Chapter 10 Electric Field, Magnetic Field, Maxwell’s Equations and Plane Waves
In a wireless environment, most of the signals, are transmitted in the form of electromagnetic waves. It is therefore important to have a basic understanding of electromagnetism. Since Maxwell’s equations are based upon electric and magnetic fields, we shall first introduce these two fields. Before doing that, we first have to introduce some vector operations, which are needed for the discussion of them..
10.1 The Dot and Cross Products of Vectors
In this chapter, we shall use two terms which are actually simple. They are scalar field and vector field. We shall not give formal mathematical definitions of these two terms. Consider a mountainous terrain. At every point characterized by x y and z, there is an altitude associated with it. This illustrates a scalar field. Similarly, in a region, every location has a temperature associated with it. This is another case of scalar field. Except this time, it may change with time. Thus the temperature can be expressed as . Now, consider the case where water flows through, say a tube. In this case, at each point, we not only have to note the speed of the flow, also the direction of the flow. This is a typical vector field. In this chapter, many parameters are vectors, such as electric and magnetic field intensities and so on. On the other hand, the voltage, for instance, is a scalar.
Since we are living in a three-dimensional space, we shall use the three-dimensional coordinate system in this chapter. In this system, there are three vectors representing the x, y and z directions. These three vectors are denoted as and representing vectors in the x, y, and z directions respectively and all of these vectors have magnitudes equal to unity. Every vector A will have three components, which are the projection of A onto x, y, and z coordinates.
Thus, a vector A may be denoted as
.
The Addition and Subtraction of Vectors
Let A and B be two vectors.
Then (10.1-1)
A schematic diagram of the addition and subtraction operations is now shown in Fig. 10.1-1.
Fig. 10.1-1 The Addition and Subtraction of Vectors. (a)A+B. (b)A-B.
Example 10.1-1.
Let and Then
and
The Dot Product of Two Vectors
The dot product of two vectors is defined as follows:
(10.1-2)
Let θ denote the angle between vectors A and B. The dot product of A and B is also equivalent to the following:
(10.1-3)
Note that the result of the dot product of two vectors is not a vector, rather a scalar.
Example 10.1-2.
Letand
Then
Another way of determining is to use Formula (10.1-3). Note that
, and Thus
The Cross Product of Two Vectors
The cross product between two vectors A and B is also a vector. It is defined as follows:
A×B (10.1-4)
It is sometimes not easy to remember the above formula. A convenient way is to use the following equivalent formula using the determinant notation:
A×B (10.1-5)
The meaning of the cross product of two vectors can be understood by the so-called right-hand rule. Note that two vectors form a plane. Let be the vector, which is normal to this plane as shown in Fig. 10.1-2. Then the direction of A×Bis in the direction of and A×B can be found by the following formula:
A×B= (10.1-6)
where is the angle between A and B.
Fig. 10.1-2 The Cross Product of Two Vectors.
Example 10.1-3.
Let and Then it can be easily found that, by using either Equation (10.1-5) or Equation (10.1-6),
A×B
Note that the direction of A×B is normal to the plane consisting of A and B.
10.2 The ▽Operator, the Gradient of a Scalar Field and the Line Integral of a Vector Field
The ▽operator is defined as follows:
▽ (10.2-1)
The Gradient of a Scalar.
Let V be a scalar field function. Then, the gradient of V is defined as follows:
▽ (10.2-2)
Note that while V is a scalar, ▽V is a vector.
Example 10.2-1.
Let .
Then ▽V=.
The Line Integral of a Vector
Let us consider a contour as shown in Fig. 10.2-1. There is a force along this contour. This force is a vector and therefore it is denoted as F. To move an object from an initial point p to a terminal point q requires energy E as follows:
.
The integral in the above equation is called a line integral. Line integrals play important roles in deriving many electromagnetic formulas.
Example 10.2-1.
Let
Since , we have
Consider the following two cases as shown in Fig. 10.2-1(a) and (b) respectively.
Fig. 10.2-1 Two Cases for Line Integral. (a)Case1. (b)Case2.
