Electromagnetic Fields & Wave Jeong Phill Kim [email protected] 010‐8918‐5344 School of EE Chung‐Ang University, Seoul 156‐756, Korea Required Knowledges Calculus ‐ Coordinate system ‐ Vector calculus ‐ Derivative & Integral ‐ Differential equations Physics ‐ Mechanics ‐ Basic Electromagnetics Programming ‐ Matlab Chung‐Ang University Plan of Study 1st semester ‐ Coordinate System ‐ Vector Calculus ‐ Electrostatics ‐ Magnetostatics 2nd semester ‐ Time‐varying Electromagnetics ‐ Maxwell’s Equations ‐ Plane Wave Propagation ‐ Transmission Line ‐ Impedance Matching Senior courses ‐ RF / Microwave Engineering ‐ Radar Engineering ‐ Antenna Engineering Graduate courses ‐ Advanced Electromagnetics ‐ RF/Microwave Circuit Design ‐ Antenna Theory & Design ‐ Wave propagation & Scattering ‐ Radar System Design ‐ Microwave Imaging ‐… References • D. K. Cheng, Field & Wave Electromagnetics, • U.S.Inan & A. S. Inan, Engineering Elctromagnetics • Markus Zahn, Electromagnetic Field Theory: a problem solving approach • Lecture notes : will be uploaded in e‐class ² How can we develope these things? of each one? Ch. 1 Intro Electronic enginnering around us: ² Now a days, so many electric or electronic devices and apparatus are around us as follows: { { { { { { { { { { { { { { Radio and TV Microphone and Speaker Smart phone Bluetus and WiFi connected Devices Computer connected to LAN T-money (simple contact or contacless trraffic card) Electric motor Dispalys (ex, Cathode-ray tube (CRT) Touch panel Maglev (Magnetic Levitation) Train Light (one of EM wave) Smart highway and car Induction heating (IH) and microwave oven Radar and Laser What is the theory of operations { Students studying electrical engieering generally meet voltages and currents in electric circuits at first. They study the characteristics of circuit elements (such as resistor, capacitor, and inductors) and the Kirchhoff's voltage and current laws, and use them to analize the circuit characteristics. { Some questions ??? 1) What are the voltage and current physically? In addition, how can we explain the operation of capacitor and inductor clearly? 2) The source signal may be direct current (DC) or alternating current (AC) one. ¡ fDC = 0, fAC 6= 0 ¡ Low frequency ! Circuit theory can be used for analysis ¡ Increasing frquency ! Electric circuit ??? 3) Let us consider 1 cm Resistor and Wavelength of 30 GHz ¡ In this case, is the current in resistor unform ??? 4) Now the wireless connection becomes common way data communication. Can you explaine the related phenomena ??? Permanent Magnet Electric signal + – Cone ² The related phenomena is strongly related with the physical phenomena of electromagnetic fields. ! This is the reason why we have to study electromagnetics. Coil Vibration Speaker 4H + – Chung‐Ang University 5V 10 8 1F 8H 15 2A λ Branches of EE department Electromagnetics ² Physics based { Electromagnetics is strongly based on the physics. ¡ Electric circuit and energy { Therefore it is needed for you to have interests in physics. ¡ Semiconductor ¡ Electromagnetics, Microwave, and optics { However, the problem is that most student think that physics