Electromagnetic Potentials and Topics for Circuits and Systems

Transcription

1 C H A P T E R 5 Electromgnetic Potentils nd Topics for Circuits nd Systems In Chpters 2, 3, nd 4, we introduced progressively Mxwell s equtions nd studied uniform plne wves nd ssocited topics. Two quntities of fundmentl importnce, resulting from Mxwell s equtions in differentil form, re the electromgnetic potentils: the electric sclr potentil nd the mgnetic vector potentil. We introduce these quntities in this chpter nd lso consider severl topics of relevnce to circuits nd systems. We begin the discussion of topics for circuits nd systems with two importnt differentil equtions involving the electric potentil nd discuss severl pplictions bsed on the solution of these equtions, including the nlysis of p-n junction semiconductor nd rrngements involving two prllel conductors. We then introduce n importnt reltionship between the (lumped) circuit prmeters, cpcitnce, conductnce, nd inductnce for infinitely long, prllel perfect conductor rrngements, nd consider their determintion. Next we turn our ttention to electric- nd mgnetic-field systems, tht is, systems in which either the electric field or the mgnetic field is predominnt, leding from qusisttic extensions of the sttic fields existing in the structures when the frequency of the source driving the structure is zero. The concepts of electric- nd mgnetic-field systems re importnt in the study of electromechnics. We shll lso consider mgnetic circuits, n importnt clss of mgnetic field systems, nd the topic of electromechnicl energy conversion. Mgnetic vector potentil 5.1 GRADIENT, LAPLACIAN, AND THE POTENTIAL FUNCTIONS In Exmple 3.9, we showed tht for ny vector A, # A = 0. It then follows from Guss lw for the mgnetic field in differentil form, # B = 0, tht the mgnetic flux density vector B cn be expressed s the curl of nother 282

2 5.1 Grdient, Lplcin, nd the Potentil Functions 283 vector A; tht is, B = A (5.1) The vector A in (5.1) is known s the mgnetic vector potentil. Substituting (5.1) into Frdy s lw in differentil form, E = -0B>0t, nd rerrnging, we then obtin Grdient E + 0 0t 1A2 = 0 or E + 0A 0t b = 0 (5.2) If the curl of vector is equl to the null vector, tht vector cn be expressed s the grdient of sclr, since the curl of the grdient of sclr function is identiclly equl to the null vector. The grdient of sclr, sy,, denoted (del ) is defined in such mnner tht the increment d in from point P to neighboring point Q is given by d = # dl (5.3) where dl is the differentil length vector from P to Q. Applying Stokes theorem to the vector nd surfce S bounded by closed pth C, we then hve 1 2 # ds = # dl LS C = C d (5.4) = 0 for ny single-vlued function. Since (5.4) holds for n rbitrry S, it follows tht =0 (5.5) To obtin the expression for the grdient in the Crtesin coordinte system, we write d = 0 0x 0 0 dx + dy + 0y 0z dz = 0 0x x + 0 0y y + 0 0z zb # 1dx x + dy y + dz z 2 (5.6)

3 284 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems Then compring with (5.3), we observe tht = 0 0x x + 0 0y y + 0 0z z (5.7) Note tht the right side of (5.7) is simply the vector obtined by pplying the del opertor to the sclr function. It is for this reson tht the grdient of is written s. Expressions for the grdient in cylindricl nd sphericl coordinte systems re derived in Appendix B. These re s follows: CYLINDRICAL = 0 0r r r 0f f + 0 0z z (5.8) SPHERICAL Physicl interprettion of grdient = 0 0r r r 0u u r sin u 0f f (5.8b) To discuss the physicl interprettion of the grdient, let us consider surfce on which is equl to constnt, sy, 0, nd point P on tht surfce, s shown in Fig. 5.1(). If we now consider nother point Q 1 on the sme surfce nd n infinitesiml distnce wy from P, d between these two points is zero since is constnt on the surfce. Thus, for the vector dl drwn from P to Q 1, [ ] P # 1 dl1 = 0 nd hence [ ] P is perpendiculr to dl 1. Since this is true for 0 P dl 3 Q 3 Q 1 dl 1 dl 2 dl Q 2 n 0 P Q 0 d n () (b) FIGURE 5.1 For discussing the physicl interprettion of the grdient of sclr function.

4 5.1 Grdient, Lplcin, nd the Potentil Functions 285 ll points Q 1, Q 2, Q 3, Á on the constnt surfce, it follows tht [ ] P must be norml to ll possible infinitesiml length vectors dl 1, dl 2, dl 3, Á drwn t P nd hence is norml to the surfce. Denoting n to be the unit norml vector to the surfce t P, we then hve [ ] P = ƒ ƒ P n (5.9) Let us now consider two surfces on which is constnt, hving vlues 0 nd 0 + d, s shown in Fig. 5.1(b). Let P nd Q be points on the = 0 nd = 0 + d surfces, respectively, nd dl be the vector drwn from P to Q. Then from (5.3) nd (5.9), d =[ ] P # dl = ƒ ƒ P n # dl = ƒ ƒ P dl cos where is the ngle between t P nd dl. Thus, n ƒ ƒ P = d dl cos (5.10) Since dl cos is the distnce between the two surfces long n nd hence is the shortest distnce between them, it follows tht ƒ ƒ P is the mximum rte of increse of t the point P. Thus, the grdient of sclr function t point is vector hving mgnitude equl to the mximum rte of increse of t tht point nd is directed long the direction of the mximum rte of increse, which is norml to the constnt surfce pssing through tht point; tht is, = d dn n (5.11) where dn is differentil length long n. The concept of the grdient of sclr function we just discussed is often utilized to find unit vector norml to given surfce. We shll illustrte this by mens of n exmple. Exmple 5.1 Finding unit vector norml to surfce by using the grdient concept Let us find the unit vector norml to the surfce y = x 2 t the point (2, 4, 1) by using the concept of the grdient of sclr. Writing the eqution for the surfce s x 2 - y = 0 we note tht the sclr function tht is constnt on the surfce is given by 1x, y, z2 = x 2 - y

