ECE 370 Engineering Electromagnetics
Farez Halim
April 27, 2017
1
Review of vectors and vector analysis
One of the main objectives of this course is to introduce Maxwell’s equations . . .
~ ·E
~ = − ∂ B~
• ∇
∂t
~ ×H
~ = J~ +
• ∇
~
∂D
∂t
~ ·D
~ =ρ
• ∇
~ ·B
~ =0
• ∇
. . . such that we will have a good physical understanding of what these equations mean and how to go about
solving them for simple electromagnetic problems.
We need to know how to:
• add and subtract vectors
• differentiate and integrate vectors
• proper coordinate systems
~ ∇·,
~ ∇×,
~ ∇
~ 2)
• basic operations of vector analysis (i.e. ∇,
1.1
1.1.1
Vectors
Cartesian coordinates
~ can be defined as:
A vector A
~ = Ax~ax + Ay~ay + Az~az
A
~ is
where ~ax , ~ay , and ~az are unit vectors, i.e. of unit length along the x, y, z axes. The magnitude of A
q
~ = A = A2x + A2y + A2z
|A|
1.1.2
Cylindrical coordinates
~ can be defined as:
A vector A
~ = Ar~ar + Aφ~aφ + Az~az
A
where ~ar , ~aφ , and ~az are unit vectors in the
q direction of increasing coordinate value and are always mutually orthog~
~
onal. The magnitude of A is |A| = A = 2 A2 + A2 + A2 .
x
φ
z
1
Figure 1: Cartesian coordinates
Figure 2: Cylindrical coordinates
1.1.3
Spherical coordinates
~ can be defined as:
A vector A
~ = Ar~ar + Aφ~aφ + Aθ~aθ
A
where ~ar , ~aφ , and ~aθ are unit vectors in the
q direction of increasing coordinate value and are always mutually orthog~
~
onal. The magnitude of A is |A| = A = 2 A2x + A2 + A2 . Note that
φ
θ
Figure 3: Spherical coordinates
1. x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ
p
z
2. |r| = 2 x2 + y 2 + z 2 , θ = arccos( √
), and φ = arctan( yz )
2
2
2
2
x +y +z
2
1.2
1.2.1
Vector multiplication
Vector dot and cross product
By definition, the dot product is:
~·B
~ = |A||
~ B|
~ cos θ = (magnitude of A
~ times )(magnitude of B
~ in the direction of A)
~ = Ai Bi + Aj Bj + Ak Bk
A
Figure 4: Vector dot product
And the cross product is defined by
a~i
~×B
~ = Ai
A
Bi
a~j
Aj
Bj
a~k
~ B|
~ sin θa~n
Ak = |A||
Bk
~ and B
~ and in the direction of advance of a right
where a~n is a unit vector normal to the plane containing A
~ to B.
~
hand "screw" turned from A
1.3
Differentiation of a vector
~ can be expressed as:
~ y, z), a small shift dA
For a vector function A(x,
~ = dAx~ax + dAy~ay + dAz~az
dA
1.4
Lines of force
Any vector field can be represented by a system of lines of force. A line of force is drawn tangent to the vector field
at each point in space. The direction of the line of force is in the direction of the vector field, and the number of
lines of force per perpendicular unit of area is proportional to the magnitude of the vector field.
~ → 0, ∆L
~ becomes tangent to the electric field E
~ at that point in space. Therefore,
Referring to Figure 5, as ∆L
~
~
∆L = k E
~ = ∆x~ax + ∆y~ay + ∆z~az = Ex~ax + Ey~ay + Ez~az
∆L
and as ∆x, ∆y, ∆z → 0, it leads to the differential equation that defines a line of force in Cartesian coordinates.
dx
dy
dz
=
=
Ex
Ey
Ez
In cylindrical coordinates,
dr
rdφ
dz
=
=
Er
Eφ
Ez
and in spherical coordinates,
dr
rdθ
r sin θdφ
=
=
Er
Eθ
Eφ
3
Figure 5: Line of force
2
2.1
The concepts of gradient, divergence and curl
The gradient
The gradient of a scalar function is defined as the direction of the maximum change in the vector.
~ = ∂f ~ax + ∂f ~ay + ∂f ~az
∇f
∂x
∂y
∂z
2.2
Divergence of a vector
A vector field can be represented by its flux lines (or lines of force), indicating the magnitude and direction of the
vector field.
If lines of force going in equals lines coming out, net flux is zero. The divergence is the spread of the magnitude
of the field lines and is related to the net flux at that point.
To find the number of field lines of a vector F~ that passes through a surface (i.e. the net flux), refer to Figure
~ = n̂dS. Taking the dot product
7. Note that n̂ is normal unit vector to the surface, and that dS
~ = |A||d
~ S|
~ cos θ
F~ · dS
~ is the number of field lines passing dS.