is one of most difficult subjects to study. ¡ High speed circuit design { Then How ??? ¡ Wireless communication (HW, but SW ???) ¡ Remote sensing Electromagnetics is the core and prerequisite subject for: { Electric circuit and energy ² Non-physics based { Semiconductor ¡ Software-based communication (Data compression and coding) { High speed circuit design ¡ Artificial Inteligence (AI) and Neural network (NN) { Microwave, optics, and RemoteSensing ¡ Internet of things (IOT) { EMI and EMC ¡ Algorithm and programming ² How about the autonomous navigation ??? EM is a Basic and Major Area of EE Department Communication & Signal processing Electric energy RF, Microwave & Optics Electromagnetics Integrated circuits Semiconductor Control & System Computer Chung‐Ang University History of Electromagnetics ² Electricity { Long ago ancient people were aware of thunder (lightening) and shocks from electric fishes. The record of electric fishes dates back to 2750 BC, where these fish were referred to as the Thunderer of the Nile. { Ancient traders around the Mediterranean had known rods of amber could be rubbed with cloth like to attract feathers. In about 600 BC, the ancient Greek philosopher Thales (his living city was Miletus) first wrote about the attraction phenomena of amber with a series of observations. { Electricity remained relatively unknown until 1600 AD, when the English scientist William Gilbert made a careful study about it using amber, and denominated it as the latin word electricus, meaning like amber. Now electricity becomes the representing terminology instead of electricus. Black Sea Balkan Peninsula Miletus Magnesia ² Magnetism Mediterranean Sea { It is known that Ancient Greeks living in Magnesia (about 20 km from Miletus) probably first observed that iron ore in mine attracted one another and also attracted small iron objects. The word magnet comes from Magnesia, the name of the place where this ore was found. Africa Wireless Transmission Maxwell’s Eq. { From around 1400 AD, the magnetic loadstone has been used as a compass for safe sailing in sea. Coulomb’s Law Ampere’s Circuital Law Marconi Maxwell Coulomb Ampere 1820 1861 1897 1866 1831 1784 Savart Bio‐Savart Biot Law Chung‐Ang University Faraday EM Induction Hertz Dipole Antenna Radiation Ch. 1: Electrostatics 1.1 Gravitational Force and Field ² Phenomena related to the electric charge, specially for ststic chrage ² Gravitational Force (Newton / Apple tree) { Earth environment ² Statics and Dynamics ( =0 @(Physical quantity) @t 6= 0 m me = mg R2 k g me Gravitational ! g = 2 = 9:8 m/s2 acceleration R 8 ¡11 > N m2 /kg2 <kg = 6:67 £ 10 R = Re + h ' Re = 6; 400 km > : me = £1024 kg Fg = kg : Static : Dynamic, Time-varying ² What is charge ??? { Charge is formed by collection of a lot of atoms with their electrons gained or lost. h { An atom consists of proton and some number of electrons ¡ Simple model (the following Figure) ¡ An electron is known to have a charge (negative). e = ¡1:6 £ 10 ¡19 mt { Meaning of g Fg Earth me ms ¡ Gravitational force for m = 1 kg ¡ Gravitational field R≃ae (C) ² Important concept of souce mass and test mass