5 286 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems The grdient of the sclr function is then given by =1x 2 - y2 = 01x2 - y2 0x = 2x x - y x + 01x2 - y2 0y y + 01x2 - y2 0z The vlue of the grdient t the point (2, 4, 1) is [2122 x - y ] = 14 x - y 2. Thus, the required unit vector is z n = 4 x - y ƒ4 x - y ƒ = x yb Electric sclr potentil Returning now to (5.2), we write E + 0A 0t = - (5.12) where we hve chosen the sclr to be -, the reson for the minus sign to be explined in Section 5.2. Rerrnging (5.12), we obtin E = - - 0A 0t (5.13) Electromgnetic potentils The quntity in (5.13) is known s the electric sclr potentil. The electric sclr potentil nd the mgnetic vector potentil A re known s the electromgnetic potentils. As we shll show lter in this section, the electric sclr potentil is relted to the source chrge density r, wheres the mgnetic vector potentil is relted to the source current density J. For the time-vrying cse, the two re not independent, since the chrge nd current densities re relted through the continuity eqution. For given J, it is sufficient to determine A, since B cn be found from (5.1) nd then E cn be found by using Ampère s circuitl lw H = J +0D>0t. For sttic fields, tht is, for 0>0t = 0, the two potentils re independent. Eqution (5.1) remins unltered, wheres (5.13) reduces to E = -. We shll consider the sttic field cse in Section 5.2. To proceed further, we recll tht Mxwell s equtions in differentil form re given by E = - H = J + 0D 0t # D = r # B = 0 0B 0t (5.14) (5.14b) (5.14c) (5.14d)

6 From (5.14d), we expressed B in the mnner 5.1 Grdient, Lplcin, nd the Potentil Functions 287 B = A (5.15) nd then from (5.14), we obtined E = - - 0A 0t (5.16) We now substitute (5.16) nd (5.15) into (5.14c) nd (5.14b), respectively, to obtin # - - 0A 0t b = r e 0 A - me 0t - -0A 0t b = mj (5.17) (5.17b) We now define the Lplcin of sclr quntity, denoted 2 (del squred ) s Lplcin of sclr 2 = # (5.18) In Crtesin coordintes, = 0 0x x + 0 0y y + 0 0z z # A = 0A x 0x + 0A y 0y + 0A z 0z so tht 2 = 0 0x 0 0x b + 0 0y 0 0y b + 0 0z 0 0z b or 2 = 02 0x y z 2 (5.19) Note tht the Lplcin of sclr is sclr quntity. Expressions for the Lplcin of sclr in cylindricl nd sphericl coordintes re derived in Appendix B. These re s follows: CYLINDRICAL 2 = 1 r 0 0r r 0 0r b r 2 0f z 2 (5.20)

7 288 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems SPHERICAL 2 = r 2 r2 0r 0r b + 1 r 2 sin u 0 0u sin u 0 0u b + 1 r 2 sin 2 u 0 2 0f 2 (5.20b) Before proceeding further, it is interesting to note tht the four vector differentil opertions tht we hve lerned thus fr in this chpter re such tht The curl of vector is vector. The divergence of vector is sclr. The grdient of sclr is vector. The Lplcin of sclr is sclr. Lplcin of vector Thus, ll four combintions of vector nd sclr re involved in the four opertions. Next, we define the Lplcin of vector, denoted 2 A s 2 A = 1 # A2 - A (5.21) Expnding the right side of (5.21) in Crtesin coordintes nd simplifying, we obtin in the Crtesin coordinte system, 2 A = 1 2 A x 2 x A y 2 y A z 2 z (5.22) Thus, in the Crtesin coordinte system, the Lplcin of vector is vector whose components re the Lplcins of the corresponding components of A. It should, however, be cutioned tht this simple observtion does not hold in the cylindricl nd sphericl coordinte systems. (See, e.g., Problem P5.6.) Using (5.18) nd (5.21), we now write (5.17) nd (5.17b) s t 1 # A2 = - r e 2 A - # A + me 0 0t b - me 0 2 A 0t 2 = -mj (5.23) (5.23b) Potentil function equtions Equtions (5.23) nd (5.23b) re pir of coupled differentil equtions for nd A. To uncouple the equtions, we mke use of theorem known s Helmholtz s theorem, which sttes tht vector field is completely specified by its curl nd divergence. Therefore, since the curl of A is given by (5.15), we re t liberty to specify the divergence of A. We do this by setting # A = -me 0 0t (5.24)

8 5.1 Grdient, Lplcin, nd the Potentil Functions 289 which is known s the Lorenz condition 1. This uncouples (5.23) nd (5.23b) to give us r -me 0t 2 = - e A A - me 0t 2 = -mj (5.25) (5.26) These re the differentil equtions relting the electromgnetic potentils nd A to the source chrge nd current densities r nd J, respectively. Before proceeding further, we shll show tht the continuity eqution is implied by the Lorenz condition. To do this, we tke the Lplcin of both sides of (5.24). We then hve or 2 1 # A2 = -me 2 0 0t # 2 0 A = -me 0t 2 (5.27) Substituting for 2 A nd 2 in (5.27) from (5.26) nd (5.25), respectively, we get # 0 2 A me 0t mjb = -me 0t me 0 2 0t 2 - r e b or 0 2 me 0t 2 # 0 A + me 0t b = m 0r # J + 0t b (5.28) Thus, by ssuming the Lorenz condition (5.24), we imply # J +0r>0t = 0, which is the continuity eqution. As pointed out erlier in this section, it is sufficient to determine A for the time-vrying cse for given J. Hence, we shll be concerned only with (5.26), which we shll refer to in Section 10.1 in connection with obtining the electromgnetic field due to n elementl ntenn. K5.1. Mgnetic vector potentil; Grdient of sclr; Physicl interprettion of grdient; Electric sclr potentil; Lplcin of sclr; Potentil function equtions. 1 Note Lorenz condition nd not Lorentz condition. In editions 2, 3, nd 4 of this book, s well s extensively in books by other uthors, this condition hs been mistkenly ttributed to Lorentz insted of Lorenz. See the note Lorentz or Lorenz, by J. Vn Bldel in IEEE Antenns nd Propgtion Mgzine, Vol. 33, No. 2, April 1991, p. 69.

9 290 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems D5.1. D5.2. Find the outwrd pointing unit vectors norml to the closed surfce 2x 2 + 2y 2 + z 2 = 8 t the following points: () 112, 12, 02; (b) (1, 1, 2); nd (c) 11, 12, 122. x + y x + y + z 12 x + 2 y + z Ans. () (b) (c) 12 ; Two sclr functions re given by 1 1x, y, z2 = x 2 + y 2 + z 2 2 1x, y, z2 = x + 2y + 2z Find the following t the point (3, 4, 12): () the mximum rte of increse of 1 ; (b) the mximum rte of increse of 2 ; nd (c) the rte of increse of 1 long the direction of the mximum rte of increse of 2. Ans. () 26; (b) 3; (c) D5.3. Find the Lplcins of the following functions: () x 2 yz 3 ; (b) 11>r2 sin f in cylindricl coordintes; nd (c) r 2 cos u in sphericl coordintes. Ans. () 2yz 3 + 6x 2 yz; (b) 0; (c) 4 cos u. Potentil difference 5.2 POTENTIAL FUNCTIONS FOR STATIC FIELDS As lredy pointed out in the preceding section, Eq. (5.13) reduces to E = - (5.29) for the sttic field cse. We observe from (5.29) tht the potentil function then is such tht the electric field lines re orthogonl to the equipotentil surfces, tht is, to the surfces on which the potentil remins constnt, s shown in Fig If we consider two such equipotentil surfces corresponding to A B A B E Equipotentil Surfces FIGURE 5.2 Set of equipotentil surfces in region of sttic electric field.