~ Therefore to get the net flux,
We can say that F~ · dS
I
~ · dS
~ ≡ net flux of the vector
A
S
Therefore, in general for a vector field F~ (x, y, z), we define the net flux as
I
~
ΨA =
F~ · dS
S
Shrink the surface to enclose a very small volume ∆V . . .
H
~
F~ · dS
Ψ
= S
≡ constant value
∆V
∆V
We define this constant as the divergence.
H
~ ~
~ · F~ = lim S F · dS
∇
∆V →0
∆V
In Cartesian coordinates,
~ · F~ (x, y, z) = ( ∂ ~ax + ∂ ~ay + ∂ ~az ) · F~ (x, y, z)
∇
∂x
∂y
∂z
A positive divergence means there is a region in space with a source of flux, while negative divergence means the
opposite.
4
Figure 6: Relationship of divergence and flux
Figure 7: Computing flux
2.3
Curl
The direction assigned to each component of the curl is given by the orientation of the normal surface, and the
sense of circulation
H
F~ · d~l
~
~
∇ × F = lim c
∆S→0
∆S
5
In Cartesian coordinates, the curl can be computed as:
~ × F~ =
∇
~ax
~ay
~az
∂
∂x
∂
∂y
∂
∂z
Fx
Fy
Fz
Note, for a conservative field, the curl is always zero.
3
3.1
Divergence and Stokes’ Theorems
The Divergence Theorem
Consider a volume enclosed by a surface S in a vector field F~ (x, y, z). The total flux Ψ =
internal walls into volumes V1 , V2 . . . VN , having surfaces S1 , S2 . . . SN results in
N I
X
N
X
I
Si
i=1
Which can also be written as
F~ · dS~i =
Vi
S
~ Dividing the
F~ · dS.
~=Ψ
F~ · dS
S
H
Si
F~ · dS~i
Vi
i=1
H
I
~=Ψ
F~ · dS
=
S
As N → ∞, Vi → 0, so
N
X
~ · F~ ) =
V i (∇
~ · F~ )dV =
(∇
V
3.2
~
F~ · dS
S
i=1
Which finally leads to the divergence theorem:
Z
I
I
~
F~ · dS
S
Stokes’ Theorem
H
~ Dividing the
Consider a contour C enclosing a surface S in a vector field F~ (x, y, z). The total flux Ψ = S F~ · dS.
surface into S1 , S2 . . . SN , having contours C1 , C2 . . . CN results in
H
I
N
X
F~ · d~li
Si Ci
F~ · d~l
=
Si
C
i=1
H
PN
~ × F~ ) ~
As N → ∞, Si → 0, so C F~ · d~l = i=1 Si (∇
in Si direction . Which then leads to the Stokes’ Theorem
I
I
~ × F~ ) · dS
~
F~ · d~l = (∇
C
S
6
4
Electrostatics
An electrostatic field is produced by a static charge distribution. We begin by studying the two fundamental laws
for electrostatic fields (Coulomb’s and Gauss’s)
4.1
Coulomb’s Law
Coulomb’s law of force states that
1. Along the line joining them
2. Directly proportional to the product Q1 Q2 of the charges
3. Inversely proportional to the square of the distance R between them.
Expressed mathematically for two charges,
kQ1 Q2
1
F~ =
~ar where k =
2
R
4π0
In general, for a distribution of N charges, the force on one charge Q is defined as
N
Q X Qk (~r − r~k )
F~ =
rπ0
|~r − r~k |3
k=1
~ is the force per unit charge placed in an electric field, E
~ =
The electric field intensity E,
4.2
~
F
Q.
Electric field due to continuous charge distribution
For a continuous charge distribution, the total charge is given by
Z
Q = ρL dl
Z
Q = ρS dS
Z
Q = ρV dV
for a line charge
for a surface charge
for a volume charge
The E field due to a continuous charge distribution is obtained from the formula for point charge by replacing
Q with dQ = ρL dl, dQ = ρS dS or dQ = ρV dV and integrating over the line, surface, or volume respectively.
~ = ρL ~ar and for an infinite sheet of charge, E
~ = ρS a~n
For an infinite line charge, E
2π0 r
20
4.3
Gauss’ Law - Maxwell’s Equation
~
~
~
The electric
R flux density D is related to the electric field intensity as D = E. The electric flux through a surface
~ · dS.
~ Gauss’s law states that the net electric flux penetrating a closed surface is equal to the total
S is Ψ = S D
charge enclosed, hence
I
Z
~
~
Ψ = D · dS = Qenc = ρV dV
or, Maxwell’s first equation
~ ·D
~
ρV = ∇
which states that the volume charge density is the same as the divergence of the electric flux density. This should
"not be surprising to us from the way we defined the divergence of a vector.
