Schaum outline of electromagnetics , Exercises of Electromagnetism and Electromagnetic Fields Theory

Download Schaum outline of electromagnetics and more Exercises Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! SCHAUM’S OUTLINE OF THEORY AND PROBLEMS OF ELECTROMAGNETICS Second Edition JOSEPH A. EDMINISTER Professor Emeritus of Electrical Engineering The University of Akron SCHAUM’S OUTLINE SERIES McGRAW-HILL New York San Francisco Washington, D.C. Auckland Bogoté Caracas Lisbon London Madrid Mexico City Milan Montreal New Dehli San Juan Singapore Sydney Tokyo Toronto Contents Chapter £ VECTOR ANALYSIS ..... 1 1.1 Introduction 1.2 Vector Notation 1.3 Vector Algebra 1.4 Coordinate Systems 1.5 Differential Volume, Surface, and Line Elements Chapter 2 COULOMB FORCES AND ELECTRIC FIELD INTENSITY . 3 2.1 Coulomb’s Law 2.2 Electric Field Intensity 2.3 Charge Distributions 2.4 Standard Charge Configurations Chapter 3 ELECTRIC FLUX AND GAUSS’ LAW ...cssssscerseneees: 3.1 Net Charge in a Region 3.2 Electric Flux and Flux Density 3.3 Gauss’ Law 3.4 Relation Between Flux Density and Electric Field Intensity 3.5 Special Gaussian Surfaces Chapter 4° DIVERGENCE AND THE DIVERGENCE THEOREM .......0ecccereee 47 4.1 Divergence 4.2 Divergence in Cartesian Coordinates 4.3 Divergence of D 4.4 The Del Operator 4.5 The Divergence Theorem Chapter 5 THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL .. soteeveneesenersees 59 5.1 Work Done in Moving a Point Charge 5.2 Conservative Property of the Electrostatic Field 5.3 Electric Potential Between Two Points 5.4 Potential of a Point Charge 5.5 Potential of a Charge Distribution 5.6 Gradient 5.7 Rela- tionship Between E and V 5.8 Energy in Static Electric Fields Chapter 6 CURRENT, CURRENT DENSITY, AND CONDUCTORS ....... 6.1 Introduction 6.2 Charges in Motion 6.3 Convention Current Densi 6.4 Conduction Current Density J 6.5 Conductivity o 6.6 Current / 6.7 Resistance R 6.8 Current Sheet Density K 6.9 Continuity of Current 6.10 Conductor-Dielectric Boundary Conditions. 16 Chapter 7 CAPACITANCE AND DIELECTRIC MATERIALS ...... wee OS 7.1 Polarization P and Relative Permittivity €, 7.2 Capacitance 7.3 Multiple- Dielectric Capacitors 7.4 Energy Stored in a Capacitor 7.5 Fixed-Voltage D and E 7.6 Fixed-Charge D and E 7.7 Boundary Conditions at the Interface of Two Dielectrics 114 Chapter 8 LAPLACE’S EQUATION ..... 2 8.1 Introduction 8.2 Poisson’s Equation and Laplace’s Equation 8.3 Explicit Forms of Laplace’s Equation 8.4 Uniqueness Theorem 8.5 Mean Value and Maximum Value Theorems 8.6 Cartesian Solution in One Variable 8.7 Cartesian Product Solution 8.8 Cylindrical Product Solution 8.9 Spherical Product Solution Chapter 9 CONTENTS AMPERE’S LAW AND THE MAGNETIC FIELD 9.1 Introduction 9.2 Biot-Savart Law 9.3 Ampere’s Law 9.4 Curl 9.5 Rela- tionship of J and H 9.6 Magnetic Flux Density B 9.7 Vector Magnetic Poten- tial A 9.8 Stokes’ Theorem Chapter 10 FORCES AND TORQUES EN MAGNETIC FIELDS 10.1 Magnetic Force on Particles 10.2 Electric and Magnetic Fields Com- bined 10.3 Magnetic Force on a Current Element 10.4 Work and Power 10.5 Torque 10.6 Magnetic Moment of a Planar Coil Chapter 17 INDUCTANCE AND MAGNETIC CIRCUITS ......... evovce 169 11.1 Inductance 11.2 Standard Conductor Configurations 11.3 Faraday’s Law and Self-Inductance 11.4 Interna\{nductance 11.5 MutualInductance 11.6 Mag- netic Circuits 11.7 The B-H Curve 11.8 Ampere’s Law for Magnetic Circuits 11.9 Cores with Air Gaps 11.10 Multiple Coifs 11.11 Parallel Magnetic Circuits Chapter 12 DISPLACEMENT CURRENT AND INDUCED EMF ... 12.1 Displacement Current 12.2 Ratio of J, to Jp 12.3 Faraday’s Law and Lenz's Law 12.4 Conductors in Motion Through Time-Independent Fields 12.5 Conductors in Motion Through Time-Dependent Fields 192 Chapter 13 MAXWELL’S EQUATIONS AND BOUNDARY CONDITIONS ........ 205 13.1 Introduction 13.2 Boundary Relations for Magnetic Fields 13.3 Current Sheet at the Boundary 13.4 Summary of Boundary Conditions 13.5 Maxwell’s Equations Chapter 14 ELECTROMAGNETIC WAVES .........+. 14.1 Introduction 14,2 Wave Equations 14.3 Solutions in Cartesian Coord- inates 14,4 Solutions for Partially Conducting Media 14.5 Solutions for Perfect Dielectrics 14.6 Solutions for Good Conductors; Skin Depth 14.7 Interface Conditions at Normal Incidence 14,8 Oblique Incidence and Snell’s Laws 14.9 Perpendicular Polarization 14,10 Parallel Polarization 14.11 Standing Waves 14.12 Power and the Poynting Vector Chapter 15 TRANSMISSION LINES .......... 15.1 Introduction 15.2 Distributed Parameters 15.3 Tncremental Model; Vol- tages and Currents 15.4 Sinusoidal Steady-State Excitation 15.5 The Smith Chart 15.6 Impedance Matching 15.7 Single-Stub Matching 15.8 Double- Stub Matching 15.9 Impedance Measurement 15.10 Transients in Lossless Lines asancacccoroossescosves 237 Chapter 16 WAVEGUIDES ... 274 16.1 Introduction 16.2 Transverse and Axial Fields 16.3 TE and TM Modes; ‘Wave Impedances 16.4 Determination of the Axial Fields 16.5 Mode Cutoff Frequencies 16.6 Dominant Mode 16.7 Power Transmitted in a Lossless Waveguide 16.8 Power Dissipation in a Lossy Waveguide Chapter 17 CONTENTS 17.1 Introduction 17.2 Current Source and the E and H Fields 17,3 Electric (Hertzian) Dipole Antenna 17.4 Antenna Parameters 17.5 Small Circular- Loop Antenna 17.6 Finite-Length Dipole 17.7 Monopole Antenna 17.8 Self- and Mutual Impedances 17.9 The Receiving Antenna 17.10 Linear Arrays 17.11 Reflectors « 35 Appendix B 317 333 CHAP. 1] VECTOR ANALYSIS 3 EXAMPLE 2. Given A=2a,+4a,-3a, and B=a,—a,, find A+ Band AXB. A+B=(2)(1) + (4(—1) + (—3)(0) = -2 a & & AxB=|2 4 -3|=—3a,—3a,-6a, 1 -1 06 14 COORDINATE SYSTEMS A problem which has cylindrical or spherical symmetry could be expressed and solved in the familiar cartesian coordinate system. However, the solution would fail to show the symmetry and in most cases would be needlessly complex. Therefore, throughout this book, in addition to the cartesian coordinate system, the circular cylindrical and the spherical coordinate systems will be used. All three will be examined together in order to illustrate the similarities and the differences. A point P is described by three coordinates, in cartesian (x, y, z), in circular cylindrical (r, $, z), and in spherical (r, @, @), as shown in Fig. 1-2. The order of specifying the coordinates is important aad should be carefully followed. The angle ¢ is the same angle in both the cylindrical and spherical systems. But, in the order of the coordinates, @ appears in the second position in cylindrical, (r, , z), and the third position in spherical, (r, 8, @). The same symbol, r, is used in both cylindrical and spherical for two quite different things. In cylindrical coordinates r measures the distance from the z axis in a plane normai to the z axis, while in the spherical system r measures the distance from the origin to the point. It should be clear from the context of the problem which r is intended. fa) Cartesian (®) Cylindricat (c) Spherical Fig. 1-2 A point is also defined by the intersection of three orthogonal surfaces, as shown in Fig. 1-3. In cartesian coordinates the surfaces are the infinite planes x=const., y=const., and z= const. In cylindrical coordinates, z=const. is the same infinite plane as in cartesian; $= const. is a half plane with its edge along the z axis; r=const. is a right circular cylinder. These three surfaces are orthogonai and their intersection locates point P. In spherical coordinates, = const. is the same half plane as in cylindrical; +=const. is a sphere with its center at the origin; @=const. is a right circular cone whose axis is the z axis and whose vertex is at the origin. Note that 6 is limited to the range Os @= 2. Figure 1-4 shows the three unit vectors at point P. In the cartesian system the unit vectors have fixed directions, independent of the location of P. This is not true for the other two systems (except in the case of a,). Each unit vector is normal to its coordinate surface and is in the direction in which the coordinate increases. Notice that all these systems are right-handed: a, Xa, =a, a, Xa, =2, &, Xig = ay 4 VECTOR ANALYSIS (CHAP. 1 r= const. @ = const. 2= const. z= const. P Y y x = const. x y* const, x @ = const, = const. (a) Cartesian (6) Cylindrical (c) Spherical Fig. 1:3 The component forms of a vector in the three systems are A=A,a, + Aya, + Aa, (cartesian) A=A,a, + Agag + A,a, (cylindrical) A=A,a,+Apig+ Aga, — (spherical) It should be noted that the components A,, A,, Ag, etc., are not generally constants but more often are functions of the coordinates in that particular system. z z z a, 4 3 a P| " | » ao 5 — % * ’ *e ” x x x (@) Cartesian (5) Cylindrical (©) Spherical Fig. 1-4 1.5 DIFFERENTIAL VOLUME, SURFACE, AND LINE ELEMENTS There are relatively few problems in electromagnetics that can be solved without some sort of integration—along a curve, over a surface, or throughout a volume. Hence the corresponding differential elements must be clearly understood. When the coordinates of point P are expanded to (x + dx, y + dy, z+dz) or (r+dr, @+ dg, z+dz) or (r+ dr, 64+ dO, @ +d@), a differential volume dv is formed. To the first order in infinitesimal quantities the differential volume is, in all three coordinate systems, a rectangular box. The value of dv in each system is given in Fig. 1-5. CHAP. 1} VECTOR ANALYSIS 5 rdp ¥ dy = dx dy dz du=rdrdodz du=r sin@ drdo de (@) Cartesian (6) Cylindrical {c) Spherical Fig. 1-5 From Fig. 1-5 may also be read the areas of the surface elements that bound the differential volume. For instance, in spherical coordinates, the differential surface element perpendicular to 2, is aS =(r d6)(r sin 0@dd) =r" sin OdO dd The differential line element, dé is the diagonal through P. Thus dé? = dx? + dy? + dz” (cartesian) df? = dr? +r? dp? + dz? (cylindrical) df= dr +r dé’ +r sin’ 6dg” (spherical) Solved Problems 1.1. Show that the vector directed from M(x, y1, 21} to N(x2, yo, Z2) in Fig. 1-6 is given by @2— xi) + (ye — dy + Ge ~ 71) NMxq, 2.72) Moxy, ¥1571) Fig. 1-6 The coordinates of M and N are used to write the wo position vectors A and B in Fig. 1-6. A=x8, + ya, +78, B= xa, + y&, + Za, Then B- A=(x2—x,)a, + (2 — yim + (22-21) 8 VECTOR ANALYSIS (CHAP. 1 As long as the vectors appear in the same cyclic order the result is the same. The scalar triple products not in this cyclic order have a change in sign. 1.9. Express the unit vector which points from z= on the z axis toward (7, ¢, 0) in cylindrical coordinates. See Fig. 1-9. Fig. 1.9 ‘The vector R is the difference of two vectors: R=ra,—ha, 1p — ha vP+R The angle ¢ does not appear explicitly in these expressions. Nevertheless, both R and a, vary with @ through a,. 