10 5.2 Potentil Functions for Sttic Fields 291 = A nd = B, s shown in the figure, the potentil difference A - B is given, ccording to the definition of the grdient, by A A A - B = d = # dl L L B B B = - # dl L Using (5.29), we obtin B A - B = E # dl L A A (5.30) (5.31) We now recll from Section 2.1 tht B 1A E # dl is the voltge between points A nd B. Thus, the potentil difference in the sttic field cse hs the sme mening s the voltge. The reson for the minus sign in (5.13) nd hence in (5.29) is now evident, since without it the voltge between A nd B would be the negtive of the potentil difference between A nd B. Before proceeding further, we recll tht the voltge between two points A nd B in time-vrying electric field is in generl dependent on the pth followed from A to B to evlute B E # dl, since, ccording to Frdy s lw, 1A Potentil difference versus voltge E # d dl = - B # ds C dt LS is not in generl equl to zero. On the other hnd, the potentil difference (or voltge) between two points A nd B in sttic electric field is independent of the pth followed from A to B to evlute B E # dl, since, for sttic fields, 1A E # dl = 0 C Thus, the potentil difference (or voltge) between two points in sttic electric field hs unique vlue. Since the potentil difference nd voltge hve the sme mening for sttic fields, we shll herefter replce in (5.29) by V, thereby writing E = -V (5.32) Let us now consider the electric field of point chrge nd investigte the electric potentil due to the point chrge. To do this, we recll tht the electric field intensity due to point chrge Q is directed rdilly wy from the point chrge nd its mgnitude is Q>4peR 2, where R is the rdil distnce from the point chrge. Since the equipotentil surfces re everywhere orthogonl to the Electric potentil due to point chrge

11 292 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems E Equipotentils Q FIGURE 5.3 Cross-sectionl view of equipotentil surfces nd electric field lines for point chrge. field lines, it then follows tht they re sphericl surfces centered t the point chrge, s shown by the cross-sectionl view in Fig If we now consider two equipotentil surfces of rdii R nd R + dr, the potentil drop from the surfce of rdius R to the surfce of rdius R + dr is 1Q>4peR 2 2 dr or the incrementl potentil rise dv is given by dv = - Q 4peR 2 dr = d Q 4peR + Cb (5.33) where C is constnt. Thus, V1R2 = Q 4peR + C (5.34) Since the potentil difference between two points does not depend on the vlue of C, we cn choose C such tht V is zero t some rbitrry reference point. Here we cn conveniently set C equl to zero by noting tht it is equl to V1q2 nd by choosing R = q for the reference point. Thus, we obtin the electric potentil due to point chrge Q to be V = Q 4peR (5.35) We note tht the potentil drops off inversely with the rdil distnce wy from the point chrge. Eqution (5.35) is often the strting point for the computtion of the potentil field due to sttic chrge distributions nd the subsequent determintion of the electric field by using (5.32). We shll illustrte this by considering the cse of the electric dipole in the following exmple.

12 5.2 Potentil Functions for Sttic Fields 293 Exmple 5.2 Electric field of sttic electric dipole vi the potentil due to the dipole As we hve lerned in Section 4.2, the electric dipole consists of two equl nd opposite point chrges. Let us consider sttic electric dipole consisting of point chrges Q nd -Q situted on the z-xis t z = d>2 nd z = -d>2, respectively, s shown in Fig. 5.4() nd find the potentil nd hence the electric field t point P fr from the dipole. First, we note tht in view of the symmetry ssocited with the dipole round the z-xis, it is convenient to use the sphericl coordinte system. Denoting the distnce from the point chrge Q to P to be r 1 nd the distnce from the point chrge -Q to P to be r 2, we write the expression for the electric potentil t P due to the electric dipole s Electric dipole V = Q + -Q = 4per 1 4per 2 Q 4pe 1 r 1-1 r 2 b For point P fr from the dipole, tht is, for r d, the lines drwn from the two chrges to the point re lmost prllel. Hence, r 1 L r - d 2 cos u r 2 L r + d 2 cos u nd 1-1 = r 2 - r 1 L d cos u r 1 r 2 r 1 r 2 r 2 z P(r, u, f) Equipotentils x Q d/2 d/2 u f Q r 1 r r 2 y Direction Lines () (b) FIGURE 5.4 () Geometry pertinent to the determintion of the electric field due to n electric dipole. (b) Cross sections of equipotentil surfces nd direction lines of the electric field for the electric dipole.

13 294 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems so tht V L Qd cos u 4per 2 = p # r 4per 2 (5.36) where p = Qd z is the dipole moment of the electric dipole. Thus, the potentil field of the electric dipole drops off inversely with the squre of the distnce from the dipole. Proceeding further, we obtin the electric field intensity due to the dipole to be 0 E = -V = - 0r Qd cos u 4per 2 b r - 1 r = Qd 4per 3 12 cos u r + sin u u 2 0 0u Qd cos u 4per 2 b u (5.37) Eqution (5.36) shows tht the equipotentil surfces re given by r 2 sec u = constnt, wheres from (5.37), it cn be shown tht the direction lines of the electric field re given by r cosec 2 u = constnt nd f = constnt. These re shown sketched in Fig. 5.4(b). Alterntive to using the eqution for the direction lines, they cn be sketched by recognizing tht (1) they must originte from the positive chrge nd end on the negtive chrge nd (2) they must be everywhere perpendiculr to the equipotentil surfces. Electrocrdiogrphy A technique in everydy life in which the potentil field of n electric dipole is relevnt is electrocrdiogrphy. This technique is bsed on the chrcteriztion of the electricl ctivity of the hert by using dipole model. 2 The dipole moment, p, referred to in medicl literture s the electric force vector or the ctivity of the hert, sets up n electric potentil within the chest cvity nd chrcteristic pttern of equipotentils on the body surfce. The potentil differences between vrious points on the body re mesured s function of time nd re used to deduce the temporl evolution of the dipole moment during the crdic cycle, thereby monitoring chnges in the electricl ctivity of the hert. We shll now consider n exmple for illustrting method of computer plotting of equipotentils when closed form expression such s tht for the electric dipole of Exmple 5.2 is not vilble. Computer plotting of equipotentils Exmple 5.3 chrges Computer plotting of equipotentils for set of two point Let us consider two point chrges Q 1 = 8pe 0 C nd Q 2 = -4pe 0 C situted t 1-1, 0, 02 nd (1, 0, 0), respectively, s shown in Fig. 5.5.We wish to discuss the computer plotting of the equipotentils due to the two point chrges. First, we recognize tht since the equipotentil surfces re surfces of revolution bout the xis of the two chrges, it is sufficient to consider the equipotentil lines in ny plne contining the two chrges. Here we shll consider the xz-plne. The equipotentil lines re lso symmetricl bout the x-xis, nd, hence, we shll plot them only on one 2 See, for exmple, R. K. Hobbie, The Electrocrdiogrm s n Exmple in Electrosttics, Americn Journl of Physics, June 1973, pp

