DAVID K CHENG FIELD WAVE ELECTROMAGNETICS PDF

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DAVID K. CHENG SYRACUSE UNIVERSITY A vv ADDISON-WESLEY PUBLISHING COMPANY Reading, Massachusetts Menlo Park, California London. Field and Wave Electromagnetics - 2nd Edition - David K. Cheng - Ebook download as PDF File .pdf) or read book online. David K. Cheng - Field and Wave Electromagnetics [With Bookmarks] - Ebook download as PDF File .pdf) or read book online. David K. Cheng - Field and.


David K Cheng Field Wave Electromagnetics Pdf

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either in the book or in this manual. Sincerely,. E. David K. Cheng. Electrical and Computer Engineering. Department. Syracuse University. Syracuse, NY along the stinky stream, past the giant billboard for Dr Humayun's Hair Transplant . O Malalai of Maiwand,.. crammed t Electromagnetic Waves and. Books Field And Wave Electromagnetics 2e David K Cheng Solution Manual Pdf cheng - field and wave electromagnetics 2ed solution manual - cheng - field.

Dallo, A. Pomerene, A. Starbuck, C. DeRose, A. Lentine, G. Rebeiz, and S. Cherifi-Hertel, H. Bulou, R. Hertel, G. Taupier, K. Dorkenoo, C. Andreas, J. Guyonnet, I. Gaponenko, K. Gallo, and P. Sanna and W. Chang, M. Pfeiffer, N. Volet, M. Zervas, J. Peters, C. Manganelli, E. Stanton, Y. Li, T. Kippenberg, and J. Huang, D. Wei, Y. Wang, Y. Zhu, Y. Zhang, X. Hu, S. Zhu, and M. D Appl. Volk, R. Gainutdinov, and H.

Godau, T. Thiessen, L. Eng, and A. Chang, Y. Li, N. Volet, L. Wang, J. Peters, and J. Rao and S. Weigel, M. Savanier, C. Starbuck, A. Lentine, V. Stenger, and S. Mackwitz, M. Berth, A. Widhalm, K. Lindgren, A. Zukauskas, V. Pasiskevicius, F. Laurell, and C.

Chen, Q. Xu, M. Wood, and R. Sandkuijl, A. Tuer, D. Tokarz, J. Sipe, and V. Choi, D. Ko, J. Ro, and N. Kaneshiro, Y.

Uesu, and T. Gui, H. Hu, M. Garcia-Granda, and W. Berth, V. Quiring, W. Sohler, and A. Wang, V. Pasiskevicius, and F. Rodriguez, R. We should note that constant-V surfaces need not coincide with any of the surfaces that define a particular coordinate system.

Point PI is on surface V, ; P ,. P, little hc: We define the vector that represents both the? Equation states that the space rate of increase of' y in the a, direction is equal to the projection the compor ent of the gradient of V iq that direction.

We can also writc Eq. Now, dVin Eq, change in position from P , to P; ; it Bn fFe is the total di rential of V as a rcsuit of a expressed in terms of the differentia1 changes in coordinates:. Loc where d l l , df,, and d t , are the components of the vector differential displacement a.

I I Equation is a useful formula for computing the gradient of a scalar, when the scalar is given as a function of space coordinates. In view of Eq.

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True, this , , definition would yield a correct answer for the gradient of a scalar. However, the.. We use Eq. This will lead to the definitions of The divergence and the curl. I' ,' 1. Both are very important in the study of electromagnetism. V hrj by directed field lines, which are called Jlux lines or streamlines.

These are directed. In other words, the number of flux lines that pass through a unit surfxe normal to a vector is a measure of the magnitude of the vector. The flux of a vector field is analogov? For a volume with an enclosed surface there 4 1 be an excess of outward or inward flow throu h 2. The net outward flow ' of the fluid per unit volullle is therefore a measure of the strength of the enclosed source. The numerator in Eq. We have been exposed to this type of surface intcgral in Examplc 2 This definition holds for any coordinate system; the expression for div A, like that for A, will, of course, depend on the choice of the coordinate system.

At the beginning of this section we intimated that the divergence of a vector is a type of spatial derivative. The rc: We shall now derive the expression for div A in Cartesian co- I1 of the ordinates. We wish to find div A at the point s o , yo, 2,.

