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DAVID K. CHENG SYRACUSE UNIVERSITY A vv ADDISON-WESLEY PUBLISHING COMPANY Reading, Massachusetts Menlo Park, California London. Then this book will help you improve the standard of your written English. It has been written Improve Your Written Eng. David K. Cheng - Field and Wave Electromagnetics - Ebook download as PDF File .pdf), Text File .txt) or read book online.

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Field and Wave Electromagnetics - 2nd Edition - David K. Cheng - Ebook download as PDF File .pdf) or read book online. Field and Wave Electromagnetics (2nd Edition)-David K. Cheng - Ebook download as PDF File .pdf) or read book online. Download Book names as Field And Wave Electromagnetics by David K. Cheng with Solution Pdf easily without any login or registration.

The book is especially dedicated to introduce basics to students. The first group takes the traditional development: The second group takes the axiomatic development: This is a deductive approach. Table of Contents: The Electromagnetic Model 2.

Vector Analysis 3. Solutions of Electrostatic problems 5. Steady Electric Currents 6. Steady Magnetic Fields 7. Plane Electromagnetic Waves 9. Wave guides And Cavity Resonators Antennas And Radiating Systems. Download Complete Book: Download Book Solution: I recommend that it will be the best book to cover the whole syllabus and will also be good to clear the basic concept of Field And Wave.

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,. 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. 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.

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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.

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. 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. 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.

Solution Manual - Field and Wave Electromagnetics - David K. Cheng.pdf

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.

For this problem,. The potential V , if desired, may be obtained from E by integration. Wc now cxaminc: I n gcncral, we classify materials according to their electrical properties into three types: Most metals belong to this group. The electrical properties of semiconductors fall between those of conductors and insulators in that they possess a relatively small number of freely movable charges.

Between these energy bands there may be forbidden regions or gaps where no eiec- n trons of t b w l f d ' s atom can reside. Conductors have an upper energy band partially filled with electrons or an upper pair of overlapping bands that are partially filled so that the electrons in these bands can move from one to another with only a small change in energy.

Insulators or dielectrics are materials with a completely filled upper band, so conduction could not normally. If the energy gap of the forbidden region is relatively small, small amounts of external energy may be sufficient to excite the electrons in the filled upper bapd to jump into the next band, causing conduction.

Such materials are semiconductors. An electric field will be let up in the conductor, the field exerting a force on the charges and making them nji,ve away from one another. This movement will continua'until all the charges reach the conductor surface and redistribute themselves ip suck a way that both thacharge and tfie field inside vanish.

Hence, I ', I. In othet: A finite limc is rc irql for thc chqrgcjs to f;pdistributc on a conductor. J surface and reach the eguilibri m State. This time qepenqp on the conductivity of the b copper, this tim is in the order of l 9 s , material. This ppinh will be elaborated in Section In order to find E,,, the norm 1 cfiflponent of E ailthe fprfacc of the conductor, we!? The I we are I c f static: Id ayd i Ps I 1 oduced. Using Eq. Summarizing the buundury condifions at the conductor surhcc, we have!

When the surface charge distribution I reaches an equilibrium, all four relations. Determine E and V as functions tor, we of the radial distance R.

Since there is I. There are three p t i n a regions: The E field is ths kame as that o f a point charge Q without the presence of the shell. The potential referring to the'point at infinity is , Because of Eq. This also means an amount of positive. The cull- ducting shell is an equipotential body. When a dielectric body is placed in an external electric field, there are no induced free charges that move to the surface and make the interior charge density and electric field vanish, as with conductors.

How- ever, since dielectrics contain bound charges, we cannot conclude that they have no effect on the electric field in which they are placed. Int PO]. Her dielectric material. The vector P, a smoothed point function, is the volume density of electric dipole -S moment. The first volume integral on the right side of Eq. Comparissn of the two integrals on the right side of Eq.