14 5.2 Potentil Functions for Sttic Fields 295 z V E E V 0 V 0 V 0.1 Q 1 1 E 0 V 0 FIGURE 5.5 For illustrting the procedure for the computer plotting of equipotentils due to two point chrges. Q 2 1 x side of the x-xis nd inside the rectngulr region hving corners t 1-4, 02, 14, 0214, 52, nd 1-4, 52. As we go from Q 1 to Q 2 long the x-xis, the potentil vries from + q to - q nd is given by The vlue of x lying between -1 nd 1 for given potentil is then given by or x = d 8pe 0 V = 4pe x2-4pe 0 4pe x2 = 1-3x 1 - x 2 V 0 = 1-3x 1 - x V V 0 2 2V 0 for V 0 Z for V 0 = 0 We shll begin the equipotentil line t this vlue of x on the x-xis for given vlue of V 0. To plot the line, we mke use of the property tht the equipotentil lines re orthogonl to the direction lines of E so tht they re tngentil to the unit vector 1E z x - E x z 2>E. We shll step long this unit vector by smll distnce (chosen here to be 0.1), nd if necessry, correct the position by repetedly moving long the electric field until the potentil is within specified vlue (chosen here to be V) of tht for which the line is being plotted. To correct the position, we mke use of the fct tht V = -E. Thus, the incrementl distnce required to be moved opposite to the electric field to increse the potentil by V is V>E, nd, hence, the distnces required to be moved opposite to the x- nd z-directions re 1 V>E21E x >E2 nd 1 V>E21E z >E2, respectively. The plotting of the line is terminted when the point goes out of the rectngulr region. V 0

15 296 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems (4, 5) 1 8 (4, 5) 1 4 (4, 0) Q 1 Q Q 1 8pe 0 C Q 2 4pe 0 C (4, 0) FIGURE 5.6 Personl computer-generted plot of equipotentils for the rrngement of two point chrges of Fig The vlues of potentils re in volts. The computer plot obtined from run of PC progrm tht crries out this procedure for vlues of potentils rnging from -2 V to 4 V is shown in Fig It should, however, be pointed out tht for complete plot, those equipotentil lines tht surround both point chrges should lso be considered. The computtion of potentil cn be extended to continuous chrge distributions by using superposition in conjunction with the expression for the potentil due to point chrge, s in the cse of electric field computtion in Section 1.5. We shll illustrte by mens of n exmple. Potentil due to line chrge Exmple 5.4 Electric potentil field of n infinitely long line chrge An infinitely long line chrge of uniform density r L0 C>m is situted long the z-xis. It is desired to obtin the potentil field due to this chrge. First, we divide the line into number of infinitesiml segments ech of length dz, s shown in Fig. 5.7, such tht the chrge r L0 dz in ech segment cn be considered s point chrge. Let us consider point P t distnce r from the z-xis, with the projection of P onto the z-xis being O. For the ske of generlity, we consider the point P 0 t distnce r 0 from O long OP s the reference point for zero potentil nd write the potentil dv t P due to the infinitesiml chrge r L0 dz t A s dv = = r L0 dz 4pe 0 1AP2 - r L0 dz 4pe 0 1AP 0 2 r L0 dz r L0 dz - 4pe 04 r 2 + z 2 4pe 04 r z 2 (5.38)

16 5.2 Potentil Functions for Sttic Fields 297 r L0 dz A z O 0 P 0 P r 0 r FIGURE 5.7 Geometry for the computtion of the potentil field of n infinitely long line chrge of uniform density r L0 C>m. We will, however, find lter tht we hve to choose the reference point for zero potentil t finite vlue of r, in contrst to the cse of the point chrge for which the reference point cn be chosen to be infinity. The potentil V t P due to the entire line chrge is now given by the integrl of (5.38), where the integrtion is to be performed between the limits z = -q nd z = q. Thus, V = L q z =-q = r L0 2pe 0 L dv = L q q z = 0 z =-q dz - 4 r2 + z 2 r L0 dz - 4pe 04 r 2 + z 2 dz 4 r z 2 b r L0 dz 4pe 04 r z 2 b (5.39) Introducing z = r tn nd z = r 0 tn 0 in the first nd second terms, respectively, in the integrnd on the right side of (5.39), we hve V = r p>2 p>2 L0 sec d - sec 2pe 0 d 0 b 0 L L = 0 = r L0 5[ln1sec + tn 2] p>2 2pe = 0 - [ln1sec 0 + tn 0 2] p>2 0 = 06 0 = r L0 1 4 r 2 + z 2 + z2r 0 cln 2pe r z 2 + z2r r L0 r = - ln 2pe 0 r 0 0 = 0 q d z = 0 (5.40) In view of the cylindricl symmetry bout the line chrge, (5.40) is the generl expression in cylindricl coordintes for the potentil field of the infinitely long line chrge of uniform density. It cn be seen from (5.40) tht choice of r 0 = q is not good choice, since then the potentil would be infinity t ll points.the difficulty lies in the fct tht infinity plus finite number is still infinity. We lso note from (5.40) tht the equipotentil

17 298 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems surfces re ln r>r 0 = constnt or r = constnt, tht is, surfces of cylinders with the line chrge s their xis.the result of Exmple 2.6 shows tht the electric field due to the line chrge is directed rdilly wy from the line chrge. Thus, the direction lines of E nd the equipotentil surfces re indeed orthogonl to ech other. Mgnetic vector potentil due to current element We shll now turn our ttention to the mgnetic vector potentil for the sttic field cse. Thus, let us consider current element of length dl situted t the origin, s shown in Fig. 5.8, nd crrying current I A. We shll obtin the mgnetic vector potentil due to this current element. To do this, we recll from Section 1.6 tht the mgnetic field due to it t point P1r, u, f2 is given by B = m 4p I dl r r 2 (5.41) Expressing B s B = m 4p I dl - 1 r b (5.42) nd using the vector identity A = A - A (5.43) we obtin mi B = - 4pr dl + mi dl 4pr (5.44) Since dl is constnt, dl = 0, nd (5.44) reduces to B = midl 4pr (5.45) z P dl u r y FIGURE 5.8 For finding the mgnetic vector potentil due to current element. x f