7
4.4
Electric Potential
~ is from the electric scalar potential Φ to be defined in this section. In a sense, this way
Another way of obtaining E
~
of finding E is easier because it is easier to handle scalars than vectors.
~ so that the work done in displacing the charge by d~l is
From Coulomb’s law, the force on Q is F~ = QE
~ · d~l
dW = −F~ · d~l = −QE
The negative sign indicates that the work is being done by an external agent. Thus the total work done, or the
potential energy required, in moving Q from A to B is
Z
B
~ · d~l
E
W = −Q
A
Dividing W by Q gives the potential energy per unit charge. This quantity, denoted by ΦAB , is known as the
potential difference between points A and B. Thus
ΦAB
W
=−
=
Q
Z
B
~ · d~l
E
A
The potential at any point is the potential difference between that point and a chosen point where the potential
is zero, usually ∞. Thus,
Z r
~ · d~l
E
Φ=−
∞
The superposition principle, which we applied to electric fields, applies to potentials. Thus for n charges
Q1 , Q2 . . . QN the potential at ~r is
N
1 X Qk
Φ(~r) =
4π0
|~r − r~k |
k=1
Therefore, for continuous charge distributions, we can replace Qk with differential charge elements.
1
Φ(~r) =
4π0
Z
1
Φ(~r) =
4π0
Z
1
Φ(~r) =
4π0
Z
L
S
V
ρL (r~0 )dl0
|~r − r~0 |
for a line charge
ρS (r~0 )dS 0
|~r − r~0 |
for a surface charge
ρV (r~0 )dV 0
|~r − r~0 |
for a volume charge
where the primed coordinates are used customarily to denote source point location and the unprimed coordinates
refer to field point (the point at which V is to be determined).
4.5
Relationship between E and Potential - Maxwell’s equation
RB
~ · d~l. If A and B are close, then dΦ = −E
~ · d~l.
Recall that ΦBA = − A E
∂Φ
∂Φ
∂Φ
~ = ~ax + ~ay + ~az , it leads to
Since ∇Φ
∂x
∂y
∂z
dΦ = (
∂Φ
∂Φ
∂Φ
~ax +
~ay +
~az ) · (dx~ax + dy~ay + dz~az )
∂x
∂y
∂z
~ · d~l
dΦ = ∇Φ
Finally, leads to:
~ = −∇Φ
~
E
8
Figure 8: An electric dipole
4.6
An Electric Dipole and Flux Lines
An electric dipole is formed when two point charges of equal magnitude but opposite sign are separated by a small
distance. Figure 8 illustrates a an electric dipople.
From Figure 8, the potential at point P (r, θ, φ), is given by
Q
1
1
Q
r2 − r1
Φ=
−
=
4π0 r1
r2
4π0
r1 r2
If r >> d, r2 − r1 ≈ d cos θ ≈ r2 , then the equation becomes
Φ=
Q d cos θ
4π0 r2
~ Thus, the expression for potential
Since d cos θ = d~ · ~ar , where d~ = d~az , we can define the dipole moment p~ = Qd.
can be written as
p~ · ~ar
Φ(~r) =
4π0 r2
If the dipole center is not at the origin, but at r~0 , the expression becomes
Φ(~r) =
p~ · (~r − r~0 )
4π0 |~r − r~0 |3
Therefore the electric field can be obtained using Maxwell’s equation
∂Φ
1 ∂Φ
p
~
~
E = −∇Φ = −
~ar +
a~θ =
(2 cos θ~ar + sin θa~θ )
∂r
r ∂θ
4π0 r3
An electric flux line is an imaginary path or line drawn in such a way that its direction at any point is the direction
of the electric field at that point.
Any surface on which the potential is the same throughout is known as an equipotential surface. The intersection
of an equipotential surface and a plane results in a path or line known as an equipotential line. No work is done in
moving a charge from one point to another along an equipotential line or surface, ΦA − ΦB = 0, and hence
Z
~ · ~(dl) = 0
E
on the line or surface. Thus we may conclude that the lines of force or flux lines are always normal to equipotential
surfaces.
9
4.7
Boundary Conditions
So far, we have considered the existence of the electric field in a homogeneous medium. If the field exists in a region
consisting of two different media, the conditions that the field must satisfy at the interface separating the media
are called boundary conditions. These conditions are helpful in determining the field on one side of the boundary if
the field on the other side is known. Obviously, the conditions will be dictated by the types of material the media
are made of. We shall consider the boundary conditions at an interface separating
• dielectric (r1 ) and dielectric (r2 )
• conductor and dielectric
• conductor and free space
H
H
~ · d~l = 0 and D
~ · dS
~=Q
To determine the boundary conditions, we need to use Maxwell’s equations E
Also we need to decompose the electric field intensity E into two orthogonal components:
~ =E
~t + E
~n
E
~ = E
~t + E
~ n are the tangential and normal components of E
~ to the interface of interest. A similar
where E
~
decomposition can be performed for D.
