In the following sections the knowledge brought in the previous chapters will be applied to the solution of the typical three cases of electric or magnetic fields. The solutions will be constructed in the cartesian, circular cylindrical and spherical coordinate systems. We shall try to show explicitly the use of field operators for the formulation of the field equations for the particular cases.

Task:

Find analytically the two dimensional distribution of the electric potential inside the system of electrodes shown in the Fig. 4.1. The boundary conditions are given through constant potentials on the two electrodes as shown in the figure Fig. 4.1.

Fig.4.1, The system of two dimensional electrodes with the voltage sourse

The boundary conditions are given through constant potentials on the two electrodes as shown in the figure.

Solution:

The electrodes consist of the four mutually perpendicular plane surfaces which coincide with the coordinate surfaces of the cartesian system. This is why we shall formulate the mathematics of this problem in that coordinate system. The first electrode coincides with the following coordinate surfaces x=0; y=0 and y=b and the second one with the surface x=a. First electrode is grounded and therefore its potential is zero while the second electrode has the potential defined by the inner voltage of the source, see Fig. 4.1. Since there are no free charges inside the "tube" (electrode system), the potential j(x,y) will be described by the Laplace equation

(4.1)

because of the two-dimensionality of the field. The boundary conditions can be written in the forms:

for

(4.2)

(4.3)

(4.4)

(4.5)

The solution of the (4.1) can be found as the linear combination of the elementary solutions of the same Laplace equation, that is

(4.6)

(4.7)

The solution of (4.7) can be seek by the Fourier method of the separation of variables. According to this method the solution for jk(x,y) can be seek in the form of the product of two functions X_k(x).Y_k(x) where X_k(x) and Y_k(x) depend only on x or y respectively. Hence

(4.8)

Putting (4.8) into (4.7) gives

(4.9)

It should be noted that (4.9) is the sum of two functions f₁(x)+ f₂(y)=0 where x and y are independent variables. It can be fulfilled only if each of these functions is constant. So we can write

(4.10)

(4.11)

or

(4.12)

(4.13)

and simultaneously

(4.14)

The solution for (4.12) and (4.13) are

(4.15)

(4.16)

where are unknown constants. If we take new constants related to the original primed according to the following relations

(4.17)

(4.18)

It is possible to write the solutions (4.15) and (4.17) in the form

(4.19)

(4.20)

It is evident that we have a set of solutions which differ mutually by constants k_x, k_y. It is necessary to choose only those of them which in linear combination can fulfill the boundary conditions.
For y = 0 and y = b we require j_k =0 and consequently we have to put B_1k=0. Then

(4.21)

Similarly if we require j_k(y=b)=0 , then k_y has to be indispensably

(4.22)

Finally we can write

(4.23)

Taking into account (4.14) makes possible to write for constant k_x

or

(4.24)

This allows to specify the solution for

(4.25)

Application of boundary condition for the left wall of the electrode x=0 ®j(x)=0 gives A_1n=0 and

(4.26)

where is some complex constant. The resulting solution for j(x,y) satisfying the boundary conditions can be written as the linear combination of elementary solutions as it was mentioned above

(4.27)

or

(4.28)

The constant can be determined using the last boundary condition:

for x=a

j(a,y)=U0

(4.29)

Combining (4.26) and (4.27) gives

(4.30)

Multiplying (4.28) by and then integrating on both sides of (4.28) one can obtain unknown coefficients in the form:

(4.31)

Final solution obtains then the form

(4.32)

where n=1, 2, 3, ...
Electric field intensity is given by the gradient of the potential j(x,y)

(4.33)

The equipotential surfaces and -stream lines of that field are outlined in the Fig. 4.2.

Fig.4.2, The distribution of the potential and electric field intensity vectors accross the two dimensional elctrode system with potential difference of 1V on the electrodes

Task:

Find the space distribution of the stationary electric vector potential inside and outside the infinitely long cylindrical electric conductor with the diameter "d" carrying the total DC electric current I. The corresponding current density is expected to be constant on the cross section of the conductor, see Fig. 4.3 . Draw the streamlines of the vector potential and the corresponding magnetic flux density .

Fig.4.3, The infinitely long electric conductor carrying homogeneously distributed total current I

Solution:

The boundary surface between the conductor and air surrounding is the cylindrical one. It is apparent that the use of circular cylindrical coordinate system is desirable. We have to be aware that the field we are going to describe is two-dimensional and cylindrically symmetrical, that is no field component will depend on the azimuthal coordinate y, what implies in .
By the definition the vector potential is given by the relation

(4.34)

where is magnetic flux density. This equation can be written in the form

(4.35)

where we have used the determinant formulation of curl operator in the cylindrical coordinate system derived in previous chapters.
Taking into account the cylindrical symmetry of the magnetic field and the fact that the current density is constant on the cross-section, we can specify the particular components on the right side of (4.35). Hence B_r=0; B_z=0 and B_y(r)¹0 will not depend on y and z. B_y(r) can be calculated by using the Ampere's circuital law.

(4.36)

Since we are interested in finding the By(r) inside and outside of the conductor as well, we have to apply (4.36) for some streamline of inside (path 1) and outside (path 2) the conductor as depicted in the Figure No.  4.4.