1.10. Express the unit vector which is directed toward the origin from an arbitrary point on the plane z=-—5, as shown in Fig. 1-10. Fig. 1-10 Since the problem is in cartesian coordinates, the two-point formula of Problem 1.1 applies. R=-~—xa, — ya, + 5a, ya + Sa, Vx ty? 425 1.11. Use the spherical coordinate system to find the area of the strip a<@< £8 on the spherical iz Shell of radius a (Fig. 1-11). What results when a=0 and fB=x? The differential surface element is [see Fig. 1-5(c)] dS =r sin @d0dp CHAP. 1] VECTOR ANALYSIS 9 Fig. 1-11 2x fs Then azf [ @sinodoag = 2a" (cos « — 00s B) When a@=0 and f=x, A=4za", the surface area of the entire sphere. 1.12. Obtain the expression for the volume of a sphere of radius a from the differential volume. oh From Fig. 1-5(c), du =,’sin @drd6d@. Then = an px pa 4 v={ [ [Ps eardoag = 4 na? Jp do do 3 1.13. Use the cylindrical coordinate system to find the area of the curved surface of a right circular t cylinder where r=2m, h=Sm, and 30°=@=120° (see Fig. 1-12). a 24/3 Fig. 1-12 The differential surface element is dS=rdpdz. Then S p23 A= [ { 2dodz lo Sis = Sm? 114, Transform A=ya,+xa, +———— a Te ay from cartesian to cylindrical coordinates. 10 1. VECTOR ANALYSIS (CHAP. 1 Referring to Fig. 1-2(b), x¥=rcos¢ y=rsing r=ave+y? Hence Ae=rsin da, +1 cos pa, +7 cos’ ga, Now the projections of the cartesian unit vectors on a,, a, and #, are obtained: a, 8, = COSH aa, = —sing asaasin gd a, -,= COs a,ta,=0 a,-a,=0 Therefore a, = 00s Ga, —sin ga, a,=sin da, + cos day a, =a, and A=2rsin ¢ cos ga, + (r cos” @ — sin? p)a, +7 cos” da, A vector of magnitude 10 points from (5, 52/4, 0) in cylindrical coordinates toward the origin a (Fig. 1-13). Express the vector in cartesian coordinates. L16. LY, 1.18. 1.19, z A | Sx/4 ¥ x Fig. 1-13 In cylindrical coordinates, the vector may be expressed as 10a,, where g@=a/4. Hence zx 10 z_ 10 A, = 10 cos— =— A, = 10sin ==—~ A,=0 4 V2 4 v2 ‘so that 10 10 A=—a,+— va" va” Notice that the value of the radial coordinate, 5, is immaterial. Supplementary Problems Given A=4a,+10a, and B=2a,+3a,, find the projection of Aon B. Ans. 12/V13 Given A=(10/V2)(a,+8,) and B=3(s,+2,), express the projection of B on A as a vector in the direction of A. Ans. 1,50(a, + a,) Find the angle between A=10a,+2a, and B=—4a,+0.5a, using both the dot product and the cross product. Ans. 161.5" Find the angle between A=5.8a,+1.55a, and B=-—6.93a,+4.0a, using both the dot product and the cross product. Ans. 135° Chapter 2 Coulomb Forces and Electric Field Intensity 2.1 CQULOMB’S LAW There is a force between two charges which is directly proportional to the charge magnitudes and inversely proportional to the square of the separation distance. This is Coulomb's law, which was developed from work with small charged bodies and a delicate torsion balance. In vector form, it is stated thus, Q2 ed? Rationalized SI units will be used throughout this book. The force is in newtons (N), the distance is in meters (m), and the (derived } unit of charge is the coulomb (C). The system is rationalized by the factor 4s1, introduced in Coulomb's law in order that it not appear later in Maxwell's equations. € is the permittivity of the medium, with the units C?/N - m? or, equivalently, farads per meter (F/m). For free space or vacuum, -9 10 € = €9= 8.854 x 107" F/m= 36x F/m For media other than free space, €=€ €,, Where €, is the relative permittivity or dielectric constant. Free space is to be assumed in all problems and examples, as well as the approximate value for €y , unless there is a statement to the contrary. For point charges of like sign the Coulomb force is one of repulsion, while for unlike charges the force is attractive. To incorporate this information rewrite Coulomb's law as follows: _ QQ. Q:02 SS a eo = 2: 4xeoRi, 4meoRi, 7 1 1 where F, is the force on charge Q, due to a second charge Q, , a2; is the unit vector directed from Q2 io Q:, and Rz,=R282, is the displacement vector from Q, to Q,. EXAMPLE 1. Find the force on charge Q, , 20 nC, due to charge Q, , —300 nC, where Q, is at (0, 1,2) m and Q, at (2, 0,0) m. Because 1 C is a rather large unit, charges are often given in microcoulombs (:C), nanocoulombs (nC), or picocoulombs (pC). (See Appendix for the SI prefix system.) Referring to Fig. 2-1, R,, = —2a, +a, +20 Ry = VAI + PFE =3 z F a (0,1, 2) Ray y 2 Q, 0,0) x Fig. 2-1 13 14 COULOMB FORCES AND ELECTRIC FIELD INTENSITY (CHAP. 2 and ay=}(-20, +4, +20) _ 0x 107*)(—300 x 107%) (2a, + a, + 2a, Then i gx(10 36)? ( 3 ) 2a, — 4, —2a, (3) The force magnitude is 6 N and the direction is such that Q, is attracted to Q, (unlike charges attract). This force relationship is bilinear in the charges. Consequently, superposition applies, and the force on a charge Q, due to n — 1 other charges Q., Q3,..., Q,, is the vector sum of the individual forces: 2105 Bn 2:01 ate 3 amegRA ane, oR This superposition extends in a natural way to the case where charge is continuously distributed through some spatial region: one simply replaces the above vector sum by a vector integral (see Section 2.3). The force field in the region of an isolated charge Q is spherically symmetric. ‘This is made evident by locating Q at the origin of a spherical coordinate system, so that the position vector R, from Q to a small test charge @,<<Q, is simply ra, . Then Mr _ 2.0 4n€ qr” showing that on the spherical surface r=constant, {F,| is constant, and F, is radial. 2.2 ELECTRIC FIELD INTENSITY Suppose that the above-considered test charge Q, is sufficiently small so as not to disturb significantly the field of the fixed point charge Q. Then the electric field intensity, E, due to Q is defined to be the force per unit charge on Q,: E=F,/Q,. For Q at the origin of a spherical coordinate system [see Fig. 2-2(a)], the electric field intensity at an arbitrary point P is, from Section 2.1, Plz, ¥a5 22) R=G, —x a, +O) —% Jay + -21)a: OGL yea) @) Spherical (0) Cartesian Fig. 2-2 CHAP. 2] COULOMB FORCES AND ELECTRIC FIELD {NTENSITY 15 In an arbitrary cartesian coordinate system [see Fig. 2-2(b)], E=—4 Are yR? * The units of E are newtons per coulomb (N/C) or the equivalent, volts per meter (V/m). EXAMPLE 2, Find E at (0, 3,4) m in cartesian coordinates due to a point charge Q = 0.5 uC at the origin. In this case R=3a, + 4a, R=5 Ap = 0.6a, + 0.88, 0.5 x 10 ~ go Paeaytay Oy £888) Thus |E|=180V/m_ in the direction a, =0.6a, + 0.84, . 2.3 CHARGE DISTRIBUTIONS Volume Charge When charge is distributed throughout a specified volume, each charge element contributes to the electric field at an external point. A summation or integration is then required to obtain the total electric field. Even though electric charge in its smallest division is found to be an electron or proton, it is useful to consider continuous (in fact, differentiable) charge distributions and to define a charge density by 2Q = ‘C/m* p=a, (hr) Note the units in parentheses, which is meant to signify that p will be in C/m* provided that the variables are expressed in proper SI units (C for Q and m* for v). This convention will be used throughout this book. With reference to volume v in Fig. 2-3, each differential charge dQ produces a differential electric field dQ ~ brregR?* aE of ° Fig. 2-3 at the observation point P. Assuming that the only charge in the region is contained within the volume, the total electric field at P is obtained by integration over the volume: pag E= |, 42€gR v 18 COULOMB FORCES AND ELECTRIC FIELD INTENSITY {CHAP. 2 Infinite Piane Charge If charge is distributed with uniform density p,(C/m) over an infinite plane, then the field is given by Ps E=jf 26," See Fig. 2-7. This field is of constant magnitude and has mirror symmetry about the plane charge. For a derivation of this expression, see Problem 2.12. Fig. 2-7 EXAMPLE 4. Charge is distributed uniformly over the plane z=10cm with a density p,= (1/3) nC/m?. Find E. y= f= (1/3x}10-° © 2€ 2(10°°/36} Above the sheet (z > 10cm), E=6a, V/m; and for z<10cm, =—6a, V/m. =6V/m Solved Problems 2.4. Two point charges, Q,=S50uC and Q,=10uC, are located at (—1,1,—3)m and xiz (3, 1,0) m, respectively (Fig. 2-8). Find the force on Q,. —4a, — 3a, 5 p= 2122 1 aneoR3, _ (50 x 10°*)(10"*) (= - 3s) 4x(10 °/36n\(5Y\ 5 = (0.18)(—0.8a, — 0.6a,)N a = z x @ G,1,0) Ry Q@, C1, 1,-3) Fig. 2-8 CHAP. 2] COULOMB FORCES AND ELECTRIC FIELD INTENSITY 19 The force has a magnitude of 0.18N and a direction given by the unit vector —0.88, —0.6a,. In component form, F, = 0.144a, — 0.108, N 2.2. Refer to Fig. 2-9. Find the force on a 100 #C charge at (0,0, 3) m if four like charges of a 20 #C are located on the x and y axes at +4m. Zz Fig. 2-9 Consider the force due to the charge at y=4, (207*)(20 x 1075) ( + ss) 4x(10-°/364)(5)* $s The y component will be canceled by the charge at y=-4. Similarly, the x components due to the other two charges will cancel. Hence F= 4(32)(2«.) = 1.738, N 23. Refer to Fig. 2-10. Point charge @Q,=300u%C, located at (1, —1, ~3)m, experiences a force F, = 8a, — 8a, + 4a,N vy G,-3, -2) d,-1,-3) Fig. 2-10 due to point charge Q, at (3, -3, —2) m. Determine Q,. Ry = —2a, + 2a, — 9, Note that, because 20 COULOMB FORCES AND ELECTRIC FIELD INTENSITY {CHAP. 2 the given force is along Rz, (see Problem 1.23), as it must be. 2:22 Re 4ne,R?