4.7.1
Dielectric-Dielectric Boundary Conditions
The tangential components of E are the same on two sides of the boundary, i.e.
E1t = E2t
and the difference of the two normal components of D is the free charge density on the interface, i.e.
D1n − D2n = ρS
The following is the law of refraction of the electric field at a boundary free of charge (since ρS = 0 is assumed
at the interface). Thus, in general, an interface between two dielectrics produces bending of the flux lines as a result
of unequal polarization charges that accumulate on the sides of the interface.
r1
tan θ1
=
tan θ2
r1
4.7.2
Conductor-Dielectric Boundary Conditions
The conductor is assumed to be perfect (i.e., σ → ∞ or ρc → 0). Although such a conductor is not practically
realizable, we may regard con- ductors such as copper and silver as though they were perfect conductors. To
determine the boundary conditions for a conductor-dielectric interface, we follow the same procedure used for
~ = 0 inside the conductor. Thus,
dielectric-dielectric interface except that we incorporate the fact that E
Et = 0
and,
Dn = ρs
Thus, under static conditions, the following conditions can be made about a perfect conductor:
1. No electric field may exist within a conductor
~ =0
ρv = 0, E
~ = −∇Φ
~ = 0, there can be no potential difference between any two points in the conductor; that is, a
2. Since E
conductor is an equipotential body
3. The electric field E can be external to the conductor and normal to its surface
10
5
Electrostatic Boundary Value Problems
5.1
Poisson’s and Laplace’s Equation
Poisson’s and Laplace’s equations are easily derived from Gauss’s law
~ ·D
~ =∇
~ · E
~ = ρv
∇
and Maxwell’s equation
~ = −∇Φ
~
E
Combining the two equations give
~ · (−∇Φ)
~
∇
= ρv
For a homogeneous medium, this becomes Poisson’s equation
∇2 Φ = −
ρv
For a special case, where ρv = 0, this becomes Laplace’s equation
∇2 Φ = 0
Laplace’s equation is of primary importance in solving electrostatic problems involv- ing a set of conductors maintained at different potentials. Examples of such problems include capacitors and vacuum tube diodes. Laplace’s
and Poisson’s equations are not only useful in solving electrostatic field problem; they are used in various other
field problems.
5.2
Procedure for solving Poisson’s or Laplace’s equation
We will use the uniqueness theorem to find the exact solutions; the uniqueness theorem states that if a solution to
Laplace’s equation can be found that it satisfies the boundary conditions, then the solution is unique.
The following general procedure may be taken in solving a given boundary-value problem involving Poisson’s or
Laplace’s equation:
1. Solve Laplace’s (if ρv = 0) or Poisson’s (if ρv 6= 0) equation using either (a) direct integration when V is
a function of one variable, or (b) separation of variables if V is a function of more than one variable. The
solution at this point is not unique but expressed in terms of unknown integration constants to be determined.
2. Apply the boundary conditions to determine a unique solution for V. Imposing the given boundary conditions
makes the solution unique.
~ using E
~ = −∇Φ
~ and D
~ from D
~ = E
~
3. Having obtained Φ, find E
5.2.1
Laplace’s Equation in two dimensions
In two dimensions, Laplace’s equation, with general boundary conditions in an object of dimensions 0 ≤ x ≤ L, 0 ≤
y ≤ H, looks like:
∂2Φ ∂2Φ
∇2 Φ =
+
=0
∂x2
∂y 2
Φ(0, y) = g1 (y)
Φ(x, 0) = f1 (y)
Φ(L, y) = g2 (y)
Φ(x, H) = f2 (y)
Notice that while the partial differential equation is both linear and homogeneous the boundary conditions are
only linear and are not homogeneous. This creates a problem because separation of variables requires homogeneous
boundary conditions. To completely solve Laplace’s equation we’re in fact going to have to solve it four times. Each
time we solve it only one of the four boundary conditions can be nonhomogeneous while the remaining three will
be homogeneous.
11
Case 1
∇ 2 Φ4 =
∂ 2 Φ4
∂ 2 Φ4
+
=0
∂x2
∂y 2
Φ4 (0, y) = g1 (y)
Φ4 (x, 0) = 0
Φ4 (L, y) = 0
Φ4 (x, H) = 0
Start by assuming that our solution will be in the form
Φ4 (x, y) = h(x)ϕ(y)
So from that problem we know that separation of variables yields the following two ordinary differential equations
that we’ll need to solve.
d2 ϕ
d2 h
2
2
dx2 − λ h = 0
dy 2 − λ ϕ = 0
h(L) = 0
ϕ(0) = 0, ϕ(H) = 0
Solving for ϕ(y) yields
ϕn (y) = cos(λy) + sin(λy)
By applying the boundary conditions, we get the eigenvalues and eigenfunctions for the boundary value problem,
nπ , with n = 1, 2, 3, . . .