The sketch in Fig. The net total ' mils. This relation as given in Eq. The net charge remaining ' within the volume V is the negative of this integral. Hence, ' when the divergence of P does not vanish, the bulk of the polarized dielectric appears ;;! However, since we started with an electrically neutral dielectric body, ' 11ca thc total charge of the body after polarization must remain zero. In particular, the divergence postulated in Eq.

The corresponding integral fyrrq ok Eq. We have: We write: Air has a dielectric constant of l. The dielectric constants gf s f! Ilr, is indcpcndcnt ofposition, ledium. The relative permittivity of a simple medium is a constant. I volume Example A positive point charge Q is at the center of a spherical dielectric shell of an inner radius Ri and an outer radius R..

The dielectric constant of the shell is c,. The geometry of this problem is the same as that of Example The conducting shell has now been replaced by a dielectric shell, but the procedure of solution is similar. Because of the spherical symmetry, we apply Gauss's law to. The situation in this region is exactly the same as that in Example We have, from Eqs.

Q 4nc,R 1 From Eqs. Dielectric shell. The plot for V in Fig. We note that DR is a con- tinuous curve exhibiting no sudden changes in going from one medium to another and that P R exists only in the dielectric region. It is instructive to compare Figs. From Eqs. Q have a ,m. Equations , ;and indicate that there is no net polarization volunle charge inside the dielectric shell.

However, negative polarization surface charges exist on the inner surface; positive polarization surface charges, on the outer! Ir , Table qic14lwf! These surface charges produce an electric fisld intensity that is directed radially inward, thus reducing tbe. Free char'ges w 11 appear. Tbis phenomenon is called a 4 dielectric breakdown. This is 'the principle upon which a. Find a the charges on the two spheres, and b the electric field intensities at the sphere surfaces.

Solution a Refer to Fig. Since the spherical conductors are at the same potential, directed we have rg,: But, i'nlancnt since: The b1 Q and Q. Thdfaces ave an area AS, and the fi: Soil1 on: See C. T Tal. Unit vectors a,, and a,, are, respectively, outward unit normals to media 1 and 2. Recapitulating, we find the boundary conditions that must be satisfied for static electric fields are as follows: Determine Ei, Di, and Pi inside the lucite. We assume that the introduction of the lucite sheet does not disturb the original uniform electric field E,.

The sitnation is depicted in Fig. I , interfaces are perpendicular to the electric field, only the normal field components need be considered.

No free charges exist. I, 1 it. There is no change in electric? The electric field intensity in mediuni 1 at the point F, has n magnitude E l and makks an angle cr, with the normal.

Deter- mine the magnitude and direction of the electric field intensity at point P , in medium 2. After Solution: Using Eqs. From Section we understand that a conductor in a static electric field 1s an equipotential body and that charges deposited on a conductor will distribute them- selves on its surface in such a way that the ilectric field inside vanishes. Suppose the potential due to a charge Q is V.

Obviously, increasing the total charge by some factor k would merely increase the surface charge density p, everywhere by the same factor, witlrout-affecting the charge distribution because the conductor remains an equipotential body in a static situation. We may conclude from Eq.

David K. Cheng - Field and Wave Electromagnetics

But, from Eq. The capacitance is the electric charge that must be added to the body per unit increase in its electric potential. Its SI unit is coulomb per volt. Of considerable importance in practice is the capacitor which consists of two conductors separated by free space or a dielectric medium. The conductors may be of arbitrary shapes as in Fig.

Several eiectric field lines originating from positive charges and terminating on negative charges are shown in Fig. Notc tha! Equation applies here if V is taken to mean the potential ditrerence between the two conductors, V, ,. That is,.

The capacitance of a capacitor is a physical property of the two-conductor system. It depends on the geometry of the conductor6 and on the permittivity of the medium between them; it does not depend on either the charge Q or the potential difference V , ,.

A capacitor has a capacitance even when no voltage is applied to it and no free charges exist on its conductors. Capacitance C can be determined from Eq. At this stage, since we have not yet. Choose an apprdpriate coordinate system for the given geometry.