Field and Wave Electromagnetics - 2nd Edition - David K. Cheng

Ay Az Eqs. Similarly, on the back face, '. I I The The evalt to ar 1. Ax - YO, ZO 'ax sq, yo. Here a Ax has been factored out fromatheH. Here the higher-order terms containthe factors Ay, Ay ', etc. For the top and bottom faces, we have.

David K. Cheng - Field and Wave Electromagnetics

Now the results from. The higher-order terms vanish as the dillerential volume Ax Ay A: The value of div A, in general, depends on the position of the point at which it is evaluated. We have dropped the notation x,, yo, z, in Eq. However, the notation V. A has been customarily used to denote div A in all coordinate systems; that is,.

In general orthogonal curvilinear coordinates jdi;e? Substituting Eq. Find V B. Let the long wire be coincident with the z-axif in a cylindrical coordinate system. The problem states that ,,. This property indicates that the mqgngtic flux lines close upqn themselves and that there are no magnetic sources or sinks.

A divergenceless fielq is called a solenoidal field. More will be said about this t e btfield later m tbe boqk. The direction of ds is always that of the outward normal, perpendicular to the surface ds and directed away from the volume. For a very small differential volume element Auj bounded by a surface s,, the definition of V A in Eq: This is depicted in Fig.

Let us now combine the contributions ofall these differential volumes to both sides of Eq.

We have. The left side of Eq. It is also known as Gauss's theorem. The surface integrals on the right side of Eq. Hence, the net contribution of the right side of Eq. The validity of the limiting processes leading to the proof of the divergence theorem requires that the vector field A, as well as its first derivatives. It converts a volimc intcgral of thc diwrgcncc of a vector to a cluscd surface integral of the vector.

The cube is situated in the first octant of the Cartesian coordinate system with one corqer at'the origin. Refer to Fig. Back face: Example Left face: Right face: Top face: Bottom face: IL1"Pnd F. Hence, six faces. At outer surface: Actually, since the integrand is independent of 0 or. Adding thc two results, we have. To find the volume integral we first determinev. F for an F that has only an SOU a rr. Since V.

F is a constant, its voluqe integral equa1s: I cL'1: There is another lmi: The net circulation or simply circulation of a veqor. We have I i. I Circulation of A arbund contour i. The meaning of circulation depends on what kind of field be vector A represents. If A is a force acting on an AS" object, its circulation will be thg wor done by the force jn moving the object once " F ;. The familiar phenomenon of walcr whirling down a sink drain is a n suarnplc of a vortex sink causing a circulation of fluid velocity.

Sincc circulation as dclincd in Eq. We definet. In words, Eq. Because the normal to an d surface. This is illustrated in Fig.

The component of V x A in any 3 called a s another. We now use Eq. Ilcfcr to i: Note that dP is the same for sides 1 and 3, but that the integration on side Iis gcing upward a Az change in z , while that on side 3 is going downward a - Az change. Combining Eqs. The H. Similarly, it may be shown. Ay Az. The entlre expression 1'0s tllc curl ol'. Comparcd to lllc cxprcssion for V.

A is a scalar. Fortunatciy Eq. However it is more involved because in curvilinear coordinates not only A but also ' -A- -0 dP changes - m magnitude,as the integration of A dP is carried out on opposite sides of a curvilinear rectangle. It is apparent from Eq.

Solut iorz , a In cylindrical coordinates th; following apply: Wc havc. The - by expressions for V x A given in Eqs. In 0ht: For an arbitrary surface S, we can subdivide it into many, say: Figure shows such a scheme with Asj as a typical differential element. Adding the contributions of all the differential areas to the flux, we have. As with the divergence t h e o r i d the validity of the limiting processes leading to the Stokes's theorem requirei thqi the vector field A, as well as its first derivatives, exist and be continuous both on S and along C.

The pcmctry in Fig. The simplest open surface would be3wo-dimensional plane or disk with its circumference as the contour. We remind ourselves here that the directions of dP and ds a, follow the right-hand rule. Let us first find the urfaje integral of V: Prom Eq. Wc citn surfice. Like and get the same result. But it would be quite wrong if the 0 to 3 range were used as Jons in the range of integration for both x and y. Do you know why? For the line integral around AB0'4 we hdve already evaluated the part around I C ill bt no 4 ',.