λn =
H
nπy ϕn (y) = sin
H
Because the coefficient of h(x) in the differential equation above is positive we know that a solution to this is,
nπ(x − L)
nπ(x − L)
h(x) = c1 cosh
+ c2 sinh
, with n = 1, 2, 3, . . .
H
H
Note: the shift by L is there for dealing with the h(L) = 0 boundary condition. Applying the lone boundary
condition to this “shifted” solution gives,
h(L) = c1 = 0
The solution to the first differential equation is now,
nπ(x − L)
h(x) = c2 sinh
, with n = 1, 2, 3, . . .
H
A product solution for this partial differential equation is,
nπy nπ(x − L)
sin
, with n = 1, 2, 3, . . .
Φn (x, y) = Bn sinh
H
H
The Principle of Superposition then tells us that a solution to the partial differential equation is,
∞
nπy X
nπ(x − L)
Φ4 (x, y) =
Bn sinh
sin
H
H
n=1
To determine the constants all we need to do is apply the final boundary condition.
∞
nπy X
nπ(−L)
Φ4 (0, y) = g1 (y) =
Bn sinh
sin
H
H
n=1
Using Fourier series,
Z
nπy 2 H
=
g1 (y) sin
dy
Bn sinh
H 0
H
Z H
nπy 2
Bn =
g1 (y) sin
dy
H
0
H sinh nπ(−L)
nπ(−L)
H
H
12
Case 2
∂ 2 Φ3
∂ 2 Φ3
+
=0
∂x2
∂y 2
∇ 2 Φ3 =
Φ3 (0, y) = 0
Φ3 (x, 0) = 0
Φ3 (L, y) = 0
Φ3 (x, H) = f2 (x)
Start by assuming that our solution will be in the form
Φ3 (x, y) = h(x)ϕ(y)
So from that problem we know that separation of variables yields the following two ordinary differential equations
that we’ll need to solve.
d2 ϕ
d2 h
2
2
dx2 + λ h = 0
dy 2 − λ ϕ = 0
h(0) = 0h(L) = 0
ϕ(0) = 0
Solving for h(x) yields
hn (y) = cos(λx) + sin(λx)
By applying the boundary conditions, we get the eigenvalues and eigenfunctions for the boundary value problem,
nπ , with n = 1, 2, 3, . . .
λn =
L
nπx hn (x) = sin
L
Because the coefficient of ϕ(x) in the differential equation above is positive we know that a solution to this is,
nπy nπy ϕ(y) = c1 cosh
+ c2 sinh
, with n = 1, 2, 3, . . .
L
L
Applying the lone boundary condition to this solution gives,
ϕ(0) = c1 = 0
The solution to the first differential equation is now,
nπy , with n = 1, 2, 3, . . .
ϕ(y) = c2 sinh
L
A product solution for this partial differential equation is,
nπy nπx Φn (x, y) = Bn sinh
sin
, with n = 1, 2, 3, . . .
L
L
The Principle of Superposition then tells us that a solution to the partial differential equation is,
Φ3 (x, y) =
∞
X
Bn sinh
n=1
nπy L
sin
nπx L
To determine the constants all we need to do is apply the final boundary condition.
∞
nπx X
nπH
Φ3 (x, H) = f2 (x) =
Bn sinh
sin
L
L
n=1
Using Fourier series,
Z
nπx 2 L
f2 (x) sin
dx
L 0
L
Z L
nπx 2
Bn =
f2 (x) sin
dx
nπH
L
L sinh L
0
Bn sinh
nπH
L
=
Same procedure is applicable for Case 3 and Case 4.
13
5.2.2
The Finite Difference Method
The method of finite differences (FDM) is one of the most successful numerical techniques used, not only in
electromagnetics but also in other scientific and technical branches. This technique consists of dividing the region
of interest into a mesh formed by parallel lines, which leads to a set of nodes spaced a length a from each other. A
numerical value is given to each node and, under some approximations, the potential value at each point may be
related with the potentials of the nearest neighbors by linear equations.