* _ = 00x 10")Q,_ (—2a, + 2a ~ a, 8a, — 8a, + 48,70 Tome 3 ) Solving, Q.=—40yC. 2.4. Find the force on a point charge of 50 uC at (0,0,5)m due to a charge of 5002 yC that is xf uniformly distributed over the circular disk r=Sm, z=OQm (see Fig. 2-11). Fy a@=pyrardp x Fig. 2-11 The charge density is 2 5007 x10 | 4 2 Pera gaye OEE 10 Che In cylindrical coordinates, R=-—ra, + 5a, Then each differential charge results in a differential force _ (50x 10-)(p,r dr dg) (= + Ss) © 4n(10-°/36n)(? + 25) \ VF 95 Before integrating, note that the radial components will cancel and that a, is constant. Hence 25 15 (50 x 10-)(0.2 x 104) 5r dr dg ‘rio */36n\(r? +25)" 17 = 908 fo © maa vn |: a, = 16.56a, N F= 2.5. Repeat Problem 2.4 for a disk of radius 2m. xb. Reducing the radius has two effects: the charge density is increased by a factor ent P2_ (5) a ayo while the integral over r becomes rdr GF 25yi2= 9.0586 tdr foe fray =0.0143 instead of { i The resulting force is F=6. 25)(ore 0143" 0. aaag 16.56, N)=25.27a, N CHAP. 2] COULOMB FORCES AND ELECTRIC FIELD INTENSITY 23 component. Referring to Fig. 2-15, R= —4a, +3a, 20x 10-° ‘ - ) = =, 448, an E IneA3) 57.6a, + 43.28, Vim [a Pe z P-2,-1, 4) R y (2,-4,2) “ * Fig. 2-15 2.41. As shown in Fig. 2-16, two uniform line charges of density p,;=4nC/m lie in the a x=0 plane at y=+4m. Find E at (4,0, 10) m. ¥ Fig. 2-16 The line charges are both parallel to a,; their fields are radial and parallel to the xy plane. For either line charge the magnitude of the field at P would be The field due to both line charges is, by superposition, E= 2 Fos 45°), = 18a, V/m 2.12. Develop an expression for E due to charge uniformly distributed over an infinite plane with a density p,. 24 COULOMB FORCES AND ELECTRIC FIELD INTENSITY [CHAP. 2 The cylindrical coordinate system will be used, with the charge in the z=0Q plane as shown in Fig. 2-17. gen Oro Ob (ate) Vre +x © anedt + 2°) Fig. 2-17 Symmetry about the z axis results in cancellation of the radial components. Ex (24% pz drdp dmedr? + 24 ™ se lale nz a,=2 a, “26, Pee 2€ This result ix for points above the xy plane. Below the xy plane the unit vector changes to —a,. The generalized form may be written using s, , the unit normal vector: als 26 The electric field is everywhere normal to the plane of the charge and its magnitude is independent of the distance from the plane, 2.13. As shown in Fig. 2-18, the plane y=3m_ contains a uniform charge distribution of density p, =(10~*/67) C/m*. Determine E at al} points. 3,2) Fig. 2-18 For y>3m, aft Eo = 30a, Vin and for y<3m, CHAP. 2] COULOMB FORCES AND ELECTRIC FIELD INTENSITY 25 2.14. Two infinite uniform sheets of charge, each with density p,, are located at x=+1 (Fig. 2-19}. Determine E in all regions. Only parts of the two sheets of charge are shown in Fig. 2-19. Both sheets result in E fields that are directed along x, independent of the distance. Then (plea, x<-l1 E,+ E,= 0 -1<x<1 {e,féom x>1 2.15, Repeat Problem 2.14 with p,on x=—-1 and—p,on x=1. 0 x<-l E, + E, =} (0,/€o)a, -1<x<1 0 x>t 2.16. A uniform sheet charge with p,=(1/37)aC/m? is located at z=Sm_ and a uniform line charge with p,=(—25/9)nC/m at z=-3m, y=3m. Find E at (x, —1, 0)m. The two charge configurations are parallel to the x axis. Hence the view in Fig. 2-20 is taken looking at the yz plane from positive x. Due to the sheet charge, Bop, & | 3] tee, Br EN, Px, -1,0) Fig. 2-20 AtP, a,=—a, and £, =—6a, V/m Due to the line charge, E, =P 28 COULOMB FORCES AND ELECTRIC FIELD INTENSITY (CHAP. 2 The radial components cancel. Therefore fl f raed "ine a Pr “Big i Note that as a, E->(p,/2e,)a,, the field due to a uniform plane sheet. Charge lies on the circular disk rsa, z=0 with density p,=p)sin’?¢ Determine E at (0, 9, h). dE= polsin’ @)r dr do ( —ra,+ a) anetr+h) Vv ah The charge distribution, though not uniform, still is symmetrical such that all radial components cancel. fof “ (sin? p)rdrdg fat -1 +3) “ane h | Pte? bes \arah kh 2.22. Charge lies on the circular disk r<4m, z=0 with density p,=(10-*/r) (C/m’). 2 Determine Eat r=0, z=3m. _ (0“fryrdr dg /—ra, + 3a, oem sned +9) aes) (Wim) As in Problems 2.20 and 2.21 the radial component vanishes by symmetry. E=(2.7x wy [" (4 rf ee =1.51%10'a, Vim or L.Sla,MV/m 2.23. Charge lies in the z=—3m_ plane in the form of a square sheet defined by -25x< 2m, —2=y<2m_ with charge density p,=2(x?+y?+9)”nC/m’. Find E at the origin. From Fig. 2-25, —xa,—ya,+3a, (m) dQ = p, dx dy = 2x? +y7 +9) X10 de dy (C) Zz Q,-2,-3) Q,2,-3) Fig. 2-25 and so. dE= 2(x? + y? +9)? x 107% dx oy (me ~ya, + 3a, vi “ane(x? + y? +9) Vitayea9 ) (ie) CHAP. 2] COULOMB FORCES AND ELECTRIC FIELD INTENSITY 29 Due to symmetry, only the z component of E exists. 9 E= ffeovee dx dy . = 8640, V/m 2t-2 4mey 2.24. A charge of uniform density p,=0.3nC/m? covers the plane 2¢-—3y+z=6m. FindE 2.25. 2.26. BOR rc on the side of the plane containing the origin. Since this charge configuration is a uniform sheet, E=p,/2€9 and E=(17.0)a,V/m. The unit normal vectors for a plane Ax+ By+Cz=D are 1, + Ba, + Ca, VA? + BP + C? Therefore, the unit normal vectors for this plane are a,=t 2a, — 3a, +a, via From Fig. 2-26 it is evident that the unit vector on the side of the plane containing the origin is produced by the negative sign. The electric field at the origin is ~2a, + 3a, “) y vis )vim = «7.0 Supplementary Problems Two point charges, Q,~=250#C and Q,=—300pC, are located at (5,0,0)m and (0,0,~5)m, respectively. Find the force on Q, Ans. r= (13.9%) N v2 Two point charges, Q,=30pC and Q,=-100pC, are located at (2,0,5)m and (-1,0,—2)m, respectively. Find the forceon@Q,. Ans. F, = (0.465(—*—™) N In Problem 2.26 find the force on Q,. Ans. —F, Four point charges, each 20 uC, are on the x and y axes at 4m. Find the force on a 100-zC point charge at (0,0,3)m. Ans. 1.73a,N 2,29, 2. 2.31. 2,32, 2.33. 2.37. 2.38, 2.40. 2,41, 2.42. 2.43. 2,44, COULOMB FORCES AND ELECTRIC FIELD INTENSITY (CHAP. 2 Ten identical charges of 500 4C each are spaced equally around a circle of radius 2m. Find the force on a charge of —20 yC located on the axis, 2m from the plane of the circle. Ans. (79.5)(~a,)N Determine the force on a point charge of 50 uC at (0,0, 5) m due to a point charge of 5002 4C at the origin. Compare the answer with Problems 2.4 and 2.5, where this same total charge is distributed over a circular disk. Ans. 28.3a,N Find the force on a point charge of 30 uC at (0,0,5}m due to a 4m square in the z=0 plane between x=+2m and y=+42m_ with a total charge of 500 uC, distributed uniformly. Ans. 4.66a,N Two identical point charges of @(C) each are separated by a distance d(m). Express the electric field E for points along the line joining the two charges. Ans. If the charges are at x=0 and x=d, then, for O0<x<d, -2 [3-g ~ 4meglx? (d~xy }. (vim) Identical charges of Q(C) are located at the eight corners of a cube with a side €(m). Show that the coulomb force on each charge has magnitude (3.2907/4ste,¢7) N. Show that the electric field E outside a spherical shell of uniform charge density p, is the same as E due to the total charge on the shell located at the center. Develop the expression in cartesian coordinates for E due to an infinitely long, straight charge 5 . a, + configuration of uniform density pe. Ans. E “ine yt Two uniform line charges of p,=4nC/m each are parallel to the z axis at x=0, y=t4m. Determine the electric field E at (+4,0,z)m. 9 Ans, +188, V/m Two uniform line charges of p-=SnC/m each are paraliel to the x axis, one at z=0, y=-2m and the other at z=0, y=4m. Find Eat (4,1,3)m. Ans. 30a, V/m Determine E at the origin due to a uniform tine charge distribution with p, =3.30nC/m located at xs3m, y=4m. Ans. —7.13a,—9.50a, V/m Referring to Problem 2.38, at what other points will the value of E be the same? = Ans. (0, 0, z) Two meters from the z axis, [E| due to a uniform line charge along the z axis is known to be 1.80 x 10*V/m. Find the uniform charge density p-. Ans. 2.0 pC/m The plane -—x+3y—6z=6m_ contains a uniform charge distribution p,=0.53nC/m?. Find E on the side containing the origin. Ans. o(*-— a+ 68.) Vim Two infinite sheets of uniform charge density p,=(10°°/6)C/m’ are located at z= —Sm and y=-—5m. Determine the uniform line charge density p, necessary to produce the same value of E at (4, 2, 2), if the line charge is located at z2=0, y=0. Ans, 0.667 nC/m Two uniform charge distributions are as follows: a sheet of uniform charge density p,= —50nC/m? at y=2m and a uniform line of pp=O0.2uC/m at z=2m, y=—1m. At what points in the region will E be zero? Ans. (x, —2.273, 2.0)m A uniform sheet of charge with p, = (—1/32}nC/m’ is located at z=5m_ and a uniform line of charge with p,=(—25/9)nC/m is located at z=-3m, y=3m. Find the electric field E at (0,-1,0)m. Ans. 8a, V/m CHAP. 3] ELECTRIC FLUX AND GAUSS’ LAW 33 Fig. 3-1 Lx. <. /\ @ @) Fig. 3-2 If in the neighborhood of point P the lines of flux have the direction of the unit vector a (see Fig. 3-3) and if an amount of flux d‘V crosses the differential area dS, which is a normal to a, then the electric flux density at P is =f 2 D=75% (C/m’) Fig. 33 A volume charge distribution of density (C/m*) is shown enclosed by surface S in Fig. 3-4. Since each coulomb of charge Q has, by definition, one coulomb of flux Y, it follows that the net flux crossing the closed surface § is an exact measure of the net charge enclosed. However, the 34 ELECTRIC FLUX AND GAUSS" LAW [CHAP. 3 density D may vary in magnitude and direction from point to point of S; in general, D will not be along the normal to S._ If, at the surface element dS, D makes an angle @ with the normal, then the differential flux crossing dS is given by aW = D dS cos @= D+ dSa,, = D- dS where dS is the vector surface element, of magnitude d§ and direction a,. The unit vector a, is always taken to point out of S, so that d'¥ is the amount of flux passing from the interior of S to the exterior of S through dS. 3.3. GAUSS’ LAW Gauss’ law states that The total flux out of a closed surface is equal to the net charge within the surface. This can be written in integra] form as $B d8~ Qer A great deal of valuable information can be obtained from Gauss’ law through clever choice of the surface of integration; see Section 3.5. 