Total charge of electrons = Multiple of e (quantized) For most matters R me ! ms ; m ! mt mt ms Fg = kg 2 = mt g R jTotal charges of protonj = jTotal charge of electronsj { An ion is an atom (or group of atoms) that has lost or gained one or more electrons, giving it a net positive or negative charge, respectively (the following Figure). g!a ¡¡¡! m1 Fg Fg m2 Fg = mt a ² Gravitational energy (Wg ) and potential (Vg ) Neutral charge Neutral charge { Energy Z Wg = ¡ 9+ h: Negative charge 9+ Fg ¢ dl ' Fg l ! mt g h : Joule (J) 4+ 1 electron gained Positive charge 4+ { Potential: Energy for mt = 1 kg Vg = 1 electron lost ² Negative mass ? Chung‐Ang University vertical height from Earth surface Electromagnetics Wg = gh mt (Now only mathematically ? 5 1.2 Expemiment on Electric Force ² Analogy: Torsion Balance apparatus ² Experiment (Charles-Augustin de Coulomb) ³q q ´ 1 2 Fe = ke R2 ke = 8:9877 £ 109 Gravitational ms Force (N) Fg = mt kg 2 R = mt g ms Field g = kg 2 R Energy Wg = mt gh Wg = gh Potential Vg = mt ² Torsion balance { The torsion balance consists of a bar suspended from its middle by a thin fiber of a very weak torsion spring. R Fe { A known charge is loaded at the end of the bar. If a test charge is located with a distance R from the charge on the bar (normal direction to the bar axis). q1 Fe = qt E qs E = ke 2 R We = qt El We Ve = = El qt Example Comparison of related forces in atom ² Parameters proton and electron in hydrogen atom qp = ¡qe = 1:6 £ 10¡19 C R = 0:53 £ 10¡10 m (Bohr radius of H atom) Charles A. de Coulomb ² Gravitational and electric forces me m = 3:6 £ 10¡47 R2 qp qe Fe = ke 2 = 8:2 £ 10¡8 R Fg = kg ² With the concept of source and test charges ³ q ´ s Fe = qt ke 2 R (N) (N) 8:2 £ 10¡8 Fe = = 2:3 £ 1039 Fg 3:6 £ 10¡47 ² Thus we can ignore the gravitational force comapred to the electric force. ² Corresponding one to graviational field g ? Chung‐Ang University qs R2 me = 9:11 £ 10¡31 kg { The sensitivity of the instrument comes from the weak spring constant of the fiber, so a very weak force causes a large rotation of the bar. E = ke Fe = qt ke mp = 1:67 £ 10¡27 kg { The equilibrium sate means that the twisting force (calculated from the measured rotated angle) equals to the applied force. ! Electric q2 { The bar will rotate, twisting the fiber, until it reaches an equilibrium. Fe = qt E Gravitational and Electric qs R2 Electromagnetics 6 ² Constant ke and ²0 1.3 Electric Force and Field in Free Space { You might have already known the following relation between ke and ²0 ² What is the meaning of Electrostatics ? 1) Charges are not moving. 2) Quanties of charges are not varying in time. Source charge at the origin 1.3.1 Discrete Source Charges F z E ² Single point charge at origin ³ q ´ s Fe = qt ke 2 ^ r = qt E r qt = 2 C qt = 1 C E qt = 1 C r { y qs x ³ q ´ s E = ke 2 ^ r r 1 4¼²0 1 ²0 = = 8:854 £ 10¡12 4¼ ke ke = qt = 1 C E ² There is no change in E field itself with the coordinate chosen, but its expression may be different depending on the chosen coordinate systems. ² Concept of source- and field-points: 8 > : Field-point vector <r = (x; y; z) 0 0 0 0 r = (x ; y ; z ) : Source-point vector > : R = r ¡ r0 : Source-to-field point vector qs ^ ´ Fe = qt ke 2 R R ³ q ´ s ^ E = ke 2 R R 8 R = r ¡ r0 > > > < r = jRj > > > :R ^ =R R qs1 R1 qs2 N X E= z r 1’ En n=1 Source charge at arbitrary point Field point E qt=1 C z ² Multiple point charges Force on qt = 1 C ³ { This relation can be clearly understood after studying a electric flux density D. { From now ke and 1=(4¼²0 ) will be used case by case. ² Single point charge at an arbitrary point r0 = (x0 ; y0 ; z0 ) ??? ² Electric field: (unit charge) Where does this expression come ? En = ke μ r 2’ qn R2n ¶ ^ Rn x 0 R2 r r n’ rN’ Rn qsn RN qsN Rn = r ¡ r0n E qt = 1 C r x 0 r’ qs R y ² Practical measurement (Caution !!!) Fe qs !0 qt E , lim Chung‐Ang University Electromagnetics 7 y 1.3.2 Continuous Distribution Example Electric Force and Field ² There are two point charges 10 nC at (1,0,2) and 5 nC at (-2,0,3). { Calculate E at (1,3,2). R1 = r ¡ r01 = (1; 3; 2) ¡ (1; 0; 2) = (0; 3; 0) R1 = 3 p p R2 = 32 + 32 + 12 = 19 R2 3 qs1 2 = 10 nC R1 ‐2 r02 R2 = r ¡ = (1; 3; 2) ¡ (¡2; 0; 3) = (3; 3; ¡1) qs2 = 5 nC z 0 x 1 3 y ² Expressions of charge distributions 8 > <½L ¢ln : line charge (C) qn = ½S ¢Sn : surface charge (C) > : ½V ¢vn : volume charge (C) 8 > <½L : line charge density (C/m) ½ = ½S : surface charge density (C/m2 ) > : ½V : volume charge density (C/m3 ) vn r n’ r y Line charge z E = E1 + E2 = (¡1:63; ¡11:62; 0:54) jEj = 11:74 (V/m) dq En = ke qn R2n ¶ y where qn = ½V ¢vn Surface charge ² This vector calculation is a quite tedious and time-consuming work !!! dE z qn ! dq = ½ dv; Rn ! R μ ¶ ½ dv ^ En ! ke R R2 μ^¶ Z R 0 ! E = ke ½(r ) dv R2 V0 (nN) 0 x ^ n; R { Exact expression with { Calculate F to the charge of ¡10 nC at (1,3,2). r ρL r’ En μ dE R N X i=1 qt = 1 C 0 x { Approximation formula of total E E' Rn qn ² Case study (volume charge distribution) Therefore the related E in V/m are μ ¶ R1 E1 = ke qs1 = (0; ¡9:99; 0) R31 μ ¶ R2 E2 = ke qs2 = (¡1:63; ¡1:63; 0:54) R32 F = qt E = (16:3; 116:2; ¡5:4) jFj = 117:4 (nN) Field point En z r R s y r’ dq x Volume charge { Any simpler methods to find the E ??? dE z ¡ Gauss' Law ¡ Electric potential approaches r R ρv r’ y dq x Chung‐Ang University Electromagnetics 8 ² Therefore dE½ and the resulting E½ become Example Finite Length Line Charge ^ ½ from ² Differential electric field at r0 = ½ small charge element ½L dz0 at z = z0 : Finite length μ ¶ z R 0 dE = ke ½L dz R3 L 8 0 0 zz r =^ > > > <r = ½ ^½ dq = L dz’ RL 0 0 >R = r ¡ r = ½ ^½¡^ zz > > p : r′ R = jRj = ½2 + z02 L ² dEÁ = 0 by axial symmetry. Then dE½ and dEz are ( ^=½ R¢½ R¢^ z = ¡z0 μ ¶ ^ R¢½ ^ = ke ½L dz0 dE½ = dE ¢ ½ R3 0 dE½ = μ = ke ½L 0 ¡z dz0 R3 R¢^ z R3 ke ½L ½ E½ = Z ¯L cos ¯ d¯ 0 ke ½L sin ¯L ½ L L =p RL ½2 + L2 ² Ez can be found similarly as ke ½L Ez = ½ dE dEz ! dE ! Z ¯L sin ¯ d¯ 0 ke ½L (cos ¯L ¡ 1) ½ ½ ½ cos ¯L = =p RL ½2 + L2 Ez = ² Sum of results from upper and lower wings ke ½L (sin ¯1 + sin ¯2 ) ½ ke ½L Ez = Ez1 + Ez2 = (cos ¯1 ¡ cos ¯2 ) ½ 8 8 L1 >sin ¯ = p L2 > > > 2 < <sin ¯1 = p 2 2 ½ + L1 ½2 + L22 ½ ½ > > > > :cos ¯2 = p 2 :cos ¯1 = p 2 2 ½ + L1 ½ + L22 E½ = E½;1 + E½;2 = μ E½ = sin ¯L = ³½ ´ 0 = ke ½L dz R3 dEz = dE ¢ ^ z = ke ½L ! ! R ΒL β r ke ½L cos ¯ d¯ ½ ¶ dz0 ¶ L1 L E = ?? ? 