From R to 0: Of course, Stokes's theorem has been established in Eq. We worked out the example above for practice on surface and line integrals. Two identities involving repeated del operations are of considerable importence in the study of electromagnetism, especially when we introduce potential functions. We shall discuss them separately below.

In words, curl of the gradient of m y scalar Jield is identically zero. The existence of V and itsfitst derivatives evcrywhcre is implied here. The combination of Eqs. Since 4 coordinate system is qot specified in the deriva- tiox the identity is a gencral Qne a i d is invariant jwith the choiccs of coordinatc I q; stems. F 4 converse statement r: Let a vector field be E.

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Then, if: The negative sign here is unimportant'as far as Identity I is concerned. It is included in Eq. We know from Section that a curl-free vector field is a conservative field: V x A PC on the left slde. Applying the divcrgqncc thcorcm, wu. The closed surface S can be spli into two open surfaces.

Since the contours. Because this is true for any arbitrary volume, the ic l a m integrand itself must be zero, as indicated by the identity in Eq. A converse statement of Identity I1 is as follows: Lct ;I vcctor held be B. We are reminded of the circling magnetic flux lines of a solenoid or an inductor. As we will see in Chapter 6, magnetic flux density B is solenoidal 2y using and can be expressed as the curl of another vector field called magnetic vector I general potential A.

A vector field F is Stokes's 1. Solenoidal and irrotational if vritn 1. I charge-li. Solenoidal but not irrotational if V.

A steady magnetic field in a current-carrying conductor. A static electric field in a charged region. Neither solenoidal nor irrotationnl ii , V-F O ' rr. The most general vcctor licld thcn has holh-a nonzero divcrgcncc: I wnzcro curl. In an unbounded region we assume that both the divergence and the curl of the vector field vanish at infinity. If the vector field is confined within a region bounded by a surface, then it is determined if its divergence and curl throughour-the region, as well as the normal componeht of the vector over the bbundirlg surface, are given.

Here we assume that the vector'functfon is single-valued and that its derivatives are finite and continuous. We have I. I 2- and I. Afarhemoricaf Jferlt s for Phydcisrs,.

Section 1. The proccdurc for obtaining F from given g and G is not obvious at this t! The fact that Fi is irrotational enables us to define a scalar potential function V, in view of identity 2- M , such that i. Helmholtz's theorem states that a general vector function F can be written as the sum of the gradient of a scalar function and the curl of a vector function. Example Given a vector function I: M Determine the scalar potential function V whose negative gradient equals F.

Each-wqlponent of V x F must vanish. Examination of Eqs. The constant is to be determined by a boundary condition or the condition at infinity. B and A x B in Cartesian coordinates? I ilp ;I. True or false'! True or false'? Truc o r Salsc'? True or false? R Explain how a general vector functiomcan be expressed in terms of a scalar potential function and a vector potential function. What is the location of the point a in Cartesian coordinates? Is this E a conservative field? V f in Cartesian coordinates.

A ds over the triangular area. C Can A be,expmsed as the gradient of a scalar? Problem Graph for. V x A s O j i by expansion in general orthogonal curvilinear coordinates.

They are the definition of basic quantities, the development of rules of operation, and the postulation of fundamental. We are now ready to introduce the fundamental postulates for the study of source-field relationships in electrostatics. In electrostatics, electric charpcs the sourccs are at rcst. The development of electrostatics in elementary physics usually be,. This law states that the force between two charged bodies, q , and q 2 , that are very small compared with the distance of separation, R l 2 , is proportional to the.

Using vector- notation, Coulomb's law can be written mathematically as. Electrostatics can proceed from Coulomb's law to define electric field intensity E, electric scalar potential, V, and electric flux density, D, and then lead to Gauss's law and other relations. We maintain, however, that Coulomb's law, though based on experimental evidence, is in fact also a postulate. Consider the two'stipulations of Coulomb's law: The question arises regarding the first stipulatjonl,How small must!

In practice the charged bodies cannot be of vanishing sizes idcal point charges , and there 'is dificuity in determining the "true" distancz between two bodies ol finite dimensions. For given body sizes. However, practical cdnsiderations weakness of force, existence of extraneous charged bodies, etc. We deiive Gauss's law and Coulomb's law from the divergepce afid curl relatiobs, and do not present them as separatc postulates. Field behaviors 'in material media will be studied and expressions for elec- trostatic energy and forces will be developed.