3.4 RELATION BETWEEN FLUX DENSITY AND ELECTRIC FIELD INTENSITY Consider a point charge Q (assumed positive, for simplicity) at the origin (Fig. 3-5). If this is enclosed by a spherical surface of radius r, then, by symmetry, D due to Q is of constant magnitude over the surface and is everywhere normal to the surface. Gauss’ law then gives =} D-d8=D f.as = D(4xr?) CHAP. 3] ELECTRIC FLUX AND GAUSS’ LAW 35 from which D=(Q/4ar*. Therefore Q 2 = Gar age But, from Section 2.2, the electric field intensity due to Q is - @ Aner? a, It follows that D= e€,E. More generally, for any electric field in an isotropic medium of permittivity €, D=cE Thus, D and E fields will have exactly the same form, since they differ only by a factor which is a constant of the medium. While the electric field E due to 2 charge configuration is a function of the permittivity €, the electric fiux density Dis not. In problems involving multiple dielectrics a distinct advantage will be found in first obtaining D, then converting to E within each dielectric. 3.5 SPECIAL GAUSSIAN SURFACES The surface over which Gauss’ law is applied must be closed, but it can be made up of several surface elements. If these surface elements can be selected so that D is either normal or tangential, and if |D{ is constant over any element to which D is normal, then the integration becomes very simple. Thus the defining conditions of a special gaussian surface are 1. The surface is closed. 2. At each point of the surface D is either normal or tangential to the surface. 3. Dis sectionally constant over that part of the surface where D is normal. EXAMPLE 3 Use a special gaussian surface to find D due to a uniform line change p, (C/m). Take the line charge as the z axis of cylindrical coordinates (Fig. 3-6). By cylindrical symmetry, D can Lia, (Clim) aQ 3.6. 3.7. ELECTRIC FLUX AND GAUSS' LAW (CHAP. 3 The charge enclosed from the plane is Q = (4m7)(40 wC/m’) = 160 pC and from the line Q= (2 m\—6 pCfm) = —12 pC Thus, Qune= = 160~ 12 = 148 pC. A point charge Q is at the origin of a spherical coordinate system. Find the flux which crosses the portion of a spherical shell described by a= 656 (Fig. 3-10). What is the result if a=0 and B=2x/2? Fig. 3-10 The total flux “W=(@ crosses a complete spherical shell of area 4x7’. The area of the strip is given by 20 6B A -| { Psin @d0d@ = 2nr*(cos a — cos f) Then the flux through the strip is A Q big =—s Qa - Vou = 97,7 2 = "5 (cos a — cos B) For a=0, B=2/2 (ahemisphere), this becomes Wye = Q/2- A uniform line charge with p-=50C/m_ lies along the x axis. What fiux per unit length, W/L, crosses the portion of the z=—3m_ plane bounded by y= +2m? The flux is uniformly distributed around the line charge. Thus the amount crossing the strip is obtained from the angle subtended compared to 2x. In Fig. 3-11, o = 2 arctan () = 1.176 rad Ww (= The —= 50({—— } = 9.3 en L 50! ) 9.36 pC/m CHAP. 3) ELECTRIC FLUX AND GAUSS’ LAW 39 3.8. A point charge, Q=30nC, is located at the origin in cartesian coordinates. Find the electric flux density D at (1,3, —4)m. Referring to Fig. 3-12, oar ad _30x 10° (* +3a,— a) 4n(26) Vv = 11 f 2: + 3a, — 4a, 2 (9.18 10 ye ] Che or, more conveniently, D = 91.8 pC/m’. z 2 y R x ,3,-4) D Fig. 3-12 3.9. Two identical uniform line charges lie along the x and y axes with charge densities p,= 20 uC/m. Obtain D at (3,3, 3) m. ‘The distance from the observation point to either line charge is 3V2m. Considering first the line charge on the x axis, D,=-2£0,= 20 pC/m (24) 2a” 2nGVIm)\ V2 and now the y axis line charge, =Ptg — 20uCim (a, +a, Pe Derg maviw ( v2 ) 40 ELECTRIC FLUX AND GAUSS’ LAW (CHAP. 3 The total flux density is the vector sum, pe man (3) =@. 2(2ae) uch? 3.10. Given that D=10xa, (C/m?), determine the flux crossing a 1-m? area that is normal to the x axis at_ x =3m. Since D is constant over the area and perpendicular to it, W = DA = (30 C/m)(1 m?) = 30€ 3.11. Determine the flux crossing a 1mm by 1mm area on the surface of a cylindrical shell vi. at r=10m, z=2m, ¢=53.2° if ey D=2xa, + 2(1t—y)a,+4za, (C/m’) At point P (see Fig. 3-13), x = 10.cos 53.2°=6 y=10sin 53.2 =8 z P(LO, 53.2°, 2) Then, at P, D = 12a, -14a,+8a, C/m* Now, on a cylinder of radius 10m, a t-mm? patch is essentially planar, with directed area aS = 10°“(0.6a, + 0.8a,) m? Then dW = D+ dS = (12a, — 14a, + 8a,)> 10-°(0.6a, + 0.8a,) = -4.0 nC The negative sign indicates that flux crosses this differential surface in a direction toward the z axis rather than outward in the direction of dS. 3.12. A uniform line charge of pe=3pC/m lies along the z axis, and a concentric circular cylinder of radius 2m has , = (—1.5/4x) ¢C/m?. Both distributions are infinite in extent with z. Use Gauss’ Jaw to find D in all regions. Using the special gaussian surface A in Fig. 3-14 and processing as in Example 3, p=fts, O<r<2 CHAP. 3] ELECTRIC FLUX AND GAUSS’ LAW 43 3.16. The volume in spherical coordinates described by ra contains a uniform charge density pe. Use Gauss’ law to determine D ani compare your results with those for the corresponding E field, found in Problem 2.54. What point charge at the origin will result in the same D field for r >a? For a gaussian surface such as E in Fig. 3-17, Que= $D- dS 3 wW’p = D(a?) and p-Fa, r=a For points outside the charge distribution, r>a 4 3 map = D(4xr} whence I€ a point charge @=4na'p is placed at the origin, the D field for r>a_ will be the same. This point charge is the same as the total charge contained in the volume. 3.17. A parallel-plate capacitor has a surface charge on the lower side of the upper plate of +p, =i, (C/m*). The upper surface of the lower plate contains —p, (C/m). Neglect fringing and a use Gauss’ law to find D and E in the region between the plates. janced All flux leaving the positive charge on the upper plate terminates on the equal negative charge on the lower plate. The statement negiect fringing insures that all flux is normal to the plates. For the special gaussian surface shown in Fig. 3-18, | D-as+{ p-as+ [ D-ds hoo — ise =0+ L. D-d8+0 or pA=D fas =DA where A is the area. Consequently, D=p.a,(Cim’) and E="e, (Vim) Both are directed from the positive to the negative plate. 3.18. 3.19, 3.20. 3.21. 3.22. 3.23. 3.24. 3.25. ELECTRIC FLUX AND GAUSS’ LAW [CHAP. 3 LA Fig. 3-18 Supplementary Problems Find the net charge enclosed in a cube 2m on an edge, parallel to the axes and centered at the origin, if the charge density is =50xtcos(2y) (che e cos G ») C/m*) Ans. 84.9 nC Find the charge enclosed in the volume 1=r<3m, 0=g~<=2/3, OSz<=2m_ given the charge density g=2zsin?@ (C/m‘). Ans. 4.91C Given a charge density in spherical coordinates eos @ Po (rf) find the amounts of charge in the spherical volumes enclosed by r=", r=5m, and r=, Ans. 3.97 por, 6.2Apor, 6.28por A closed surface S contains a finite line charge distribution, 0<€<2m, with charge density € Pe=— posing (Cim) What net flux crosses the surface S? Ans. ~2p) (C) Charge is distributed in the spherical region r=2m_ with density —200 pra (uCfm*) What net flux crosses the surfaces r=I1m, r=4m, and r=500m? Ans. ~B002 uC, -16002 pC, — 16002 pC A point charge @ is at the origin of spherical coordinates and a spherical shell charge distribution at r=a_ has a total charge of Q'—Q, uniformly distributed. What flux crosses the surfaces r= k for k<a and k>a? Ans. Q, Q A uniform line charge with g,=34C/m lies along the x axis. What flux crosses a spherical surface centered at the origin with r=3m? Ans. 18 uC If a point charge Q is at the origin, find an expression for the flux which crosses the portion of a sphere, Bra centered at the origin, described by w= = 8. Ans. in @ CHAP. 3] ELECTRIC FLUX AND GAUSS’ LAW 45 3.26. 3.27. 3.28. 3.29. 3.30. 3.31. 3.32. 3.33. 3.34, 3.35, 3.36. 3.37. A point charge of Q (C} is at the center of a spherical coordinate system. Find the flux Y which crosses an area of 4x m? on a concentric spherical shell of radius 3 m. Ans. Q/9 (C) An area of 40.2m? on the surface of a spherical shell of radius 4m is crossed by 10 uC of flux in an inward direction. What point charge at the origin is indicated? Ans. —SO0pC A uniform line charge p; lies along the x axis. What percent of the flux from the line crosses the strip ofthe y=6 planehaving ~1£2=1? Ans. 5.26% A point charge, Q@=3nC, is located at the origin of a cartesian coordinate system. What flux crosses the portion of the z=2m_ plane for which -—45x=4m and —4=y=4m? Ans, 0.50C A uniform line charge with p,=5yC/m_ lies along the x axis, Find D at (3,2, 1) m. Ans. (o3s9(==*) Chm? A point charge of +Q is at the origin of a spherical coordinate system, surrounded by a concentric uniform distribution of charge on a spherical shell at_ 7=a for which the total charge is —Q. Find the fiux Y crossing spherical surfaces at. <a and >a. Obtain D in all regions. + < Ans. w= 4270 =[ Qo 1sa 0 r>a Given that D=500e~°"a, (C/m’), find the flux YW crossing surfaces of area 1 m* normal to the x axis and located at x=1m, x=5m, and x= 10m. Ans. 452 uC, 303 uC, 184 pC Given that D=5x’a,+10za, (C/m?), find the net outward flux crossing the surface of a cube 2m on an edge centered at the origin. The edges of the cube are parallel to the axes. = Ans. 80C Given that D= ea, ~2 i a, (C/m’) in cylindrical coordinates, find the outward flux crossing the right circular cylinder described by r=2b, z=0, and z=Sb(m). Ans. 129b*(C) Given that sing D=2 ae COs Gay 3p a, in cylindrical coordinates, find the flux crossing the portion of the z=0 plane defined by r=a, Os o=2/2. Repeatfor 3n/2=@=22. Assume flux positive in the a, direction. a Ans. 3, wis In cyclindrical coordinates, the disk r=@, z=0 carries charge with nonuniform density p.