Case 1:
(10.2-3)
For the first integral in Equation (10.2-3), and for the second integral, . Thus
(10.2-4)
Case 2:
For the first integral, y=x and in the second integral, x=0. Therefore, we have
(10.2-5)
Let A be a scalar function. Let A be the gradient of A. Let c denote a closed tour. Then we can show that the following holds:
(10.2-6)
Equation (10.2-6) also indicates that if the line integral of a vector A is zero, the vector A can always be expressed as the gradient of a scalar functionA.
Example 10.2-3.
Fig. 10.2-2 A Tour for Example 10.2-3.
Consider . Let the closed tour be the one shown in Fig. 10.2-2. This tour c consists of three tours, namely and . First, we compute the gradient of A.
A=▽
For tour , .
The first integral of the above equation disappears because for this tour.
For tour , we can show that
For tour , we can use the same kind of reasoning to show that
.
Note that in the above integrals, since we integrate along the line , x is substituted by y and y is substituted by x. By adding up the values corresponding to the three tours, we conclude that
10.3 The Divergence of a Vector Field
In the above section, the ▽operator was applied to a scalar field and the result is a vector. In this section, we shall introduce the result of applying the ▽operator to a vector field. LetA be a vector. The operator, ▽, called the divergence of A, is defined as follows:
▽ (10.3-1)
Note that the divergence of a vector is a scalar.
The physical meaning of divergence is explained as follows. Consider Fig. 10.3-1 in which there is a vector field. The flux of this vector field A through a surface s is defined as follows:
. (10.3-2)
Fig. 10.3-1 Flux through a Surface S
Since flux goes in and out of a closed surface, there is a net outward flux through a closed surface s defined as follows
(10.3-3)
Equation (10.3-3) tells us whether there is a sink or a source. If the amount of flux going into the surface is equal to the amount of flux going out of the surface, the net outward flux is equal to 0. Let us pay attention to the case where the enclosed surface reduces to zero. In doing so, the associated volume is also reduced to zero. We are therefore interested in a point. The net outward flux is:
Now, it can be shown that the divergence of A, as defined in Equation (10.3-1), is related to the above term as follows:
▽ (10.3-4)
The proof of Equation (10.3-4) will not be given in this book. It suffices for the reader to remember, from Equation (10.3-4), that the divergence of a vector field A is the net outward flux A per unit volume as the volume reduces to zero.
Example 10.3-1.
Let
Then
▽
Consider the following two cases: and . In the first case, the divergence is 0 and in the second case, it is 1. The reader is encouraged to see why they are so by using Equation (10.3-4). In both cases, flux is in the x-direction. For the first case, although the flux changes with respect to y, in the x-direction, when a flux goes into the volume, the same flux goes out of it without any change. But, for the second case, it seems that the amount of flux is increased as it goes through the volume. This is why the divergence is not 0 in this case.
Given a vector field A and a closed surface s, the net outward flux Ψ of A through the surface s is defined in Equation (10.3-3). We can use Equation (10.3-3) to find Ψ directly. But, based upon Equation (10.3-4), we can prove the following Divergence Theorem:
. (10.3-5)
We shall not give a formal proof of the above theorem. The reader can gain some feeling about the above equation through the two examples given below:
Example 10.3-2.
Let and s be the surface as illustrated in Fig. 10.3-2.
Fig. 10.3-2 The Closed Surface for Example 10.3-2.
can be computed as follows:
If we use Equation (10.3-5), we first compute ▽as follows:
▽=0.
From Equation (10.3-5), we conclude that =0.
Example 10.3-3.
Let and s be the same surface as used in Example 10.3-2.
It can be easily seen that ▽=3 and .
The physical meaning of ▽can be best understood by assuming the case where ▽=0. Under such a condition, according to the Divergence Theorem expressed in Equation (10.3-5), . As we shall see later, this result is quite significant.
10.4 The Curl of a Vector Field
Let A be a vector. The ▽operator is called the curl of A, and is defined as follows:
▽ (10.4-1)
In some sense, which we shall not elaborate here, ▽, is related to "circulation". If ▽=0, this indicates that A has no circulation; otherwise, A has some circulation. This concept will become useful as we study the electrostatic field intensity which has no circulation and magnetic field intensity which has circulation.