18 5.3 Poisson s nd Lplce s Equtions 299 Compring (5.45) with (5.1), we see tht the mgnetic vector potentil due to the current element situted t the origin is given by A = mi dl 4pr (5.46) It follows from (5.46) tht for current element I dl situted t n rbitrry point, the mgnetic vector potentil is given by A = mi dl 4pR (5.47) where R is the distnce from the current element.thus, it hs mgnitude inversely proportionl to the rdil distnce from the element (similr to the inverse distnce dependence of the electric sclr potentil due to point chrge) nd direction prllel to the element. We shll mke use of this result in Section K5.2. Potentil difference; Potentil due to point chrge; Computtion of potentil due to chrge distributions; Electric dipole; Plotting of equipotentil lines; Mgnetic vector potentil due to current element. D5.4. In region of sttic electric field E = yz x + zx y + xy z, find the potentil difference V A - V B for ech of the following pirs of points: () A(2, 1, 1) nd B(1, 4, 0.5); (b) A(2, 2, 2) nd B(1, 1, 1); nd (c) A(5, 1, 0.2) nd B(1, 2, 3). Ans. () 0 V; (b) -7 V; (c) 5 V. D5.5. Three point chrges re locted s follows: 30pe 0 C t (3, 4, 0), 10pe 0 C t 13, -4, 02, nd -40pe 0 C t 1-5, 0, 02). Find the following: () the potentil t the point (0, 0, 3.2); (b) the coordinte x to three deciml plces of the point on the x-xis t which the potentil is mximum; nd (c) the potentil t the point found in (b). Ans. () 0 V; (b) m; (c) V. D5.6. For ech of the following rrngements of point chrges, find the first significnt term in the expression for the electric potentil t distnces fr from the origin 1r d2: () Q t (0, 0, d), 2Q t (0, 0, 0), nd Q t 10, 0, -d2 nd (b) Q t (0, 0, d), -2Q t (0, 0, 0), nd Q t 10, 0, -d2. Ans. () Q>pe (b) 1Qd 2 >4pe 0 r cos 2 0 r; u POISSON S AND LAPLACE S EQUATIONS In Section 5.2, we introduced the sttic electric potentil s relted to the sttic electric field in the mnner Poisson s eqution E = -V (5.48) Substituting (5.48) into Mxwell s divergence eqution for D given by # D = r (5.49)

19 300 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems we obtin where e is the permittivity of the medium. Using the vector identity we cn write (5.50) s - # ev = r # A = # A + A # (5.50) (5.51) e# V + e # V = -r or e 2 V + e# V = -r (5.52) If we ssume e to be uniform in the region of interest, then e = 0 nd (5.52) becomes 2 V = -r e (5.53) This eqution is known s Poisson s eqution. It governs the reltionship between the volume chrge density r in region of uniform permittivity e to the electric sclr potentil V in tht region. Note tht (5.53) lso follows from (5.25) for 0>0t = 0 nd =V. In Crtesin coordintes, (5.53) becomes 0 2 V 0x V 0y V 0z 2 = - r e (5.54) which is three-dimensionl, second-order prtil differentil eqution. For the one-dimensionl cse in which V vries with x only, 0 2 V>0y 2 nd 0 2 V>0z 2 re both equl to zero, nd (5.54) reduces to 0 2 V 0x 2 = d2 V dx 2 = - r e (5.55) We shll illustrte the ppliction of (5.55) by mens of n exmple. p-n junction semiconductor Exmple 5.5 Solution of Poisson s eqution for p n junction semiconductor Let us consider the spce chrge lyer in p-n junction semiconductor with zero bis, s shown in Fig. 5.9(), in which the region x 6 0 is doped p-type nd the region x 7 0 is doped n-type. To review briefly the formtion of the spce chrge lyer, we note tht since the density of the holes on the p side is lrger thn tht on the n side, there is tendency for the holes to diffuse to the n side nd recombine with the electrons. Similrly, there is tendency for the electrons on the n side to diffuse to the p side nd recombine with the holes. The diffusion of holes leves behind negtively chrged cceptor toms, nd the diffusion of electrons leves behind positively chrged donor toms. Since these

20 5.3 Poisson s nd Lplce s Equtions 301 Spce Chrge Acceptor Ion Hole E Electron x 0, p-type x 0 x 0, n-type Donor Ion () r e N D d p 0 d n x e N A (b) E x d p 0 d n x e N A d p e (c) e N A 2 (d p + d n d p ) 2e V 2 e N A d p 2e d p 0 d n x FIGURE 5.9 For illustrting the ppliction of Poisson s eqution for the determintion of the potentil distribution for p-n junction semiconductor. (d) cceptor nd donor toms re immobile, spce chrge lyer, lso known s the depletion lyer, is formed in the region of the junction, with negtive chrges on the p side nd positive chrges on the n side. This spce chrge gives rise to n electric field directed from the n side of the junction to the p side so tht it opposes diffusion of the mobile crriers cross the junction, thereby resulting in n equilibrium. For simplicity, let

21 302 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems us consider n brupt junction, tht is, junction in which the impurity concentrtion is constnt on either side of the junction. Let N A nd N D be the cceptor nd donor ion concentrtions, respectively, nd d p nd d n be the widths in the p nd n regions, respectively, of the depletion lyer. The spce chrge density r is then given by r = e - ƒeƒn A for -d p 6 x 6 0 ƒeƒn D for 0 6 x 6 d n (5.56) s shown in Fig. 5.9(b), where ƒeƒ is the mgnitude of the electronic chrge. Since the semiconductor is electriclly neutrl, the totl cceptor chrge must be equl to the totl donor chrge; tht is, ƒeƒn A d p = ƒeƒn D d n (5.57) We wish to find the potentil distribution in the depletion lyer nd the depletion lyer width in terms of the potentil difference cross the depletion lyer nd the cceptor nd donor ion concentrtions. Substituting (5.56) into (5.55), we obtin the eqution governing the potentil distribution to be d 2 V dx 2 ƒeƒn A = d e ƒeƒn D - e for -d p 6 x 6 0 for 0 6 x 6 d n (5.58) To solve (5.58) for V, we integrte it once nd obtin ƒeƒn A x + C dv dx = d e 1 for -d p 6 x 6 0 ƒeƒn D - x + C e 2 for 0 6 x 6 d n where C 1 nd C 2 re constnts of integrtion. To evlute C 1 nd C 2, we note tht since E = -V = -10V>0x2 x, 0V>0x is simply equl to -E x. Since the electric field lines begin on the positive chrges nd end on the negtive chrges, nd in view of (5.57), the field nd, hence, 0V>0x must vnish t x = -d p nd x = d n, giving us dv dx = d ƒeƒn A 1x + d e p 2 for -d p 6 x 6 0 ƒeƒn D - 1x - d e n 2 for 0 6 x 6 d n (5.59) The field intensity, tht is, -dv>dx, my now be sketched s function of x s shown in Fig. 5.9(c). Proceeding further, we integrte (5.59) nd obtin ƒeƒn A 2e 1x + d p2 2 + C 3 for -d p 6 x 6 0 V = d ƒeƒn D - 2e 1x - d n2 2 + C 4 for 0 6 x 6 d n