{r, @). Use appropriate special gaussian surfaces to find approximate values of D on the z axis (a) very close to the disk (0<z~<€a), (b) very far from the disk (z> a). Ans, (0) EO, py 2 were 9=[" [ar ordre A point charge, Q=2000pC, is at the origin of spherical coordinates. A concentric spherical distribution of charge at r=1m has a charge density p,=40x pC/m’. What surface charge density on a concentric shell at_ r=2m would resultin D=0 for r>2m? xh -¥ Ans. ~71.2 pC/m? 48 DIVERGENCE AND THE DIVERGENCE THEOREM [CHAP. 4 In Fig. 4-2 the cube is turned such that face 1 is in full view; the x components of A over the faces to the left and right of 1 are indicated. Since the faces are small, [ A+ dS=—A,(x) Ay Az heft face [ A> dS~A,(x + Ar) Ay Az right face aA, | Ax | Ay A: ox yee AQ) A, + Ax) aS as so that the total for these two faces is 2A Ay Ay Az ox The same procedure is applied to the remaining two pairs of faces and the results combined. OA, | JA, 4 OAL A-dS= ( 4 Sy =) f ox ay * oz Ax Ay d2 Dividing by Ax Ay Az= Av and letting Av—0, one obtains OA, | GA, | BA; yy SO div A= at ay * az (cartesian) The same approach may be used in cylindrical (Problem 4.1) and in spherical coordinates. div A=-—(rA,) + -——* oA, (cylindrical iv A= Ze. +t 36 tae cylindrical) 1a, t 1 aAy . divA= 23 (PA) + Fein 600 2a, sin 6) + in 6 (spherical) EXAMPLE 1. Given the vector fied A= sx%(sin =e. find divA at x= 1. of. ™\z _ max 5 Na = = 52°(cos) 5 Ox sin 3 * cos + 10% sin and divAl,.1~=10. EXAMPLE 2. In cylindrical coordinates a vector field is given by A=rsin ga, +r? cos oa, +2re “a, Find div A at (3, 1/2, 0). div A=12 (r'sin 0415 (Peos $)+ 2 re") =2sing —rsin g— 10re* 1 7 and div Alcina.) = 2 sin -3 5aine - 10(5 Je =-5 CHAP. 4] DIVERGENCE AND THE DIVERGENCE THEOREM 49 EXAMPLE 3. In spherical coordinates a vector field is given by A=(5/r*)sin Oa, +r cot Oag+ rsin O cos pa,. Find div A. diva = 43 (sin +8 2 (sin 8 cot @) + 13 a canb ay (HP 0008 4) = -1 sing 4.3 DIVERGENCE OF D From Gauss’ law (Section 3.3), pp-as +dS Av oe In the limit, D-dsS QO lim —— =divD = lim == amo vu _ ano au e This important result is one of Maxwell’s equations for static fields: dvD=p and divE=£ € if € is constant throughout the region under examination (if not, divé€E=). Thus both E and D fields will have divergence of zero in any isotropic charge-free region. EXAMPLE 4. In spherical coordinates the region ra contains a uniform charge density p, while for r>a the charge density is zero. From Problem 2.54, E=E,a,, where E,=(pr/3€o} for rsa and E,=(pa*/3€or7} for r>a. Then, for r<a, and, for _r>a, 44 THE DEL OPERATOR Vector analysis has its own shorthand, which the reader must note with care. At this point a vector operator, symbolized ¥, is defined in cartesian coordinates by aX), HM), HD vera ut ay + ae a, In the calculus a differential operator D is sometimes used to represent d/dx. The symbols Vand J are also operators; standing alone, without any indication of what they are to operate on, they look strange. And so ¥, standing alone, simply suggests the taking of certain partial derivatives, each followed by a unit vector. However, when V is dotted with a vector A, the result is the divergence of A. Be Ahr Ara ay a oy oz a @ a VeAan(Satse tse a): (A,a, + A, a, + Aa.) = Hereafter, the divergence of a vector field will be written V- A +O DIVERGENCE AND THE DIVERGENCE THEOREM (CHAP. 4 Warning! The del operator is defined only in cartesian coordinates. When V - A is written for the divergence of A in other coordinate systems, it does not mean that a del operator can be defined for these systerns. For example, the divergence in cylindrical coordinates will be written as 1aAg , 34, v aT A + 36 t a2 (see Section 4.2). This does not imply that vat2e ete, 2O,, in cylindrical coordinates. In fact, the expression would give false results when used in VV (the gradient, Chapter 5) or V A (the curl, Chapter 9). 4.5 THE DIVERGENCE THEOREM Gauss’ law states that the closed surface integral of D~ dS is equal to the charge enclosed. If the charge density function p is known throughout the volume, then the charge enclosed may be obtained from an integration of p throughout the volume. Thus, fD-as= | padv = Qa But p=V-D, and so pp-as~| (v-D)dv This is the divergence theorem, also known as Gauss’ divergence theorem. It is a three-dirmmensional analog of Green’s theorem for the plane. While it was arrived at from known relationships among D, Q, and p, the theorem is applicable to any sufficiently regular vector field. divergence theorem fa-as=[(v-A)dv Is by Of course, the volume v is that which is enclosed by the surface S. EXAMPLE 5. The region <a in spherical coordinates has an electric field intensity Examine both sides of the divergence theorem for this vector field. For $, choose the spherical surface r= bea. gE-as fo-wa eb )- 2 =52(r8 pry Le ice (b?sin 640d a,) VE=35 f\-2 ae pm op a pm =f [Bsn odoag then { [ [Besa odrdoas lh ty 3: b ln he _ Anpb* _ Anpb* ~ 3e " 3e The divergence theorem applies to time-varying as well as static fields in any coordinate CHAP. 4] DIVERGENCE AND THE DIVERGENCE THEOREM 53 4.5. 4.6, sh = 4.7. 4.8. 49. 4.10. is. ro 4.11, Given A=x’a, + yza,+xya,, find V-A. a ae 2 aya VeAns +5, Uz) +5 Gy) =a +z Given A=(x?+y7)"a,, find V-A at (2, 2,0). VeA=—S (xt ty 2x) and = V- Alaz) = 8.84 x 107? Given A=rsin pa, + 2r cos ga, +2z7a,, find V- A. 18,4. 1a a = Ta sin &) + 55 (2r cos 6) + 5 22”) =2sin @ —2sin @ +42 = 42 Given A= 10sin? ga, + ra, + [(z7/r) cos* pa,, find V« A at (2, , 5). 10 sin’ ¢ +2z cos’ r VeA= and V+ Algae =5 Given A=(5/r?)a, + (10/sin @)a,—r?p sin Oa,, find V- A. 1 3 1 ao . rain 6.36 1) * 7 sin Bag (7 @8i0 8) = —F 14 VrAR= 5, G+ Given A=Ssin @ag+Ssinga,, find V+ A at (0.5, 2/4, 2/4). __i a a 1 Fae — ap S0s 8 COs FAR 3d 8 Ot Gag FIMO 1O +5 og and V+ Altoscare.aray = 24.14 Given that D=p za, in the region —1<z<1 in cartesian coordinates and D= (Poz/|z|)a, elsewhere, find the charge density. v-D=p For -1=2z<=1, a P= 5, (Poz) = Po and for z<—-l or z>1, 3 pa (Fa)=0 The charge distribution is shown in Fig. 4-5. 4.12. 4.3. 4.14. 4.15. 4.16. DIVERGENCE AND THE DIVERGENCE THEOREM (CHAP. 4 Given that D=(10r?/4)a,(C/m’) in the region O<r=3m_ in cylindrical coordinates and D=(810/4r)a, (C/m?) elsewhere, find the charge density. For 0<7<3m, =e ee 4 and for r>3m, 10 p75, (8l0/4)=0 Given that Q cee] (1-cos 37)a, in spherical coordinates, find the charge density. a1 8f a @ ae | a2, p -32[r agi ll 008 36) | =F sin Br In the region O<r=ilm, D=(—2x10-‘/r)a, (C/m*) and for r>1m, D=(-4x 10~*/r*)a, (C/m?), in spherical coordinates. Find the charge density in both regions. For 0<r=1m, -2x10 4 (2x10) = (Cim’) le i 2 p= Q ir and for r>1m, 19 4) pa55, (4x =o In the region r=2, D=(S5r7/4)a, and for r>2, D=(20/r?)a,, in spherical coordinates. Find the charge density. For r=2, = i & 4 = 0= 55, Gr'/4)=5r and for r>2, u 1 Pp 3 Fay %)-0 e ia Given that D=(10x7/3)a, (C/m?), evaluate both sides of the divergence theorem for the volume of a cube, 2m on an edge, centered at the origin and with edges parallel to the axes. fv-as=[ {¥>D) dv Nol Since D has only an x component, D+ dS is zero on all but the faces at x=1m and x=-1m (see Fig. 4-6). fo-as=[ fe aden eff 10D syle (-a,) CHAP. 4] DIVERGENCE AND THE DIVERGENCE THEOREM 55 Le er Ts Fig. 4-6 Ye Now for the right side of the divergence theorem. Since ¥+D= 10x’, [o-ma~f f° faa) aay de = ff f fot} war=Be hada 4.17. Given that A=30e~'a,—2za, in cylindrical coordinates, evaluate both sides of the “ divergence theorem for the volume enclosed by r=2, z=0, and z=5 (Fig. 4-7). fa-ds={(v-ayav Fig. 4-7 It is noted that A,=0 for z=0 and hence A- dS is zero over that part of the surface. fa-as= [ [mete -2aeacas [[/ —2(5)a, +r dr dpa, = 60e"(24)(5) — 10(2)(2) = 129.4 For the right side of the divergence theorem: Varo Core) 4 2-22) = 2 _ 396-72 and for-arau= [i [" [ (P-s0e-—2)pdrdpds = 129.4 4.18 Given that D=(10r7/4)a, (C/m?) in cylindrical coordinates, evaluate both sides of the xb. divergence theorem for the volume enclosed by r=im, r=2m, z=0 and z=10m - (see Fig. 4-8). $0-as= | (v-D) av 58 4.37, 4,38, 4,39. 4.40. 4Al. 4.45, DIVERGENCE AND THE DIVERGENCE THEOREM [CHAP. 4 and for _r>b, Bw D= onl e s For r<a, D=0. Find pin all three regions. Ans. 0, po, 0 In the region 0<r<2 (cylindrical coordinates), D=(4r°'+2e-°" + 4r'e")a,, and for r> 2, D=(2.0S7/r)a,. Find p in both regions. Ans. —e-°*, 0 In the region r<2 (cylindrical coordinates), D=(10r+(r/3)Ja,, and for r>2, D= (3/(128r)la,. Find p in both regions. Ans. 20-+r,0 Given D=10sin @a,+2cos 6a,, find the charge density. Ans. seus + 2cot? 6} Given __3r “Paai™ in spherical coordinates, find the charge density. Ans. 3(r? + 3)/(? + 1) Given 10 -_ 2 D=Gll-e (1+ 2 + 27)Ia, in spherical coordinates, find the charge density. Ans, 40e"* In the region r=1 (spherical coordinates), andfor r>1, D=(5/(63r’)Ja,. Find the charge density in both regions. Ans. 4~7,0 The region +=2m_ (spherical coordinates) has a field E=(5r x 10-%/e,)a, (V/m). Find the net charge enclosed by the shell r=2m. Ans. 5.03 X 107°C Given that D=(5r?