Fig.4.4, To the application of the Amper's circuital law.

The application of (4.36) to path 1 or 2 gives for B_y(r) the following relations

(4.37)

(4.38)

The different dependences of the magnetic flux density on r inside and outside the conductor are caused by the different values of current flux density that is constant inside and equal to zero outside the conductor.
Combining the (4.37), (4.38) and (4.35) gives two differential equations for potential inside and outside the conductor as follows

(4.39)

(4.40)

where

The integration of these equations gives

(4.41)

(4.42)

The integration constants K₁ and K₂ can be determined by requiring the potential to be continuous on the boundary and by putting it equal zero on the same boundary. Hence we can write

(4.43)

and consequently it results in

(4.44)

(4.45)

The graphical illustrations of (4.44) and (4.45) are in the Fig. 4.5

Fig.4.5, The distribution of Azin(r) an Azout(r) with plane x=0

Remark:

In the case we would be given the potential first, we could determine the space distribution of magnetic flux density using the .

Task:

Find the space distribution of electromagnetic field vectors (electric field intensity and magnetic flux density) in the open space around the two electric point charges +q, -q, which oscillate harmonically along the z axis, see Fig. 4.6, so that corresponding electric dipole is described by the relation and , where l is the maximum distance of charges and w is the angular frequency of oscillation. We propose that l is very small.

Fig.4.6, Realization of elementary electric dipole

Solution:

The oscillating point charges represent some alternating electric current defined by

(4.46)

In the case of harmonic time dependence it is suitable to represent the particular magnitudes by their complex functions - phasors. For harmonic vector functions we shall use their complex representations phasor-vectors. Taking the real parts of them we obtain real functions we need. Hence the phasor of electric current I_p of our elementary dipole is given by

(4.47)

where is the phasor representing the real dipole .
Having introduced the phasors or phasor-vectors we can imagine instead of the dipole, consisting of two oscillating point charges, the small and short elementary antenna carrying the complex harmonic electric current. Since the antenna is very short we can consider the current being constant along the antenna. In the following we shall consider the radiation of a small harmonic antenna as shown in the Fig. 4.7.

Fig.4.7, The small harmonic antenna as a representation of oscillating point charges.

In the infinite free space the complex electric vector potential generated by an elementary sections of electric conductor with current flowing along the z axis (see Fig. 4.7) can be described by integral

(4.48)

Relation (4.48) can be found as a solution of Poisson differential equation for the vector potential generated by current in the free space. Due to the simplicity let us investigate the radiated field in the far region only. It allows us to approximate the calculation of the distance between the element of antenna and the point P (see Fig. 4.7) as follows

(4.49)

where . The primed coordinates denote the position of the current element of the antenna. The integration gives

where the term if goes to zero. It is true only for l<<l, where l is the wave length of the radiated field.
So finally the vector potential of the small antenna is described by

(4.50)

In spherical coordinate system the unit vector is given by and can be rewritten into the form

(4.51)

One can see that vector potential has only two components different from zero - . The component and consequently the field is rotationally symmetrical around the z axis. Having the expression for the vector potential we can calculate the required fields and .
and we can use the relation (3.27) for curl in the spherical coordinate system.

In our case A_y = 0;

Hence



As a result we can write

(4.52)

One can se that only B_y ¹ 0 and B_J, B_r ¹ 0. The field is of solenoidal form where

(4.53)

(4.54)

represents the radiation field, whereas , due to dependence, decreases very fast with r. It represents the field in the middle zone. In order to calculate the electric field intensity we have to use Maxwell equations. From the first Maxwell equation in the complex representation

(4.55)

we can find as follows

(4.56)

In the spherical coordinate system, having the phasor-vector defied by (4.51) one can perform a similar procedure for the calculation of as it was done above in the case of the calculation of .

(4.57)

This result can be written in a simpler form. If we perform the following substitution

and

where a is the propagation constant and Z₀ is the wave impedance for vacuum, we can write (4.58) in the form

(4.58)

It is possible also to express (4.58) in terms of the complex amplitude of the electric dipole . We can write

and consequently

(4.59)

The particular terms in (4.59) can be grouped with respect to the dependences . In this case it is possible to write

(4.60)

(4.61)

E₁, E₂, E₃ represent the radiation field (far field), the field in the middle zone and the field in the near zone, respectively. The density of electric power radiated by the dipole is given by Poynting vector

(4.62)

where and are the radiation components given by (4.53) and (4.61). Then it is possible to write (4.62) in the form

or

(4.63)

The total power radiated by the dipole can be calculated by the integration through the whole surface of the sphere in far field.

(4.64)

where we have introduced the effective radiation resistance R_effrad by the formula

(4.65)

where due to preservation of the physical dimension the symbol W (1 Ohm) was kept through derivation of (4.65).
The simplified sketch of the and streamlines of the elementary dipole field are outlined in the Fig. 4.8.

Fig.4.8, The simplified sketch of the and streamlines of the elementary electric dipole. Vectors , in the piont P1 are shown. is the Pointing vector at point P1