Figure 9: a Grid and nodes corresponding to the nearest neighbors. b Mesh in a rectangular domain
Let us suppose we want to know the potential Φ at point P of coordinates (x, y), approximately. For relating
the potential at this node with the potentials at P1 , P2 , P3 , and P4 , if the function representing Φ is continuous, we
can expand the potential in Taylor series as follows
Φ(x + a, y) ≈ Φ(x, y) +
∂Φ(x, y)
1 ∂ 2 Φ(x, y) 2
a+
a + ···
∂x
2 ∂x2
and for the point on the left
1 ∂ 2 Φ(x, y) 2
∂Φ(x, y)
a+
a + ···
∂x
2 ∂x2
In the same manner, for the neighbors above and below
Φ(x − a, y) ≈ Φ(x, y) −
Φ(x, y + a) ≈ Φ(x, y) +
∂Φ(x, y)
1 ∂ 2 Φ(x, y) 2
a+
a + ···
∂y
2 ∂y 2
Φ(x, y − a) ≈ Φ(x, y) −
∂Φ(x, y)
1 ∂ 2 Φ(x, y) 2
a+
a + ···
∂y
2 ∂y 2
and
The addition of the two first equations leads to
∂ 2 Φ(x, y)
Φ(x + a, y) + Φ(x − a, y) − 2Φ(x, y)
≈
∂x2
a2
and same for the last two
∂ 2 Φ(x, y)
Φ(x, y + a) + Φ(x, y − a) − 2Φ(x, y)
≈
∂y 2
a2
From these results the approximate expression of the Laplace equation at point P is
∇2 Φ(x, y) =
∂ 2 Φ(x, y) ∂ 2 Φ(x, y)
Φ(x + a, y) + Φ(x − a, y) + Φ(x, y + a) + Φ(x, y − a) − 4Φ(x, y)
+
≈
=0
∂x2
∂y 2
a2
And therefore, the potential
Φ(x + a, y) + Φ(x − a, y) + Φ(x, y + a) + Φ(x, y − a)
4
By this procedure we have reduced the solution of the Laplace differential equation to a set of algebraic equations
that allow us to estimate the potential at each point of the net.
Φ(x, y) ≈
14
5.2.3
Method of Images
A Point Charge Above a Grounded Conducting Plane
Consider a point charge Q placed at a distance h from a perfect conducting plane of infinite extent as in 10(a). The
image configuration is in 10(b). The electric field at point P(x, y, z) is given by
Figure 10: The method of images
~ = E~+ + E~− = Qr~1 + −Qr~2 = 1 x~ax + y~ay + (z − h)~az − x~ax + y~ay + (z + h)~az
E
4πr13
4πr23
4π [x2 + y 2 + (z − h)]3/2
[x2 + y 2 + (z + h)]3/2
The potential at P is easily obtained for z ≥ 0
1
Q
1
Φ=
−
4π [x2 + y 2 + (z − h)]3/2
[x2 + y 2 + (z + h)]3/2
and Φ = 0 for z < 0. The surface charge density of the induced charge can also be obtained from the electric field,
ρs = Dn = E|z=0 =
The total induced charge on the conducting plane is
Z
Z ∞Z
Qi = ρs dS =
−∞
By changing variables r2 = x2 + y 2 and dxdy = rdrdφ
Z
Z
Qi = ρs dS =
Qh
2π[x2 + y 2 + h2 ]3/2
∞
−∞
2π
Z
0
0
Qhdxdy
2π[x2 + y 2 + h2 ]3/2
∞
Qhrdrdφ
2π[r2 + h2 ]3/2
Solving this integral gives Qi = −Q as expected, because all flux lines terminating on the conductor would have
terminated on the image charge if the conductor were absent.
In general, when the method of images is used for a system consisting of a point charge between two semiinfinite
conducting planes inclined at an angle φ (in degrees), the number of images is given by
360◦
N=
−1
φ
15
6
6.1
Magnetostatics
Ampere’s Law
~ around a dosed path is the same as the
Ampere’s law states that the line integral of the tangential component of H
net current Ienc enclosed by the path.
I
~ · d~l = Ienc
H
By applying Stoke’s theorem to the left-hand side, we obtain
I
I
Z
~ · d~l = (∇
~ × H)
~ · d~s = Ienc =
~
H
J~ · dS
L
S
S
Therefore, we derive Maxwell’s third equation,
~ ×H
~ = J~
∇
6.2
Magnetic Flux
An isolated magnetic charge does not exist. Thus the total flux through a closed surface in a magnetic field must
be zero; that is,
I
~ · dS
~=0
B
By applying the divergence theorem, we get
I
~ · dS
~=
B
Z
~ · BdV
~
∇
=0
v
Therefore, we derive Maxwell’s fourth equation:
~ ·B
~ =0
∇
6.3
Magnetic Vector Potential
~ = 0, B
~ bust be the curl of some other vector which we define as
Since from Maxwell’s fourth equation, vec∇ · B
the magnetic vector potential:
~ =∇
~ ×A
~
B
Analogous to electrostatic fields, the magnetic vector potential also has a Poisson’s equation:
~ = −µJ~
∇2 A
and
µ
Ax =
4π
6.4
6.4.1
Z
V
Jx (r~0 )dV 0
and similar for Ay and Az
|~r − r~0 |
Boundary conditions for magnetostatic fields
H at a boundary
Ht1 − Ht2 =
6.4.2
I
∆l
B at the boundary
Bn1 = Bn2
16
7
Maxwell’s Equations in the Time Varying Domain
7.1
Electromagnetic induction
1. Faraday’s Law: the induced electromotive force in a circuit is equal to the negative of the rate at which the
magnetic flux through the circuit is changing
emf = −
dΨm
dt
2. Lenz’s Law: the current induced in the circuit will appear in such a direction that it opposes the change that
produced it
From Faraday’s law, we can say
I
~ · d~l = − dΨm = − d
E
dt
dt
Z
~ · dS
~
B
Next, use Stokes’ theorem and Leibnitz rule:
Z
~ × E)
~ · dS
~=−d
(∇
dt
s
Z
~ · dS
~=−
B
Z
s
~
∂B
∂t
!