/4)a, in spherical coordinates, evaluate both sides of the divergence theorem for the volume enclosed between r=1 and r=2. Ans. 715” Given that D=(10r'/4)a, in cylindrical coordinates, evaluate both sides of the divergence theorem for the volume enclosed by r=2, z=0, and z=10. Ans. 8000 Given that D=10sin 6a,+2cos @a,, evaluate both sides of the divergence theorem for the volume enclosed by the shell r=2. Ans. 40x? Chapter 5 The Electrostatic Field: Work, Energy, and Potential 5.1 WORK DONE IN MOVING A POINT CHARGE A charge O experiences a force F in an electric field E. In order to maintain the charge in equilibrium a force F, must be applied in opposition (Fig. 5-1): F=QE F, =—-QE ha 9 oF Fig. 5-1 Work is defined as a force acting over a distance. Therefore, a differentia! amount of work dW is done when the applied force F,, produces a differential displacement dl of the charge; i.e. moves the charge through the distance dé=|dll. Quantitatively, dW =F,+dl=—QE- dl Note that when @Q is positive and dl is in the direction of E, dW =-QEd@€<0, indicating that work was done by the electric field. {Analogously, the gravitational field of the earth performs work on a (positive) mass M as it is moved from a higher elevation to a lower one.} On the other hand, a positive dW indicates work done against the electric field (cf. lifting the mass M). Component forms of the differential displacement vector are as follows: dl= dra, + dya, + dza, (cartesian) di= dra, +r doa, + dza, (cylindrical) di= dra,+rd@ag+rsin Pda, (spherical) The corresponding expressions for df were displayed in Section 1.5. EXAMPLE 1. An electrostatic field is given by E=(x/2+2y)a,+2za, (V/m). Find the work done in moving a point charge Q=—-204C (a) from the origin to (4, 0,0) m, and (6) from (4,0,0) m to (4,2,0)m. (a) The first path is along the x axis, so that dl= dx a,. dW =-QE-di= (20x 10 G+ 2y) ax W=(20x wf (E+ 2y) dx = 8043 , (6) The second path is in the a, direction, so that dl = dya,. 2 w= cox 10-9 f 2x dy = 320 uJ 59 oO THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL [CHAP. 5 5.2 CONSERVATIVE PROPERTY OF THE ELECTROSTATIC FIELD The work done in moving a point charge from one location, B, to another, A, in a static electric field is independent of the path taken. Thus, in terms of Fig. 5-2, Le ~[eea or f Bano where the last integral is over the closed contour formed by () described positively and @ described negatively. Conversely, if a vector field F has the property that $F-dl=0 over every closed contour, then the value of any line integral of F is determined solely by the endpoints of the path. Such a field F is called conservative; it can be shown that a criterion for the conservative property is that the curl of F vanish identically (see Section 9.4). eA @ © Fig. 5-2 EXAMPLE 2. For the & field of Example 1, find the work done in moving the same charge from (4, 2, 0) back to (0, 0, 0) along a straight-line path. 10.8.0) W = (20x 10" [(Z+2y}a, +2 Ma.2.0) (dx a, + dy a,) (00.0) 7g = (20x 10° f (5+2y) ae +2edy toa.) The equation of the path is y=x/2; therefore, dy=4de and W = (20x 10-9 f’teae = —400 ps , From Example 1, 80-+320= 4904] of work was spent against the field along the outgoing, right-angled path. Exactly this much work was returned by the field along the incoming, straight-line path, for a round-trip total of zero (conservative field). §3 ELECTRIC POTENTIAL BETWEEN TWO POINTS The potential of point A with respect to point B is defined as the work done in moving a unit positive charge, Q,,, from B to A. Ww A Vaa=o= -f E-di (J/C or V) It should be observed that the initial, or reference, point is the lower limit of the line integral. Then, too, the minus sign must not be omitted. This sign came into the expression by way of the force F,=-QE, which had to be applied to put the charge in equilibrium. Because E is a conservative field, Vaa = Vac — Vac CHAP. 5) THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL 63 From the calculus, the change in V from M to AN is given by ev ov du dV =— ax +—dy+~ d: ox tay Ta? Now, the del operator, introduced in Section 4.4, operating on V gives wv oa wy a a OY ax ay * ae It follows that dV =VV-dr The vector field VV (also written grad V) is called the gradient of the scalar function V. It is seen that, for fixed |dr|, the change in V in a given direction dr is proportional to the projection of VV in that direction. Thus VV lies in the direction of maximum increase of the function V. Another view of the gradient is obtained by allowing the points M and N to lie on the same equipotential (if V is a potential) surface, V(x, y,z)=c, [see Fig. 5-5(b)]. Then dV = 0, which implies that VV is perpendicular to dr. But dr is tangent to the equipotential surface; indeed, for a suitable location of N, it represents any tangent through M. Therefore, VV must be along the surface normal at M. Since VV is in the direction of increasing V, it points from V(x,y,z)=c, to V(x, y,z)=c2, where ¢,>c,. The gradient of a potential function is a vector field that is everywhere normal to the equipotential surfaces. The gradient in the cylindrical and spherical coordinate systems follows directly from that in the cartesian system. It is noted that each term contains the partial derivative of V with respect to distance in the direction of that particular unit vector. av av av W-=—at+—a,t+— i Vv 3x + 3y a,+ 3; (cartesian) av. av av a wW=s-a, +ragt +3 me (cylindrical) ave av WV =—— a, +2 ty + i ar +738" tr ainaapte (spherical) While VV is written for grad V in any coordinate system, it must be remembered that the del operator is defined only in cartesian coordinates. 5.7. RELATIONSHIP BETWEEN E AND V From the integral expression for the potential of A with respect to B, the differential of V may be written dvV=~-E-dl On the other hand, dV =VV dr Since dl=dr_ is an arbitrary small displacement, it follows that E=-W The electric field intensity E may be obtained when the potential function V is known by simply taking the negative of the gradient of V. The gradient was found to be a vector normal to the equipotential surfaces, directed to a positive change in V. With the negative sign here, the E field is found to be directed from higher to lower levels of potential V. 64 THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL [CHAP. 5 EXAMPLE 4. In spherical coordinates and relative to infinity, the potential in the region r>0 surrounding a point charge Qis V=Q/4meqr. Hence, E=~W= -2(;2). g ar \aregr! Aner” in agreement with Coulomb’s law. (V is obtained in principle by integrating E; so it is not surprising that differentiation of V gives back ¥.) 5.8 ENERGY IN STATIC ELECTRIC FIELDS Consider the work required to assemble, charge by charge, a distribution of n=3 point charges. The region is assumed initially to be charge-free and with E=0 throughout. Referring to Fig. 5-6, the work required to place the first charge, Q,, into position 1 is zero. Then, when Q, is moved toward the region, work equal to the product of this charge and the potential due to Q, is required. The total work to position the three charges is We= W,+ Wi + W, = 0+ (Q2¥2.2)+ (QaVar + OsVa2) The potential V,, must be read “the potential at point 2 due to charge Q, at position 1.” (This rather unusual notation will not appear again in this book.) The work W, is the energy stored in the electric field of the charge distribition. (See Problem 5.17 for a comment on this identification.) Now if the three charges were brought into place in reverse order, the total work would be We = W,+ W,+ W, =0+(Q2V23) + (QiMis + QMi2) When the two expressions above are added, the result is twice the stored energy: 2We = Qi(Vi.2 + Vi.3) + Q2(V2.1 + Vos) + Os(Va.1 + Vs.2) The term Q,(V,,2 + Vi.3) was the work done against the fields of Q, and Q;, the only other charges in the region. Hence, Vi2+Vi3=¥,, the potential at position 1. Then 2We = O1V, + QiV2+ OsVs le and We=5 > nV 2 at for a region containing m point charges. For a region with a charge density p (C/m’) the summation becomes an integration, 1 We = 2 { pV dy Other forms (see Problem 5.12) of the expression for stored energy are 1 if oy 1;D? We =; [ D-Eav Wem3 [ef du We=5 oa a Fig. 5-6 CHAP. 5] THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL 65 In an electric circuit, the energy stored in the field of a capacitor is given by W, = 40V =4CV” where C is the capacitance (in farads}), V is the voltage difference between the two conductors making up the capacitor, and Q is the magnitude of the total charge on one of the conductors. EXAMPLE 5. A parailel-plate capacitor, for which C= eA/d, has a constant voltage V applied across the plates (Fig. 5-7). Find the stored energy in the electric field. Fig. 5-7 With fringing neglected, the field is E=(V/d)a, between the plates and E=O0 elsewhere. 1 We= 3 J eF? du ev =3(@) fe _€AV? ~~ 2d 1 2 =3cv As an alternate approach, the total charge on one conductor may be found from D at the surface via Gauss’ law {Section 3.3). Vv poo, VA Q=|D|A= d 1 1feav)_1 Then w=50v=3( 7 \=5ev Solved Problems 5.1, Given the electric field E=2xa,—4ya, (V/m), find the work done in moving a point charge +2C (a) from (2,0,0)m to (0,0,0) and then from (0,0,0) to (0, 2,0); (6) from (2, 0, 0) to (0, 2, 0) along the straight-line path joining the two points. (See Fig. 5-8.) fenes (a) Along the x axis, y=dy=dz=0, and dW = -2(2xa,) + (de a.) = —ax de 68 THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL (CHAP. 5 (2,3,-4) Fig. §-10 With the line charge along the x axis, the x coordinates of the two points may be ignored. m= V9+16=5m ty = V25 + 144= 13m Then "4 Var = - Pe tp = Pt A |, Der ne," Te 6.88V 5.8. Find the potential at :,=5m with respect to rg=15m due to a point charge Q= 500 pC at the origin and zero reference at infinity. fh a Due to a point charge, - ra OTR To find the potential difference, the zero reference is not needed. Vip =—— AP anes O0V ___500x 10-7 (- 2). AB an(10*/36x) (515. The zero reference at infinity may be used to find V; and Ys. =2 (j= -2 aro 5) 0.90 Ms" res Then Vas = Vs— Vis=0.60V 598. Forty nanocoulombs of charge is uniformly distributed around a circular ring of radius 2m. Find the potential at a point on the axis 5 m from the plane of the ring. Compare with the result where all the charge is at the origin in the form of a point charge. With the charge in a fine, pedé 4n€oR 40x i0-" _10~* Here Or Data) oe I and (see Fig. 5-11) R=V29m, dé=(2m)d¢. Ve c _G0"F/2)Q) dp 4x(10-~ 4n(10-°/362)V29 CHAP. 5] THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL 69 dQ =p,rdo Fig. 5-11 If the charge is concentrated at the origin, _40x 10° = =72.0V re o(5) §.10. Five equal point charges, @=20nC, are located at x=2,3,4,5,6m. Find the poten- ei, tial at the origin. 