22 5.3 Poisson s nd Lplce s Equtions 303 where C 3 nd C 4 re constnts of integrtion. To evlute C 3 nd C 4, we first set the potentil t x = -d p rbitrrily equl to zero to obtin C 3 equl to zero. Then we mke use of the condition tht the potentil be continuous t x = 0, since the discontinuity in dv/dx t x = 0 is finite, to obtin or ƒeƒn A 2e d ƒeƒn p 2 D = - 2e d n 2 + C 4 C 4 = ƒeƒ 2e 1N A d p 2 + N D d n 2 2 Substituting this vlue for C 4 nd setting C 3 equl to zero in the expression for V, we get the required solution ƒeƒn A 2e 1x + d p2 2 for -d p 6 x 6 0 V = d ƒeƒn D - 2e 1x2-2xd n 2 + ƒeƒn A 2e d p 2 for 0 6 x 6 d n (5.60) The vrition of potentil with x s given by (5.60) is shown in Fig. 5.9(d). We cn proceed further nd find the width d = d p + d n of the depletion lyer by setting V1d n 2 equl to the contct potentil, V 0, tht is, the potentil difference cross the depletion lyer resulting from the electric field in the lyer. Thus, V 0 = V1d n 2 = ƒeƒn D 2e d n 2 + ƒeƒn A 2e d p 2 = ƒeƒ 2e N D 1N A + N D 2 N A + N D = ƒeƒ N A N D 1d 2 2e N A + N n + d 2 p + 2d n d p 2 D = ƒeƒ N A N D d 2 2e N A + N D d 2 n + ƒeƒ N A 1N A + N D 2 2 d 2e N A + N p D where we hve mde use of (5.57). Finlly, we obtin the result tht 2eV 0 d = 1 + B ƒeƒ N A 1 N D b which tells us tht the depletion lyer width is smller, the hevier the doping is. This property is used in tunnel diodes to chieve lyer widths on the order of 10-6 cm by hevy doping s compred to widths on the order of 10-4 cm in ordinry p-n junctions. We hve just illustrted n exmple of the ppliction of Poisson s eqution involving the solution for the potentil distribution for given chrge distribution. Poisson s eqution is even more useful for the solution of problems in

23 304 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems Lplce s eqution which the chrge distribution is the quntity to be determined given the functionl dependence of the chrge density on the potentil. We shll, however, proceed to the discussion of Lplce s eqution. If the chrge density in region is zero, then Poisson s eqution (5.53) reduces to 2 V = 0 (5.61) This eqution is known s Lplce s eqution. It governs the behvior of the potentil in chrge-free region chrcterized by uniform permittivity. In Crtesin coordintes, it is given by 0 2 V 0x V 0y V 0z 2 = 0 (5.62) Lplce s eqution is lso stisfied by the potentil in conductors under the stedy-current condition. For the stedy-current condition, 0r>0t = 0, nd the continuity eqution given for the time-vrying cse by reduces to # Jc + 0r 0t = 0 # Jc = 0 (5.63) Replcing J c by se = -sv, where s is the conductivity of the conductor nd ssuming s to be constnt, we obtin or # se = s # E = -s # V = -s 2 V = 0 2 V = 0 The problems to which Lplce s eqution is pplicble consist of finding the potentil distribution in the region between two conductors, given the chrge distribution on the surfces of the conductors, or the potentils of the conductors, or combintion of the two. The procedure involves the solving of Lplce s eqution subject to the boundry conditions on the surfces of the conductors. We shll illustrte this by mens of n exmple involving vrition of V in one dimension. Prllel-plte rrngement, cpcitnce Exmple 5.6 cpcitor Solution of Lplce s eqution for prllel-plte Let us consider two infinite, plne, prllel, perfectly conducting pltes occupying the plnes x = 0 nd x = d nd kept t potentils V = 0 nd V = V 0, respectively, s shown by the cross-sectionl view in Fig. 5.10, nd find the solution for Lplce s eqution in the region between the pltes. The rrngement my be considered n ideliztion of

24 5.3 Poisson s nd Lplce s Equtions 305 x d, V V 0 Equipotentil E x 0, V 0 FIGURE 5.10 Cross-sectionl view of prllel-plte cpcitor for illustrting the solution of Lplce s eqution in one dimension. prllel-plte cpcitor with its pltes hving dimensions very lrge compred to the spcing between them. The potentil is obviously function of x only, nd hence (5.62) reduces to 0 2 V 0x 2 = 02 V 0x 2 = 0 Integrting this eqution twice, we obtin V1x2 = Ax + B (5.64) where A nd B re constnts of integrtion. To determine the vlues of A nd B, we mke use of the boundry conditions for V; tht is, giving us V = 0 for x = 0 V = V 0 for x = d 0 = A102 + B or B = 0 V 0 = A1d2 + B = Ad or A = V 0 d Thus, the prticulr solution for the potentil here is given by V = V 0 d x for 0 6 x 6 d (5.65) which tells us tht the equipotentils re plnes prllel to the conductors, s shown in Fig Proceeding further, we obtin 0V E = -V = - 0x V 0 x = - d x for 0 6 x 6 d (5.66)

25 306 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems This field is uniform nd directed from the higher potentil plte to the lower potentil plte, s shown in Fig The surfce chrge densities on the two pltes re given by ev [r S ] x = 0 = [D] x = 0 # 0 x = - d # ev 0 x x = - d ev [rs] x = d = [D] x = d # 0 1-x 2 = - d # x 1- x 2 = ev 0 d The mgnitude of the surfce chrge per unit re on either plte is Q = ƒr S ƒ112 = ev 0 d Finlly, we cn find the cpcitnce C per unit re of the pltes, defined to be the rtio of Q to V 0. Thus, C = Q = e per unit re of the pltes V 0 d (5.67) The units of cpcitnce re frds (F). Prllel-plte rrngement, conductnce If the medium between the pltes in Fig is conductor, then the conduction current density is given by J c = se = - sv 0 d x The conduction current from the higher potentil plte to the lower potentil plte per unit re of the pltes is I c = ƒj c ƒ112 = sv 0 d The rtio of this current to the potentil difference is the conductnce G (reciprocl of resistnce) per unit re of the pltes. Thus, G = I c V 0 = s d per unit re of the pltes (5.68) Cylindricl nd sphericl cpcitors The units of conductnce re siemens (S). We hve just illustrted the solution of Lplce s eqution in one dimension by considering n exmple involving the vrition of V with one Crtesin coordinte. In similr mnner, solutions for one-dimensionl Lplce s equtions involving vritions of V with single coordintes in the other two coordinte systems cn be obtined. Of prticulr interest re the cse in which V is function of the cylindricl coordinte r only, pertinent to the geometry of cpcitor mde up of coxil cylindricl conductors, nd the cse in which V is