Assume chargks 4 Q and --Q on the conductors. Find E from Q by Eq. I volt, or 4. Find V I 2by evaluating. The space between the plates is filled elween with a dielectrlc of a constant permittivity E. Determine the capacitance. A cross section of the capacitor is shown in Fig.

For this problem we-could have started by assuming a potential difference V12 between the upper and lower plates. The electric field intensity between the plates is uniform and equals. The space between the conductors is filled with a dielectric of permittivity c, and the length of the capacitor is L Deter- mine the capacitance of this capacitor.

W e use cylindrical coordinates for this problem. Note that Eq. Since the conductor surfaces are not planes here, the E field is not constant in the dielectric and Eq. A cylindrical capacitor Example 3- Again we neglect the fringing effect of the field hear the edges of the conductors. The potential difference Between the inner and outer conductors is. We could not soive this problem from an assumed Vuhbecause the electric field is not uniform between the inner and outer conductors.

Thus we wouid not know how lo cxprcss li and Q in lcrms of Vuhuntil wc l c m c d how to solve such a boundary- value problem. Example A sphcriwl capacitor consists of an inncr conducting spllcrc nl radius R iand an outer conductor with a spherical inner wall of radius R,.

The space in-between is filled with a dielectric of permittivity s. Applying Gauss's law to a spherical. Capacitors are often combined in various ways in electrk-hcuits. The two basic ways are series and parallel connections. In the series, or head-to-tail. In the parallel connection of capacitors, the external terminals are connected to the conductors of all the capacitors as in Fig. When a potential difference V is applied to the terminals, the charge cumulated on a capacitor depends on its capacitance.

The total charge is the sum of all the charges. A DC voltage of V is applied to the external terminals a-b. Determine the following: The total equivalent capacitance C, is then. We have four unknowns: Q,, Q,, Q,, and Q,. Four equa- tions are needed for their determination. Series connection of C , and C 2: Q3 Q4. Using the given values of C,, C,.

I 11 In Section I. In order to bring a charge Q, slowly, so that kinetlc energy and radiation effects may be neglected from lnfinlty againrt the field of a charge Q, in free space to a distance R12, the amount of work required is. Now suppose adother charge Q, is brought from.

The sum of AW in hq. W,, stored in the asserhbiy of the three charges Q,, Q,, and Q,. That is. V2 and V, are the potentials, respectiyely. The purpose of the subscript a on of an electric nature. We have CC; is to denote that the energy is 1I 1. First, We can be negative. For instance, W2 in Eq. In that case, work is done by the field not against the field established by Q , in moving Q, from Infinity.

Example Find the energy required to assemble a uniform sphere of charge of radius b and volume charge density p. Because of symmetry. In terms of the total tharge 11; in Eq. The sphere of charge in Fig. Let the for discrete charges must be modified. Without going through a separate proof, wz e potential replace Q, by p dv and the summation by an integration and obtain.

In Eq. Example Solvc the problcm in Example 3- 19 b y using Eq. Since p is a constant, it can be taken out of the integral sign. For a spherically symmetrical problem,. To find V at R, we must find the negative of the line integral of E in two regions: Note that y,in Eq. As a matter of fact. A, the radius h approaches zero, the self-energy of a mathematical point charge of a given Q is infinite see Eq.

The self-energies of pomt charges Q, are not included in Eq. Of course, there are, strictly, no I point charges inasmuch as the smallest charge unit, the electron, is itselfa distrrbution I of charge. T o this end, we substitute V D for p in Eq. Hence the surface integral in Eq. As a and r-cncrgy of ncrgy uf a: If-energies strictly, no. However, this definition of energy density is artificial because a physical justification has not been found to localize energy with an electric field; all we know is that the volume integrals in Eqs.

Example In Fig.A b Parallelogram rule. Si Units and Universal Constants. David K. Antennas And Radiating Systems. Again we neglect the fringing effect of the field hear the edges of the conductors.

Ii I 8 Equations and arq polnt relations; that is, tbeyhold at every point in space. Q have a ,m. But, from Eq.

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