Wilhqfls, J.

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Faller, and H: We need only consider one of the four fundamental vector field quantities of the electromagnetic model discussed in Section. Furthermore, only the permittivity of free b's law: The test charge q, of course, cannot be zero in practice; as a matter of bct, it cannot be less tlian the charge on an electron.

However, the finiteness of the test charge ivould not maks the measured E differ appreciably from its calculated value if the test charge is small enough not to disturb the charge distribution of the source.

They are. Thasc iwo postulales arc concise, simple, and indepenhent of any c is. See Eq. Ii I 8 Equations and arq polnt relations; that is, tbeyhold at every point in space. They are referred to as thq diffeiential form of fhe pqstulates of electrostatics, since both divergence and curl operations involve spatial derivatives.

In practical applications we are usually interested in the total field ofan aggregate or a distribution of charges. This is more conveniently 'obtained by a;? Taking the volume integral of both sides of Eq. Gauss's law is one i f the most important relations in cicctrostatics. Wc will discuss it furthcr in Scctiol! An integral form can also be 06tained for the hurl relation in Eq. We have.. As a yatte?

P -ch; r.. Since a polnt charge has no preferred directions, its electric field must be everywhere radial 2nd bas the sicintansity: Equation tells us that ille electric jirld intensity o j u point charge is is the ourward radial direction and bqs a magairude propoytional ro the charge and incersely proportional to the square of the distance born the charge.

This is a very important basic formula in electrostatics. It i s readily verified: Let thc position vector of q be R' and that of a field poiht P be R, as shown in Fig. Then, from Eq. Example Determine the electric fieid intensity a1 P All dimensions are in meters. The position. The permittivity of air is essentially the same as that of the free space. When a point charge y, is placed in the ficld of another point chargc 1, 'lt the origin, a force F,, is experienccd by 4, due to electric field intensity E,, of q , at y2.

Equation is a mathcmatical form of Coulomb's luw alrcady statcd in Scction 3- 1 in conjunction with Eq. Note that the exponent on R is exactly 2, which is a consequence of the fundamental postulate Eq. Deflection -f. I P ,, Fig. Example The electrostatjc de1: The field. Edexerts a force on the electrons each carryng a charge - e, causing a deflection in the y direction.

Integrating again, we have. Note that the electrons have a parabolic trajectpry between the aeflection plates. During that time there is an additional vertical deflection. Supposc a n clac1rost: Since electric field intensity is a linear func. Although Eq. Let the center of the dipole coincide with the origin of a spherical coordicate System.

Similarly, for the second term on the right side of Eq. Substitution of Eqs. The derivation and intcrprolati. The electric dipole is an irqporkant entity in tGe study of the electric field in dielectric media. We see that E of a dipole is inversely propoirional to the cube of the distance R. I Except for some especially sikp14 ases, the vectdr triple ibtegral in Eq. For a line charge, we have Let us assume that tlje. We are perfectly frce. On the ather hapd. Gauss's law would not be of much help.

TLessence of applying Gauss's law lies first in the recognition of symmetry conditions, and second in thqsdtable choice of a surface over which the normal component of E resulting fro? Gauss's law cc uld nbt help in the Ycrivqtion of Eq. This problem was solved in Example by using Eq. With the obvious cylindrical symmetry, we construct q'cylindrical Gaussian surface of a radius r and an arbitrary lenith L with the line charge as its axis, as shown in Fig. Infinitely long. There is no contribution from the top or the bottom-face of the cylinder because on Ihc top f x c 11s This result is, of course, the same as that given in Eq.

We note that the length, L, of the cylindrical Guassian surface does not appear in the final expression; hence we could have chosen a cylinder of a unit length.

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Example Determine the electric field intensity of an infinite planar charge with I. Gauss's law can be used to much advantage here. Applying Gauss's law to anhfinite planar charge surface charge, p , Example We choose 3s the Gaussian ourSac,c n rectangular box with top and bottoln ticcs of an arbitrary area A equidistant from the planar charge, as shown in Fig The sides of the box are perpendicular to the charged sheet.

If the charged sheet coincides with the xy-plane, then on the top face. Example Determine the E field caused by a spherical cloud of electrons with. First we recognize that the given source condition has spherical symmetry. The proper Gaussian surfaces must therefore be concentric spherical surhces.