Example 10.4-1.
Let
Then, we have
▽
We like to emphasize here that ▽is a vector.
What is the physical meaning of ▽? In Section 10.2, we introduced the concept of line integral of a vector field. Let c denote a closed tour and let A be a vector field defined all along this tour. This tour c, since it is closed, is associated with a surface s, as shown in Fig. 10.4-1.
Fig. 10.4-1 An Illustration of ▽
. We can prove that ▽is related to the integral as follows:
▽. (10.4-2)
Note that ▽is a vector and its direction, denoted as , is perpendicular to the surface Δs determined by the right hand rule. A formal proof of Equation (10.4-2) will not be given. Based upon Equation(10.4-2), it is easy to prove the Stoke's Theorem, as expressed below:
(10.4-3)
The physical meaning of ▽can be best understood by considering the special case where ▽=0. In this special case, we conclude immediately that or A has no circulation. Later, after we have introduced the electrostatic field intensity E, we will prove that . This is quite significant.
The term ds in Equation (10.4-3) is a vector and the direction of it is that of the normal perpendicular to it and follows the right hand rule. In the following, we shall give two examples to give the reader some feeling about Equation (10.4-3).
Example 10.4-2.
Let A be the same as that in Example 10.3-2. That is, . Let the tour c be the one illustrated in Fig. 10.4-2.
Fig. 10.4-2 The Tour for Example 9.4-2.
▽.
.
.
.
If we use Equation (10.4-3), we can conclude now that . In fact, this can be verified by computing directly.
From Fig. 10.4-2, we can see that the tour consists of four tours, denoted as and . The line integrals along these four tours are as follows:
.
Example 10.4-3.
In this example, let A be the one used in Example 10.3-3. That is, . We will again use the same closed tour shown in Fig. 10.4-2. In this case, we can easily see that
▽=0.
Thus, we can immediately conclude that . Again, this can be proved by computing directly as follows:
From the above two examples, we can see that the curl of a vector does give us valuable information about a line integral. It is often easier to obtain the line integral of a vector along a closed tour by using the surface integral curl of the vector on the surface associated with the closed tour, as demonstrated in the above two examples.
Up to now, we have introduced three terms, namely gradient, denoted as ▽A, divergence, denoted as ▽, and curl, denoted as ▽×A. For each of these three terms, there is an equation related to it. These equations are Equation (10.2-6), related to gradient, Equation (10.3-5), related to divergence and Equation (10.4-3), related to curl. Let us summarize and relabel these equations as below:
(10.4-4)
(10.4-5)
. (10.4-6)
10.5 The Electrostatic Fields
In this section, we shall study various aspects of the electric field. We call this electrostatic field because it will not change with time. The electric field which changes with time will be discussed later.
Electric Field
First of all, let us consider a positive charge Suppose that there is another positive charge , which is a test charge, in certain sense. Then there is a repulsive force between these charges as follows according to Coulomb’s Law:
F (Coulomb’s Law) (10.5-1)
where F = the repulsion force between and , in Newtons
= charge 1, in coulombs
charge 2, in coulombs
r = the distance between charge 1 and charge 2, in meters
= a unit vector in the direction linking and , dimensionless
ε = the permittivity (or dielectric constant), in Farads per meter.
Fig. 10.5-1 depicts the Coulomb’s law.
Fig. 10.5-1 Coulomb’s Law.
Coulomb’s Law was invented by Charles Coulomb, a French army colonel (1736-1706). He retired from the army because of the French revolution. His elaborate experiments showed a force, either repulsion or attraction, exists between two charges. Coulomb’s Law is quite similar to Newton’s gravitational law.
The permittivity ε for vacuum, denoted as , is equal to For air at atmospheric pressure, ε is equal to 1.0006. For dielectric materials, ε >.
Having presented Coulomb’s law, we can now introduce the electric fieldE. Given a charge , for any location in space and a second test charge located there, there will be a force forced by which is
.
The electric fieldE, or also called electric field intensity, defined as force per unit charge, is thus
(10.5-2)
where E = electric field in Newton per Coulomb.