~ +B
~·
· dS
~
∂S
∂t
~
Leading to Maxwell’s equation for a time varying B
~
~ ×E
~ = − ∂B
∇
∂t
7.2
Maxwell’s equations in a point form
Z
Z
~
~ × E)
~ · dS
~=−d
~ · dS
~
~ ×E
~ = − ∂ B and (∇
B
∇
∂t
dt
s
I
Z
Z
~
∂D
∂
~
~
~
~
~
~
~
~ · dS
~
∇×H =J +
and
H · dl = J · dS +
D
∂t
∂t s
s
Z
I
~
~
~
~
D · dS = ρdV
∇ · D = ρ and
v
Is
~ · dS
~=0
~ ·B
~ = 0 and
∇
B
s
7.3
Wave Propagation in Perfect Dielectric
~ ×E
~ = − ∂ B~ predicts that ∂ B~ generates an electric
In a perfect dielectric, such as free space, J~ = 0 and ρ = 0. ∇
∂t
∂t
~
~
~ ×H
~ = ∂D
~ ×E
~ = − ∂ B~ ,
field. Furthermore, ∇
predicts that ∂ D generates a magnetic field. Consider the equation ∇
∂t
∂t
∂t
2~
~
~ × (∇
~ × E)
~ = −∇
~ × ∂ B ⇒ ∇2 E
~ = µ ∂ E ←− vector wave equation of E
~
⇒∇
∂t
∂t2
~ ×H
~ =
We can also show by taking the curl of ∇
~ = µ
∇2 H
~
∂D
∂t ,
we get
~
∂2H
~
←− vector wave equation of H
∂t2
Taking the one dimensional solution of the wave equation for the electric field, we get
~ = Ex~ax Em cos ωt − ωz
E
v
17
where v is the velocity of propagation
1
v =√
µ
~ consider:
To find the corresponding H,
~
~
~ ×E
~ = − ∂ B = −µ ∂ H
∇
∂t
∂t
1
~ = ~ay Em cos ωt − ωz
⇒H
µv
v
Therefore,
Hy =
1
Ex
µv
or,
Ex
=
Hy
r
µ
=η
which is a constant we define as η, which is the intrinsic impedance of a medium. For free space,
ηo = 377Ω
Figure 11: Our electromagnetic wave
We call this type of wave (Figure 11) as a uniform plane wave. The surfaces of constant phase are planes and in
any of these planes, the fields are uniform with respect to x and y. Note that the direction of propagation is alone
~ ×H
~
the vector direction of E
7.4
Poynting’s Theorem
By conservation of energy, the net rate at which energy flows into a volume V is equal to the rate at which energy
is stored, plus the rate for dissipation. The rate at which energy is flows into V is defined by
I
~
R1 = − P~ · dS
s
18
The rate at which energy is stored in V is defined by
Z
R2 =
v
~
~
~ · ∂B + E
~ · ∂D
H
∂t
∂t
!
dV
And finally, the rate at which energy is dissipated in V is,
Z
~
R3 = (J~ · E)dV
v
By conservation of energy, R1 = R2 + R3
I
~=−
P~ · dS
s
Z
v
~
~
~ · ∂D
~ · ∂B + E
H
∂t
∂t
!