1 2 L2 ² With a change of variables z0 = ½ tan ¯ (Fig. XXX) ( dz0 = ½ sec2 ¯ d¯ R = ½ sec ¯ cos ¯ d¯ ½ dz0 ½2 sec2 ¯ d¯ = = 3 R ½3 sec3 ¯ ½ 0 0 2 2 sin ¯ d¯ ½ tan ¯ sec ¯ d¯ z dz = = 3 3 3 R ½ sec ¯ ½ Chung‐Ang University Electromagnetics 9 Case a ² For a ¯eld point on the z-axis (jzj > L=2), ¯1 = ¯2 = ¼=2, Ri = jz § L=2j E½ = 0; d E P { Field point for a maximum Ez can be found by @ ³z´ @Ez =0 =0 ! @z @z R3 Ez =??? ² For a ¯eld point with ¯2 = ¯1 L §a ! a2 ¡ 2z2 = 0 ! zm = p 2 L { This can be solved with N-polygon approximation (Appendix). Ez = 0 ke ½L sin ¯1 ½ L=2 ke ½L p =2 2 ½ ½ + (L=2)2 E½ = 2 ² Surface Charge Distribution Circular surface Case b L/2 { Ez along the z-axis can be found as L P Example Uniform Line Charge on In¯nite Line E L/2 ² With L ! 1 E½ = 2 ke ½L ½ Case c The integrand is not a function of Á. ∞ Z L { Along the z-axis, E = ^ zEz with I ^ R¢^ z dl Ez = ke ½L 2 R dl = adÁ z z ^ ¢^ R z = cos ° = = p 2 R a + z2 az Cc z = k e ½L 3 3 R R Cc = 2¼a (circle's circumference) Chung‐Ang University E ∞ ½z d½ 3 0 R ½ = z tan °; d½ = z sec2 ° d° ¸ Z °0 · 3 z tan ° sec2 ° d° Ez = ke ½S (2¼) z3 sec3 ° 0 Z °0 = ke ½S (2¼) sin ° d° (0,0,z) s Ez 0 b 0 y dq x 0 z dE x = ke ½S (2¼) (1 ¡ cos °0 ) z cos °0 = p b2 + z2 (0,0,z) { Ez is simpli¯ed (Due to the axial symmetry) Ez = ke ½L (2¼) Circular disk charge b Ez = ke ½S (2¼) Example Uniform line charge on Circular line ² Calculate the electric ¯eld on the z-axis. dq = ½S dS = ½S ½ d½ dÁ ¶ Z b Z 2¼ μ ^ R¢^ z Ez = ke ½S ½ d½ dÁ R2 0 0 z R 0 ′ r′ y L dq Infinite planar charge { For an infinite plane (b ! 1) zm 0 Ez (0,0,z) Ez Ez = ke ½s (2¼) ‐zm { For a ring surface charge with inner and outer radius a and b, respectively, s y 0 x (Appendix) Electromagnetics 10 Example Spherical cloud of charge ² Charges are collected (uniformly distributed) in a form of spherical cloud whose radius is a. Spherical shell ??? Inside and outside z z=z0 v a y { E for z0 < a and z0 > a ??? x Ezi Q1 = 4¼z30 ½v 3 i= 1; 2; Q2 = 4¼a ½v 3 { Two conducting hemispheres are clamped together. { The outer conductor is momentarily short. { Unclamping the outer conducting hemispheres, and then measure the induced charge Qout . The observed experimental result is Flux line z D ! x D»E ² Proportional constant ke ??? { Comparison of the expression of ²0 and ke 1 = 8:854 £ 10¡12 4¼ ke 1 ke = 4¼²0 ²0 = Flux line Gaussian surface { With replacing ke with 1=(4¼²0 ), we can obtain the familiar formula: μ ¶ qt qs ^ F= R 4¼²0 R2 μ ¶ qs ^ E= R 4¼²0 R2 Electromagnetics E Same direction y qs { A proportional constant ²0 is introduced and called permittivity of the air. Gaussian surface Qout = ¡Qin Chung‐Ang University 8 qs <E = ke 2 ^r r :D = qs ^r 4¼ r2 D = ²0 E D Gaussian surface ² Experiment by Michael Faraday { Inner conducting sphere is charged (Qin ) E { For a point charge at the origin in a free space 1.4 Electric Flux and Gauss' Law { For validating his idea on flux concept 1.4.1 Electric Flux and its Density Flux line How and Why ??? ² Introducing a flux density D, the following mathematical formulation called Gauss' Law for flux can is valid: I D ¢ dS = Q (all charges inside S) ² Relation between D and E for z0 < a for z0 > a 3 ; qs S ( Qi = ke 2 ; z0 Flux lines ª = Qin Ez ² Let us consider the case of circular disk carrying uniform charge distribution (Appendix) E=^ zEz { Concept of flux line ª was suggested to be equal to total charge in the inner sphere. 