If there is more than one charge, for any location in the space, each charge causes an electric field intensity vector and the summation of the vectors is the resulting electric field at that location.
Example 10.5-1.
Fig. 10.5-2 The Electric Field Resulting from Two Charges.
Consider Fig. 10.5-2. Two charges are located at (0,0) and (0,2) respectively. For the point (1,1), there will be two vectors as shown and it is clear that the addition of these two vectors creates a vector in the positive y-axis direction.
Example 10.5-2.
Fig. 10.5-3 The Electric Field Induced by a Line Charge.
In Fig. 10.5-3, there is a line charge starting from (-1,0) to (1,0). The charge density of this line charge is denoted as ρ. Consider an infinitesimal section dx. This section contains charge and causes an electric field at (0,y) as follows:
As shown in Fig. 10.5-3, there is also an electric field symmetrical to the above electric field, which cancels out the electric field in the x-direction. Thus, only the y-direction component electric field exists. The combined y-direction electric field is
. (10.5-3)
To find the solution of Equation (9.5-3), let Then
(10.5-4)
Suppose the line charge extends from (-a,0) to (a,0),it can be proved that in this case,
. (10.5-5)
If ∞, Equation (8.2-5) becomes
(10.5-6)
Compare Equations (10.5-6) and (10.5-2). We now can see that the electric field intensity for a line charge is inversely to the distance while that for a point charge is inversely to the square of the distance. This is expected intuitively.
Electric Flux Density
Having introduced the concept of electric field, or electric field intensity, we can now introduce another related concept, called, the electric flux density. Note that the electric field intensity is inversely proportional to the permittivity. In other words, it is dependent of the medium. The electric flux density is something independent of the medium.
The famous British scientist, Michael Faraday (1791-1867), performed a very interesting experiment. We shall not give details of this experiment. But it can be briefly illustrated as follows: Consider Fig. 10.5-4. In Fig. 10.5-4, there are two concentric spheres. Between these two spheres, there is an insulation material. The surface of the inner sphere is charged positively. The outer sphere was temporarily grounded so that it was free of charge. Then Faraday found that the outer sphere was charged negatively. It was also observed that the amount of negative charge induced on the outer sphere was exactly the same as that in the inner sphere.
Fig. 10.5-4 An Experiment with Two Concentric Spheres
Somehow, we can visualize that there are lines emanating from the inner sphere to the outer sphere. This leads to the concept of electric flux. Denote the flux by . We have
We may imagine that the inner sphere shrinks to a point charge. Then the electric flux density D, at any point r meters from the center, as illustrated in Fig. 8.2-5, is defined as the flux per square meter as follows:
, (10.5-7)
where D = electric flux density in Coulomb per m2.
Fig. 10.5-5 The Electric Flux.
Comparing Equation (10.5-7) and (10.5-2), we now have
, (10.5-8)
where D = electric flux density in Coulomb per m2
E = electric field in Newton per Coulomb
ε = permittivity in Faraday per m.
Gauss Law
Let s denote a closed surface enclosing some charges. Let Q denote the total charges enclosed by s. We can show that the following is true:
. (10.5-9)
This is referred to as the Gauss' law. By using the divergence theorem expressed in Equation (10.3-5), we have
. (10.5-10)
Let denote the volume charge density. That is,
. (10.5-11)
Then, we have
▽. (10.5-12)
Equation (10.5-12) is also called the differential form of Gauss' law.
Electrostatic Potential
Consider Fig. 10.5-6. Assume that there is an electric field E and a charge q is moved along a tour from a to b. At any point, there is a force qE inserted on the charge q. The total work required by performing this action is
(10.5-13)
Fig. 10.5-6. The Moving of a Charge along a Tour.
If the tour is a closed one, as shown in Fig. 10.5-7, the total work done is zero. That is,
(10.5-14)
Equation (10.5-14) shows that the electrostatic field is a conservative field.
Fig. 10.5-7 The Moving of a Charge along a Closed Tour.
As discussed at the end of Section 10.2, since the line integral of E along a closed tour is zero, E can be expressed as the gradient of a scalar function. It is customaryto define the gradient of E as the potential function Vas follows:
(10.5-15)