Z
dV −
~
(J~ · E)dV
v
Z h
Z
i
~ · (∇
~ × E)
~ −E
~ · (∇
~ ×H
~ − J)
~ dV − (J~ · E)dV
~
=−
H
v
v
Z h
i
~ · (∇
~ × E)
~ −E
~ · (∇
~ × H)
~ dV
H
=
Zv
~ · (H
~ × E)dV
~
= ∇
v
Z
~ × H)
~ · dS
~
= (E
(by vector identity)
(divergence theorem)
s
~ ×H
~
P~ = E
(the Poynting vector)
The Poynting Vector, P~ , is also known as the intensity of the EM wave. It is the time-dependent power flow
density. The average power density is therefore,
1 2
E ~az
< P~ >≡ P~avg =
2η m
7.5
Maxwell’s Equations for time periodic case
We will develop a method that is similar to that for phasors in AC steady state circuits. The only difference is that
we will have a vector with magnitude, direction, and phase. The general electric field is
~
~
E(x,
y, z, t) = E(x,
y, z) cos(ωt + φ(x, y, z))
In exponential notation,
h
i
~
~
E(x,
y, z, t) = Re E(x,
y, z)ejφ ejωt
~
We define E(x,
y, z)ejφ as our phasor vector, therefore,
h
i
~
~ s ejωt
E(x,
y, z, t) = Re E
Therefore, for our plane wave
~ = −~ax Em cos ωt − ωz + φo
E
v
i
h
~ = ~ax Re Em ej(φo − ωz
v ) ejωt
E
~ s = ~ax Em ejφo e−j ωz
v
E
ωz
~ s = ~ax Esm e−j v
E
19
This allows us to transform Maxwell’s equations into the frequency domain:
~ ×E
~ s = −jω B
~s
∇
~ ×H
~ s = J~s + jω D
~s
∇
~ ·D
~ s = ρs
∇
~ ·B
~s = 0
∇
7.6
The Loss Tangent
This is an important parameter to characterize materials such as dielectrics or conductors. Consider,
~ ×H
~ s = J~s + jωE
~s
∇
~ s , so by substituting that in, we get
We know that J~s = σ E
~ ×H
~ s = σE
~ s + jωE
~s
∇
~ s and ∇
~ ×H
~ s as δ
Drawing a phasor diagram for this, one can define the angle between jωE
tan δ =
σ
ω
If tan δ >> 1, we have a very good conductor, while if tan δ << 1 we have a very good dielectric.
7.7
Plane wave propagation in lossy media
A lossy dielectric is a medium in which an EM wave loses power as it propagates due to poor conduction. Consider
~ ×E
~ s = −jω B
~ s,
a linear, isotropic, homogeneous, lossy dielectric medium that is charge free. Take the equation ∇
and consider the curl on both sides
~ ×∇
~ ×E
~ s = −jωµ∇
~ ×H
~s
∇
~ ∇
~ ·E
~ s ) − ∇2 E
~ s = −jωµ(J~s + jω D
~ s)
∇(
~ s = jωµ(σ + jω)E
~s
∇2 E
~ s = −ω 2 µ 1 − j σ E
~s
∇2 E
ω
~ s = −ω 2 µ (1 − j tan δ) E
~s
∇2 E
The solution do this is
√
Esx = Esm e±jω µ(1−tan δ)z
Where we can define:
p
γ = α + jβ = jω µ(1 − tan δ)
r
µ √
α=ω
( 1 + tan δ − 1) m−1
2
r p
µ
β=ω
( 1 + tan2 δ + 1) rad · m−1
2
complex propagation constant
attenuation constant
phase constant
Converting the solution back into time domain, we get
Ex = Em e−αz cos(ωt − βz)
The wavelength of this wave is given by
2π
λ=
ω
q
√
µ
2 (
1 + tan2 δ + 1)
20
Figure 12: EM wave in lossy media
The velocity of the wave is given by
v=
ω
1
=q √
β
µ
( 1 + tan2 δ + 1)
2
We can also show that Ex and Hy are no longer in phase.
r
Ex
µ
=η=
Hy
(1 − j tan δ)
Furthermore, the average power density is given by:
1
~s × H
~ x∗ ]
P~avg = Re[E
2
1 2 −2αz
1
~
Pavg = Esm e
Re
2
η
21
8
Reflection and transmission of EM waves
Consider Figure 13, we can derive equations for the incident waves, reflected waves, and the transmitted waves
Figure 13: A plane wave incident normally on an interface between two different media
Incident wave:
The equations for the incident wave is given by:
Esx,i = Esm,i e−jβ1 z
Esm,i −jβ1 z
Hsy,i =
e
η1
Reflected wave:
The equations for the reflected wave is given by:
Esx,r = Esm,r ejβ1 z
Esm,r jβ1 z
e
Hsy,r =
−η1
Note the positive sign in the exponential, denoting propagation in the negative z direction.
Transmitted wave:
The equations for the transmitted wave is given by:
Esx,t = Esm,t e−jβ2 z
Esm,t jβ2 z
Hsy,t =
e
η2
At the interface, z = 0,
~ is continuous
1. Tangential component of E
Esm,i + Esm,r = Esm,t
~ is continuous
2. Tangential component of H
Esm,i
Esm,r
Esm,t
+
=
η1
−η1
η2
22
Solving the two equations yield:
Esm,r
η2 − η1
=
Esm,i
η2 + η1
Esm,t
2η2
τ=
=
Esm,i
η2 + η1
Γ=
reflection coefficient
transmission coefficient
23