1 Ane 2x10 /1 1.1.1.2 sett tee) ezsiv Gtitatste) $ Ry 4n€ §.11, Charge is distributed uniformly along a straight line of finite length 2 (Fig. 5-12). Show that for two external points near the midpoint, such that 7, and r, are small compared to the length, the potential V,z is the same as for an infinite line charge. ef Le aox Fa o-L Fig. 5-12 70 5.12. THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL [CHAP. 5 The potential at point 1 with zera reference at infinity is « ad. ya2[ pe hy 4re(z* + ryt —_2Pe Ane, [in + V2? + FDS Pe 2ney [in (L 4 VEP +r) - Inn] Similarly, the potential at point 2 is Po in(L + VET +A) — Inn} 2nen Nowif L>r, and Ln, fe Vieq - pag, O2L— Ine) Be Ad In 2h — " ong Fe (in 2b — Inn) Then Waa Ye Poin? 2néy he which agrees with the expression found in Problem 5.7 for the infinite line. Charge distributed throughout a volume u with density p gives rise to an electric field with energy content I pV du Show that an equivalent expression for the stored energy is 1 Wy 3 | &E’ du Figure 5-13 shows the charge-containing volume v enclosed within a large sphere of radius R. Since pg vanishes outside v, 1 1 1 weet { ede=t { =F . af 0 FF ealY 3D een PH volume woh Sphere Fig. 5-13 CHAP. 5) THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL 73 5.19. $.20. 52. 5.22. 5.23. 5.24. 5.25. potential represents 0 r<a E--wWe= (ate, r>a ae pm pm yas? We=4 { cok*a 04 | ff () sin 0 drd0.d = 2n€,V3a 2 2h bi \r Note that the total charge on the shell is, from Gauss’ law, €oVoa! 2 7 \tana’) = 4e€yVoa o-va-( while the potential at the shell is V=Vy. Thus, W,=30V, the familiar result for the energy stored in a capacitor (in this case, a spherical capacitor with the other plate of infinite radius). Supplementary Problems Find the work done in moving a point charge @Q=-~20uC from the origin to (4, 2, 0) m in the field E=2(x + 4y)a, + 8xa, (V/m) along the path x?=8y. = Ans. 1.60mJ Repeat Problem 5.2 using the direct radial path. Ans, -39.35 uJ (the nature of the singularity along the z axis makes the field nonconservative) Repeat Problem 5.2 using the path shown in Fig. 5-14. Ans, -117.9 J z (2, 1/4, 1/2) Find the work done in moving a point charge Q=3yC from (4m,2,0) to (2m, 2/2, 2m), cylindrical coordinates, in the field E=(10°/ra,+10°za, (V/m). Ans. —0.392 Find the difference in the amounts of work required to bring a point charge Q=2nC from infinity to r=2m_ and from infinity to +=4m, in the field E=(10'/r)a, (V/m)_ Ans. 1.39107] A uniform line charge of density p-=1nC/m is arranged in the form of a square 6m on a side, as shown in Fig. 5-15. Find the potential at (0, 0,5) m. Ans. 35.6V Develop an expression for the potential at a point d meters radially outward from the midpoint of a finite line charge L meters long and of uniform density p,(C/m). Apply this result to Problem 5.24 as a check. Lj2+ Vd? + L7/4 Pea, A: a t ) Ans. hen d 74 5.26. $8.27. 5.28. 5.29. 5.30. 51. $.32. 5.33. 5.34, 5.35. 5.36, 5.37, THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL [CHAP. 5 Show that the potential at the origin due to a uniform surface charge density p, over the ring z=0, R<r<R+1_ is independent of R. A total charge of 160 nC is first separated into fours equal point charges spaced at 90° intervals around a citcle of 3m radius. Find the potential at a point on the axis, Sm from the plane of the circle. Separate the total charge into eight equal parts and repeat with the charges at 45° intervals, What would be the answer in the limit’ p,= (160/61) nC/m? Ans. 247 V In spherical coordinates, point A is at a radius 2m while B is at 4m. Given the field E= (-16/r}a, (V/m), find the potential of point A, zero reference at infinity. Repeat for point B. Now express the potential difference V,—V, and compare the result with Problem 5.6. Ans. Vs=2Vp=-8V If the zero potential reference is at r=10m and a point charge Q=0.5nC is at the origin, find the potentials at r=5m and r=15m. At what radius is the potential the same in magnitude as that at r=5m__ but opposite in sign? Ans. 0.45 V, -0.15 V, © A point charge @=0.4nC is located at (2,3,3}m in cartesian coordinates. Find the potential difference V,,, where point A is (2,2, 3) m and Bis(-2,3,3)m. Ans. 2.70V Find the potential in spherical coordinates due to two equal but opposite point charges on the y axis at y=+d/2. Assume r>d. Ans. (Qdsin 0)/(4re9r”) Repeat Problem 5.31 with the charges on the z axis. Ans. (Qd cos 6)/(4r€ or?) Find the charge densities on the conductors in Problem 5.14. +60€, a _ 9 ~W€o a (C/m’} on p =0, = Ans. (C/m*) on @ =f A uniform line charge p,=2nC/m liesinthe z=0 plane parallel to the x axisat_ y=3m. Find the potential difference V,» for the points A(2 m, 0, 4m) and B(O, 0, 0) Ans. —18.4V A uniform sheet of charge, p,=(1/6x)nC/m’, is at x=0 and a second sheet, p,= (-1/6n) nC/m?, is at x=10m. Find Vag, Vac, and Vac for A(10m,0,0), B(4m,0,0), and C(O,0,0). Ans. —36V, -24V, -60V Given the cylindrical coordinate electric fields E=(5/r)a, (V/m) for 0<r<2m and E= 2.5a,V/m for r>2m, find the potential difference Vag for A(1 m, 0, 0) and B(4m, 0, 0). Ans, 8.47V A parallel-plate capacitor 0.5m by 1.0m, has a separation distance of 2cm and a voltage difference of 10 V. Find the stored energy, assuming that € = € . Ans. 11.1nJ CHAP. 5} THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL 75 $.38. 5.39, 5.40. 5.42, The capacitor described in Problem 5.37 has an applied voltage of 200 V. (a) Find the stored energy. (6) Hold d, (Fig. 5-16) at 2cm and the voltage difference at 200V, while increasing d; to 2.2cm. Find the final stored energy. [Hint: AW, = (AC)V"] Ans. (a) 4.4 p); (b) 4.2 pl Find the energy stored in a system of three equal point charges, Q=2nC, arranged in a line with 0.5 m separation between them. Ans. 180n) Repeat Problem 5.39 if the charge in the center is —2 nC. Ans. -108nI Four equal point charges, @=2nC, are to be placed at the corners of a square } m on a side, one at atime. Find the energy in the system after each charge is positioned. Ans. 0, 108 nJ, 292 nJ, 585 n3 Given the electric field E=—Se~'“a, in cylindrical coordinates, find the energy stored in the volume described by r2a and O=z=Se Ans. 7.89% 10°"a* Given a potential V =3x?+4y? (V), find the energy stored in the volume described by Osx= Im, O<y<1m, and O0<z<1m. Ans. 147 pl 78 CURRENT, CURRENT DENSITY, AND CONDUCTORS (CHAP. 6 EXAMPLE 1. What electric field intensity and current density correspond to a drift velocity of 6.0 x 10-* m/s in a silver conductor? For silver, g=61.7MS/m and p=5.6x10 ?m?7/V-s. U_ 60x10 * Bata" -1.97x10' BT 5 6IG ET HOT 10S V/m J = 0E =6.61 X 10° A/m* 6.5 CONDUCTIVITY o In a liquid or gas there are generally present both positive and negative ions, some singly charged and others doubly charged, and possibly of different masses. A conductivity expression would include all such factors. However, if it is assumed that all the negative ions are alike and so too the positive ions, then the conductivity contains two terms as shown in Fig. 6-4(a). In a metallic conductor, only the valence electrons are free to move. In Fig. 6-4(b} they are shown in motion to the left. The conductivity then contains only one term, the product of the charge density of the electrons free to move, p, , and their mobility, 4. ~O O- —& @- +0 ss i ~O J -O O- : @- beSzo =O~O} } © ~O =O O a ~-O =O =O o=p wtp, GT pelle = Delle + Py ity, (@) Liquid or gas {b) Conductor {c} Semiconductor Fig. 6-4 A somewhat more complex conduction occurs in semiconductors such as germanium and silicon. [n the crystal structure each atom has four covalent bonds with adjacent atoms. However, at room temperature, and upon influx of energy from some external source such as light, electrons can move out of the position called for by the covalent bonding. This creates an electron-hole pair available for conduction. Such materials are called intrinsic semiconductors. Electron-hole pairs have a short lifetime, disappearing by recombination. However, others are constantly being formed and at all times some are available for conduction. As shown in Fig. 64{c), the conductivity o consists of two terms, one for the electrons and another for the holes. In practice, impurities, in the form of valence-three or valence-five elements, are added to create p-type and n-type semiconductor materials. The intrinsic behavior just described continues, but is far overshadowed by the presence of extra electrons in n-type, or holes in p-type, materials. Then, in the conductivity g, one of the densities, p, or p, , will far exceed the other. EXAMPLE 2. Determine the conductivity of intrinsic germanium at room temperature. At 300K there are 2.5 10" electron-hole pairs per cubic meter. The electron mobility is j= 0.38m7/V-s and the hole mobility isu, =0.18m’/V-s. Since the material is not doped, the numbers of electrons and holes are equal. 