(
∂Aρ
∂z− ∂Az
∂ρ
)
φ +
(
1
ρ
∂(ρAφ)
∂ρ− 1
ρ
∂Aρ
∂φ
)
z
CURL IN SPHERICAL COORDINATES
We state without proof that the curl in spherical coordinates is given by
~∇× ~A =1
r sin θ
(
∂(Aφ sin θ)
∂θ− ∂Aθ
∂φ
)
r +1
r
(
1
sin θ
∂Ar
∂φ− ∂(rAφ)
∂r
)
θ +1
r
(
∂(rAθ)
∂r− ∂Ar
∂θ
)
φ
APPLICATIONS
(1.) Stokes’ Theorem
Consider a surface in space which has been subdivided into incremental pieces as
shown.
Rearranging equation (1.55), for a particular surface increment we get
∮
C∆S
~A · d~L =(
~∇× ~A)
·(
n lim∆S→0
∆S)
(1.56)
If the line integral here is carried out for every such region in the whole surface, S,
there will be cancellation along all the cell sides except those which constitute the
outside boundary, C, of S. Thus, equation (1.56) becomes, on integrating over the
whole region,∮
C
~A · d~L =∫
S
(
~∇× ~A)
· d~S (1.57)
Here, we have used the fact that lim∆S→0
∆S = dS and ndS = d~S. The very significant
result in equation (1.57) is known as Stoke’s Theorem. The following numerical
example will illustrate the geometry involved.
55
Example: A portion of a spherical surface is specified by r = 4, 0 ≤ θ ≤ 0.1π, 0 ≤
φ ≤ 0.3π. The path segments forming the perimeter of this surface are given by