26 5.3 Poisson s nd Lplce s Equtions 307 Cylinders Spheres e f e z r b r b V V 0 V V 0 V 0 () V 0 (b) FIGURE 5.11 Cross-sectionl views of cpcitors mde up of () coxil cylindricl conductors nd (b) concentric sphericl conductors. function of the sphericl coordinte r only, pertinent to the geometry of cpcitor mde up of concentric sphericl conductors. These two geometries re shown in Figs. 5.11() nd (b), respectively. The vrious steps in the solution of Lplce s eqution nd subsequent determintion of cpcitnce for these two cses re summrized in Tble 5.1, which lso includes the prllel plne cse of Fig TABLE 5.1 Summry of Vrious Steps in the Solution of Lplce s Eqution nd Determintion of Cpcitnce for Three One-Dimensionl Cses Geometry Prllel plnes Coxil cylinders Concentric spheres Figure () 5.11(b) V = V 0, r = V = V 0, r = Boundry conditions e V = 0, x = 0 V = V 0, x = d V = 0, r = b V = 0, r = b Lplce s eqution Generl solution Prticulr solution Electric field Surfce chrge densities µ 0 2 V 0x 2 = 0 V = Ax + B x V = V 0 d V 0 - d x ev 0 - d, x = 0 ev 0 d, x = d 1 0 r 0r r 0V 0r b = 0 V = A ln r + B ln 1r>b2 V = V 0 ln 1>b2 V 0 r ln 1b>2 r ev 0 ln 1b>2, -ev 0 b ln 1b>2, r = r = b e 2pe Cpcitnce per unit re per unit length d ln 1b> V r 2 r2 0r 0r b = 0 V = A r + B 1>r - 1>b V = V 0 1> - 1>b V 0 r 2 11> - 1>b2 r ev > - 1>b2, r = -ev 0 b 2 11> - 1>b2, r = b 4pe 1> - 1>b

27 308 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems K5.3. D5.7. D5.8. Poisson s eqution; p-n junction; Lplce s eqution in one dimension; Prllel-plte rrngement; Cpcitnce; Conductnce; Cylindricl nd sphericl cpcitors. The potentil distribution in simplified model of vcuum diode consisting of cthode in the plne x = 0 nd node in the plne x = d nd held t potentil V reltive to the cthode is given by V = V 0 1x>d2 4>3 0 for 0 6 x 6 d. Find the following: () V t x = d>8; (b) E t x = d>8; (c) r t x = d>8; nd (d) r S on the node. Ans. () V (b) (c) -16e 0 V 0 >9d 2 0 >16; -12V 0 >3d2 x ; ; (d) 4e 0 V 0 >3d. Find the following: () the spcing between the pltes of prllel-plte cpcitor with dielectric of e = 2.25e 0, nd hving cpcitnce per unit re equl to 1000 pf; (b) the rtio of the outer rdius to the inner rdius for coxil cylindricl cpcitor with dielectric of e = 2.25e 0, nd hving cpcitnce per unit length equl to 100 pf; nd (c) the rdius of n isolted sphericl conductor in free spce for which the cpcitnce is 10 pf. Ans. () 1.99 cm; (b) ; (c) 9 cm. 5.4 CAPACITANCE, CONDUCTANCE, AND INDUCTANCE Coxil cylindricl rrngement In the previous section, we introduced the cpcitnce nd conductnce by considering the solution of Lplce s eqution in one dimension. Specificlly, we derived the expressions for the cpcitnce per unit re nd the conductnce per unit re of prllel-plte rrngement, the cpcitnce per unit length of coxil cylindricl rrngement, nd the cpcitnce of concentric sphericl rrngement. Let us now consider the three rrngements shown in Fig. 5.12, ech of which is cross-sectionl view of pir of infinitely long coxil perfectly conducting cylinders with mteril medium between them. In Fig. 5.12(), the E b r J c b r b r f V V 0 e V V 0 s m H V 0 () V 0 (b) (c) FIGURE 5.12 Cross sections of three rrngements, ech consisting of two infinitely long, coxil, perfectly conducting cylinders. The medium between the cylinders is perfect dielectric for (), conductor for (b), nd mgnetic mteril for (c).

28 5.4 Cpcitnce, Conductnce, nd Inductnce 309 mteril medium is dielectric of uniform permittivity e; in Fig. 5.12(b), it is conductor of uniform conductivity s; nd in Fig. 5.12(c), it is mgnetic mteril of uniform permebility m. In () nd (b), potentil difference of V 0 is pplied between the conductors, wheres in (c), current I flows with uniform density in the +z-direction on the inner cylinder nd returns with uniform density in the - z-direction on the outer cylinder. We know from the discussion in Section 5.3 tht the rrngement of Fig. 5.12() is tht of coxil cylindricl cpcitor nd from Tble 5.1 tht its cpcitnce (C) per unit length, defined s the mgnitude of the chrge per unit length on either conductor to the potentil difference between the conductors, is given by Cpcitnce per unit length, c c = C l = 2pe ln 1b>2 (5.69) the units of C being frds (F) nd, hence, those of c being F/m. Just s in the cse of the prllel-plte rrngement of Exmple 5.7, replcing the dielectric in Fig. 6.4() by conductor s in Fig. 5.12(b) would result in conduction current of density Conductnce per unit length, g in the medium nd, hence, current per unit length 2p 2p sv I c = J c # 0 r df r = L L r ln 1b>2 r df f = 0 = 2psV 0 ln 1b>2 J c = se = sv 0 r ln 1b>2 r from the inner cylinder to the outer cylinder. Thus, the rtio of the current per unit length from the inner to the outer cylinder to the potentil difference between the cylinders, tht is, the conductnce (G) per unit length of the rrngement, is given by f = 0 g = G l = 2ps ln 1b>2 (5.70) the units of G being siemens (S) nd, hence, those of g being S/m. Turning now to Fig. 5.12(c), we know from the ppliction of Ampere s circuitl lw in integrl form tht the current flow on the cylinders results in mgnetic field between the cylinders s given by Inductnce per unit length, l

29 310 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems H = The mgnetic flux density is then given by I 2pr f for 6 r 6 b B = mh = mi 2pr f for 6 r 6 b The mgnetic flux linking the current per unit length of the conductors is b c = B # dr f L r = = L b r = = mi 2p ln b mi 2pr dr We now define the inductnce (L) per unit length of the rrngement to be the rtio of the mgnetic flux linking the current per unit length of the rrngement to the current. Thus, l = L l = m 2p ln b (5.71) The units of L re henrys (H) nd, hence, those of l re H/m. An exmintion of (5.69), (5.70), nd (5.71) revels tht g c = s e (5.72) nd lc = me (5.73) Thus, only one of the three prmeters c, g, nd l is independent, with the other two obtinble from it nd the mteril prmeters. Although this result is deduced here for the coxil cylindricl rrngement, it is generl result vlid for ll rrngements involving two infinitely long, prllel perfect conductors embedded in homogeneous medium ( medium of uniform mteril prmeters). Expressions for the three quntities c, g, nd l re listed in Tble 5.2 for some common configurtions of conductors hving cross-sectionl views shown in Fig The coxil cylindricl rrngement is repeted for the ske of completion.