2 = Nee(ue + pn) = (2.5% 10)(1.6 x 10-")(0.38 + 0.18) = 2.24 S/m CHAP. 6) CURRENT, CURRENT DENSITY, AND CONDUCTORS no 6.6 CURRENT IF Where current density J crosses a surface S, as in Fig. 6-5, the current / is obtained by integrating the dot product of J and d8. di=5-d8 i= [a-as Is Of course, J need not be uniform over 5 and 5 need not be a plane surface. Fig. 6-5 EXAMPLE 3. Find the current in the circular wire shown in Fig. 6-6 if the current density is J= 15(l—e""™Ja, (A/m’). The radius of the wire is 2mm. A cross section of the wire is chosen for S. Then di=3-ds = 15(1—e Ja, +rdrdga, 2 70.002 and if i 15(1 — er dr dg 5 = 133 x 10 *A=0.133 mA Any surface S$ which has a perimeter that meets the outer surface of the conductor all the way around will have the same total current, £=0.133mA, crossing it- 80 CURRENT, CURRENT DENSITY, AND CONDUCTORS [CHAP. 6 6.7 RESISTANCE R If a conductor of uniform cross-sectional area A and length ¢, as shown in Fig. 6-7, has a voltage difference V between its ends, then Vv Ex, and Jo # assuming that the current is uniformly distributed over the area A. The total current is then oAV [=JA=—— € Since Ohm’s law states that V=JR, the resistance is é R cA (2) (Note that 1S-'=1@; the siemens was formerly known as the mho.) This expression for resistance is generally applied to ail conductors where the cross section remains constant over the length & However, if the current density is greater along the surface area of the conductor than in the center, then the expression is not valid. For such nonuniform current distributions the resistance is given by =v faces [ oe-as If E is known rather than the voltage difference between the two faces, the resistance is given by R The numerator gives the voltage drop across the sample, while the denominator gives the total current 1. xt EXAMPLE 4. Find the resistance between the inner and outer curved surfaces of the block shown in Fig. 6-8, where the material is silver for which 0 = 6.17 x 107 S/m. If the same current J crosses both the inner and outer curved surfaces, yoke, and Enka r or CHAP. 6] CURRENT, CURRENT DENSITY, AND CONDUCTORS 83 material. Suppose, however, that through a temporary imbalance a region within a solid conductor has a net charge density py at time *=0. Then, since J=cE=(o/e)D, go v--—D=-— € a Now, the divergence operation consists of partial derivatives with respect to the spatial coordinates. If o and € are constants, as they would be in a homogeneous sample, then they may be removed from the partial derivatives. 2-pD)=- 2 og _ oa Po 8 or *e + £ p=0 The solution to this equation is P= poe Thus p decays exponentially, with a fime constant 1t=e€/a, also known as the relaxation time. At t=, p has decayed to 36.8% of its initial value. For a conductor r is extremely small, on the order of 107". This confirms that free charge cannot remain within a conductor and instead is distributed evenly over the conductor surface. EXAMPLE 6. Determine the relaxation time for silver, given that o = 6.17 x 10’S/m. If charge of density Po is placed within a silver block, find p after one, and also after five, time constants. Since € ~ €, e 10°"36n soe w tr Oo 617 X1O 1.43x10 Ys Therefore 1 P= poe '=0.368p0 St: p=poe *=6.74x 10 “p, at tT at 6.10 CONDUCTOR-DIELECTRIC BOUNDARY CONDITIONS Under static conditions all net charge will be on the outer surfaces of a conductor and both E and D are therefore zero within the conductor. Because the electric field is a conservative field, the line integral of E- dl is zero for any closed path. A rectangular path with corners 1, 2, 3, 4 is shown in Fig. 6-11. 2 3 4 1 fe-asf E- ats [ E- asf E- 1 2 y 4 If the path lengths 2 to 3 and 4 to 7 are now permitted to approach zero, keeping the interface between them, then the second and fourth integrals are zero. The path from 3 to 4 is within the i 2 Dielectric ETM a Conductor Fig. 6-11 84 CURRENT, CURRENT DENSITY, AND CONDUCTORS (CHAP. 6 conductor where E must be zero. This leaves 2 2 [e-a=[ eae=0 1 ! where E, is the tangential component of E at the surface of the dielectric. Since the interval 7 to 2 can be chosen arbitrarily, at each point of the surface. To discover the conditions on the normal components, a small, closed, right circular cylinder is placed across the interface as shown in Fig. 6-12. Gauss’ law applied to this surface gives fD- dS = Qn or [ v-as+ p-as+ { p-ds=| »,4s lop Lice ls tom The third integral is zero since, as just determined, D,=0Q on either side of the interface. The second integra} is also zero, since the bottom of the cylinder is within the conductor, where D and E are zero. Then, [ p-as=[ D, as =$ pa op lop le which can hold only if Conductor Fig. 6-12 EXAMPLE 7. The electric field intensity at a point on the surface of a conductor is given by E= 0.2a, —0.3a,—0.2a, (V/m). Find the surface charge density at the point. Supposing the conductor to be surrounded by free space, Dy. = €oEn = Ps E,= +|E} = £0.412 V/m = 10” = 2 p= (Fez )e40-412) = £3.64 pC/m The ambiguity in sign arises from that in the direction of the outer normal to the surface at the given point. In short, under static conditions the field just outside a conductor is zero (both tangential and normal components} unless there exists a surface charge distribution. A surface charge does not imply a net charge in the conductor, however. To illustrate this, consider a positive charge at the origin of spherical coordinates. Now if this point charge is enclosed by an uncharged conducting CHAP. 6] CURRENT, CURRENT DENSITY, AND CONDUCTORS 85 spherical shell of finite thickness, as shown in Fig. 6-13(a), then the field is still given by +2 E= ane’ except within the conductor itself, where E must be zero. The coulomb forces caused by +O attract the conduction electrons to the inner surface, where they create a p,, of negative sign. Then the deficiency of electrons on the outer surface constitutes a positive surface charge density P:2- The electric flux lines ¥, leaving the point charge +Q, terminate at the electrons on the inner surface of the conductor, as shown in Fig. 6-13(b). Then electric flux lines originate once again on the positive charges on the outer surface of the conductor. It should be noted that the flux does not pass through the conductor and the ez charge on the conductor remains zero. @ @) Fig. 6-13 Solved Problems 6.1. An AWG #12 copper conductor has an 80.8mil diameter. A 50-foot length carries a 2 current of 20A. Find the electric field intensity E, drift velocity U, the voltage drop, and the resistance for the 50 foot fength. Mathosd ; Since a mil is 7m inch, the cross-sectional area is 0.0808 in\ (2 <2 ny? A= of CSee\ ee) =3.31x 10° m? 2 Tin 12 Then J=-=-— = 6. 2 ATS w 1 e004 10 Ale Fos copper, o=5.8x10’S/m. Then _ J 6.04 x 106 ' £2 =n h04 10 Win V = E€=(1.04 x 1071)(50)(12)(0.0254) = 1.59 V v_ 159 Re-=—=7. “2 jap nT 795x107 The electron mobility in copper is =0.0032m*/V-s, and since o=pp, the charge density 5.8 x 10" 0.0032 pats = LBL x 10 C/m* 6.10. 6.12. 6.13. 6.14, CURRENT, CURRENT DENSITY, AND CONDUCTORS [CHAP. 6 Then from R= ¢/oA, In a cylindrical conductor of radius 2mm, the current density varies with the distance from the axis according to J=10'e” (Afm’) Find the total current 1. 2m f.002 t= [s-as=[sas= | f 1'e'r drag hy ew {-400)° 0.002 = ane) (4007 - vl =751mA Find the current crossing the portion of the y= plane defined by -O.1sx= 01m and —0.002=z=0.02m if J= 107 jxja, (A/m’) x2 701 I faeas=f" f 10? xl a, «dx dza, =4mA nee Fo Find the current crossing the portion of the x= plane defined by ~2x/4sy= af4m and ~-6.01=z=0.01m if 3=100cos2ya, (A/m?) OL pats 1=fa-as=f J 100 cos 2ya, + dy dza,=2.0A ot dm Given 3=10°sin @a, A/m’ in spherical coordinates, find the current crossing the spheri- cal shell r= 0.02 m. Since J and dS =P sin 640 doa, are radial, an px in| f 10°(0.02)? sin? 640 dg =3.95 A Show that the resistance of any conductor of constant cross-sectional area A and length ¢ is given by R= ¢/oA, assuming uniform current distribution. A constant cross section along the length @ results in constant E, and the voltage drop is v= | E-dl=E¢€ If the current is uniformly distributed over the area A, I= [3-d8=JA=0EA where o is the conductivity. Then, since R=V/I, R= aA CHAP. 6] CURRENT, CURRENT DENSITY, AND CONDUCTORS 89 6.15. Determine the resistance of the insulation in a length ¢ of coaxial cable, as shown in Fig. 6-14. 6.16. 6.17, fer => Fig. 6-14 Assume a total current J from the inner conductor to the outer conductor. Then, at a radial distance 7, A 2nré I cd = and so E=Te The voltage difference between the conductors is then t 1b Va = -[ nore” “Frat a and the resistance is 1b T inot "a A current sheet of width 4m lies in the z=0 plane and contains a total current of 10 A in a direction from the origin to (1,3,6)m. Find an expression for K. At each point of the sheet, the direction of K is the unit vector and the magnitude of K is? A/m. Thus As shown in Fig. 6-15, a current /, follows a filament down the z axis and enters a thin conducting sheet at z=0. Express K for this sheet. fe hi, Fig. 6-15 90 CURRENT, CURRENT DENSITY, AND CONDUCTORS (CHAP. 6 Consider a circle inthe z=0 plane. The current J; on the sheet spreads out uniformly over the circumference 277. The direction of K is a,. Then Fig. 6-16 However, integration is not necessary, since for uniformly distributed current a 30° segment will contain 30°/360° or 1/12 of the total. 6.19. A current /(A) enters a thin right circular cylinder at the top, as shown in Fig. 617. Express K if the radius of the cylinder is 2. cm. ; cate Fig. 6.17 On the top, the current is uniformly distributed over any circumference 2zr, so that k=-0g (Am inr™ Down the side, the current is uniformly distributed over the circumference 27(0.02 m), so that i K=O ggg (a) (Abn) 6.20. A cylindrical conductor of radius 0.05 m with its axis along the z axis has a surface charge density , = Pyo/z (C/m?). Write an expression for E at the surface. Since D,=p,, E,=p,f€o. At (0.05, , 2), £3, (Vim) £0: E=E,a,= 6.21. A conductor occupying the region x25 has a surface charge density Vy ez? Write expressions for E and D just outside the conductor.