We must find the E field in two regions. On this surface, E is radial and has a constant magnitude. The total outward E flux is. I Example Substitution into Eq. We obtain i h e same expression lor jL0E ds as in case a. We observe that aurside the charged claud the E field is exactly the same as though the total chafige is concentrated on a single point charge at the center. The variation of ER versus R is plotted in Fig.

This ir a hopelevly involved process. The moral is: This induccs us to '. The reason lor the inclusion of a negative sign in Eq. The proper Gaussian surfaces must therefore be concentric spherical surfaces. On this surface. E is radial and has a constant magnitude. In faces 'ig. Electric potential does have physical significance, and it is related to the work done in carrying a charge from one point to another. In Section we defined rhe , electric field intensity as the force acting on a unit test charge.

Therefore, in moving a unit charge from point P , to point P , in an electric field, work must be done against ' aard the the,jield and is equal to I-. This would be contrary to the principle of conservation of energy.

We have nlrcndy alluded to the path-independence nature of the scalar line integral of the irrotationni f- conservative E field when we discussed E q LYl, the difference in electric potential energy of a unit charge between point P, and point c at the P I.

Denoting the electric potential energy per unit charge by V, the electric potenrial, ion cvcn wc have. Direction of iwrcming V ' Fig. What we have defined in Eq, is a ptentid difference elcrrrorroric i. It makes no moresense l;o talk about the absolute potential of a point than about the absolpte phasc of a, phasor ar thc absolutc altitude of a geographical location: We want to make two mqre about Eq; For instance. The E field is directed from pdsitive to negative chitrges. Charge Distribution!

However, the concentric circles spheres passing through P, and P, are equipotential lines surfaces and Vp2- 1. Since this is a scalar sum, it is, in general. The distances from the charges to a field point P are.

The potential at P can be written down directly:. The "approxima! The E field can be obtaineq fro'rn - VV. In spherical coordinates we have: Equation is the shme as eq. The equation of an eqiipotential surfade of q charge distribution is ob- tained by setting the expressiop for V to equal a constant. Since q, d, and E, in Eq.

Hence the equation' for an equipotential surface is 1.

I where c, is a constant. By plotting R versus 8 for various values of c,, we draw the solid equipotential lines in Fig. We set -,. R h cE sinZ8, They are rotqtion llysfmmetrical abdut thq z-axis independent of 0 and are everywhere normal to! The electric potential due to a ; c p h u o u s qistributiqn of charge confined in a I eiven'region is obtained by intdgrafirlg the contribpdon of an element of charge over ;he charged region.

Although the disk has circular symmetry, we cannot visualize a surface around it over which the normal component of E has a constant magnitude; hence Gauss's law is not useful for the solution of this problem. Working I with cylindrical coordinates indidated in Fig. For very large z, it is convenient to expand the second term in Eqs. Hence, when the point of observation is very far away from the charged disk, thpE field approximatelj follows the inverse square law as if the total charge were coqcentrated at a point.

I, - Example Obtain a fo;rnuln b r the electric field iltensity along the axis of a uniform line charge of length 4. The uniform Iioe-dmrgfi'densit is p,. For an infinitely lopg line charge, the E fieldtpn be determined readily by applying Gauss's law. Bs iq the solution to J3xamplp However, for a line charge of finite length, as showq in Fig.

Instead, we use Eq. The distance R from the charge elebent. L 2 The E field at P is the negative gradient of V with respect to the unprimed field b coordinates. Textbook Solutions. Get access now with. Get Started. Select your edition Below by. David K Cheng.

How is Chegg Study better than a printed Field and Wave Electromagnetics student solution manual from the bookstore? Can I get help with questions outside of textbook solution manuals? How do I view solution manuals on my smartphone?Zukauskas, V. A di'splacement de by a conducting body would result in a changc in total elcctrostatic energy and require the sources to transfer charges to the conductors in order to keep them at their fixed potentials.

Specify its location a in Cartesiali coordinates, and b in. Each-wqlponent of V x F must vanish. How is Chegg Study better than a printed Field and Wave Electromagnetics student solution manual from the bookstore? Of course, Stokes's theorem has been established in Eq. Dick Morton i.: Butterworth, V. Liu, and K.

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