30 5.4 Cpcitnce, Conductnce, nd Inductnce 311 TABLE 5.2 Conductnce, Cpcitnce, nd Inductnce per Unit Length for Some Structures Consisting of Infinitely Long Conductors Hving the Cross Sections Shown in Fig Cpcitnce Conductnce Inductnce Description per unit length, c per unit length, g per unit length, l Prllel-plne conductors, Fig () Coxil cylindricl conductors, Fig (b) Prllel cylindricl wires, Fig (c) Eccentric inner conductor, Fig (d) Shielded prllel cylindricl wires, Fig (e) w e d 2pe ln 1b>2 pe cosh -1 1d>2 2pe cosh b 2 - d 2 b 2b pe ln d1b2 - d 2 >42 1b 2 + d 2 >42 w s d 2ps ln 1b>2 ps cosh -1 1d>2 2ps cosh b 2 - d 2 b 2b ps ln d1b2 - d 2 >42 1b 2 + d 2 >42 m d m w m 2p ln b p cosh-1 d m 2p cosh b 2 + d 2 b 2b m p ln d1b2 - d 2 >42 1b 2 + d 2 >42 w d d w b () (b) d (b ) d (b 2 4 d 2 ) 2 b 2 b 2 2d d d (c) (d) (e) FIGURE 5.13 Cross sections of some common configurtions of prllel, infinitely long conductors. Exmple 5.7 Cpcitnce, conductnce, nd inductnce per unit length for prllel-wire line It is desired to obtin the cpcitnce, conductnce, nd inductnce per unit length of the prllel-cylindricl wire rrngement of Fig. 5.13(c). In view of (5.72) nd (5.73), it is sufficient to find one of the three quntities. Hence, we choose to find the cpcitnce per unit length. Here we shll do this by considering the electric potentil field of two prllel, infinitely long, stright-line chrges of c, g nd l of prllelcylindricl wire rrngement

31 312 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems equl nd opposite uniform chrge densities nd showing tht the equipotentil surfces re cylinders hving their xes prllel to the line chrges. By plcing conductors in two equipotentil surfces, thereby forming prllel-wire line, we shll obtin the expression for the cpcitnce per unit length of the line. Let us first consider n infinitely long, stright-line chrge of uniform density r L0 C>m situted long the z-xis, s shown in Fig. 5.14(), nd obtin the electric potentil due to the line chrge. The symmetry ssocited with the problem indictes tht the potentil is dependent on the cylindricl coordinte r. Thus, we hve 2 V = 1 0 r 0r r 0V 0r b = 0 for r Z 0 V = A ln r + B (5.74) where A nd B re constnts to be determined. We cn rbitrrily set the potentil to be zero t reference vlue r = r 0, giving us B = -A ln r 0 nd V = A ln r - A ln r 0 = A ln r r 0 (5.75) To evlute the rbitrry constnt A in (5.75), we find tht the electric-field intensity due to the line chrge is given by 0V E = -V = - 0r A r = - r r The electric field is thus directed rdil to the line chrge. Let us now consider cylindricl box of rdius r nd length l coxil with the line chrge, s shown in Fig. 5.14(), nd pply Guss lw for the electric field in integrl form to the surfce of the box. For the cylindricl surfce, L D # ds = - ea r 12prl2 z z r l y b b y r 2 x x r 1 r L0 r L0 r L0 () (b) FIGURE 5.14 () Infinitely long line chrge of uniform density long the z-xis. (b) Pir of prllel, infinitely long line chrges of equl nd opposite uniform densities.

32 5.4 Cpcitnce, Conductnce, nd Inductnce 313 For the top nd bottom surfces, 1 D # ds = 0, since the field is prllel to the surfces. The chrge enclosed by the box is r L0 l. Thus, we hve ea - r 12prl2 = r L0l or A = - Substituting this result in (5.75), we obtin the potentil field due to the line chrge to be r L0 2pe V = - r L0 2pe ln r = r L0 r 0 2pe ln r 0 r (5.76) which is consistent with (5.40). Let us now consider two infinitely long, stright-line chrges of equl nd opposite uniform chrge densities r L0 C>m nd -r L0 C>m, prllel to the z-xis nd pssing through x = b nd x = -b, respectively, s shown in Fig. 5.14(b). Applying superposition nd using (5.76), we write the potentil due to the two line chrges s V = r L0 2pe ln r 01 - r L0 r 1 2pe ln r 02 r 2 where r 1 nd r 2 re the distnces of the point of interest from the line chrges nd r 01 nd r 02 re the distnces to the reference point t which the potentil is zero. By choosing the reference point to be equidistnt from the two line chrges, tht is, r 01 = r 02, we get V = r L0 2pe ln r 2 r 1 (5.77) From (5.77), we note tht the equipotentil surfces for the potentil field of the line-chrge pir re given by r 2 r 1 = constnt, sy, k (5.78) where k lies between 0 nd q. In terms of Crtesin coordintes, (5.78) cn be written s 1x + b2 2 + y 2 1x - b2 2 + y 2 = k2 Rerrnging, we obtin or x 2-2b k k 2-1 x + y2 + b 2 = 0 x - b k k 2-1 b + y 2 2 2k = b k 2-1 b This eqution represents cylinders hving their xes long x = b k k 2-1, y = 0

33 314 Chpter 5 Electromgnetic Potentils nd Topics for Circuits nd Systems r 2 r 1 r L0 r L0 d b b d x 0, k 1 FIGURE 5.15 Cross sections of equipotentil surfces for the line chrge pir of Fig 5.14(b). Thick circles represent cross section of prllel-wire line. nd rdii equl to b[2k>1k 2-12]. The corresponding potentils re 1r L0 >2pe2 ln k. The cross sections of the equipotentil surfces re shown in Fig We cn now plce perfectly conducting cylinders in ny two equipotentil surfces without disturbing the field configurtion, s shown, for exmple, by the thick circles in Fig. 5.15, thereby obtining prllel-wire line. Letting the distnce between their centers be 2d nd their rdii be, we hve k ;d = b k 2-1 2k = b k 2-1 Solving these two equtions for k nd ccepting only those solutions lying between 0 nd q, we obtin k = d ; 4 d2-2 The potentils of the right 1k 7 12 nd left 1k 6 12 conductors re then given, respectively, by V + = r L0 2pe ln d + 4 d2-2 V - = r L0 2pe ln d - 4 d2-2 r L0 = - 2pe ln d + 4 d2-2