11 1.4.2 Applications of Gauss' Law ² Why is this formula called the Gauss' Law ? Gauss is known to provide a mathematica formula, which can also be applied to the many physical flux quantities such as light, wave, etc.. Therefore our case is one special one. Example Uniform planar charge distribution S2 +2 nC +4 nC +3 nC ‐1 nC S3 ‐3 nC +1 nC S1 ² Charge is assumed to be uniformly distributed on an infinite plane xz-plane. Then D = ^ yDy due to the symmetry. ² If the surace of centered cube the Gaussian surface S, 8 > <Dy = Dy 2S = ½s S ! > :Ey = ² Since the Gauss' law is valid for any closed surface, I D ¢ dS = 2 nC S1 I D ¢ dS = ¡1 nC S I2 D ¢ dS = 0 ½s 2 ½s 2²0 y x d Dy s/2 z 4¼r3 ½v Dr 4¼r = 3 2 { Outside the cloud, 8 <Dr 4¼r2 = Q 3 :Q = 4¼a ½v 3 z ρ ρL L/2 S S S 2 S 1 y a x { Inside the cloud, Ex. Uniform line charge distribution y 0 ² There are two cases of a field point; the inside and the other outside of the cloud. { Constant En on S (some part En = 0 is allowed) Chung‐Ang University is chosen as ² The charge is uniformly distributed, then D=^ rDr due to the symmetry. { Symmetric charge distribution (cube / cylinder / sphere) { D½ , E½ are found to be proportional to 1=½. Area A Example Uniform spherical cloud of electron ² For meaningful use of Gauss' Law, S should be chosen appropriately: { Charge is assumed to be uniformly distributed along z-direction. Then D = ½^ D½ due to the symmetry. If the surface of the centered cylinder is chosen as the Gausian surace S, 8 ½L > <D½ = 2¼½ D½ 2¼½ L = ½L L ! ½L > :E½ = 2¼²0 ½ ρs ² It is very interesting that Dy and Ey become constant along y. S3 S z ! ½v r Dr = 3 D r D0 a ! Dr = Q 4¼r2 y x { Continuity of Dr at r = a, and if it is D0 , ‐L/2 D0 = D ! 0 Electromagnetics Q a = ½v 2 4¼a 3 8 μ 2¶ a > <D0 r2 Dr = ³ ´ > :D r 0 a 12 r 1.5 Curl and Divergence of Electric Field ² Divergence of E field (r ¢ E) Vb surface z ² As we know, in order to charaterize the E field thoroughly, its divergence and curl should be known. qt=1 C b C=C1+C2 a C2 y C1 qs ² Curl of E field (r £ E) { First way: No electric energy is required to move a charge along any closed contour C. I W = F ¢ dl = 0 x Va surface C Because F = q E, I Z E ¢ dl = 0 ! (r £ E) ¢ dS = 0 C ! r£E=0 S where S is the open surface formed by closed contour. { Second way: E for a single point source charge at r0 { r ¢ E is expressed as · Z μ^¶ ¸ R 0 r ¢ E = r ¢ ke dv ½(r ) R2 V0 · μ ^ ¶¸ Z R 0 = ke ½(r ) r ¢ dv R2 V0 E=? r1 r1 qs y x { Therefore Z ½(r0 )±(r ¡ r0 ) dv = 4¼ke ½(r) r ¢ E = 4¼ke V0 r¢E = Using the following prperty and null identity 8 μ ¶ μ ¶ μ^¶ ^ 1 R > > 1 R = ¡r > < R2 = ¡r £ r =0 ! r£ R 2 R R r £ (rf) = 0 > > > : r £ E1 = 0 ! f : scalar function z ^ 2 can be expressed as { Because R=R (Appendix) μ ¶ ^ R 1 = ¡r R2 R μ ¶ μ^¶ 1 R 2 = ¡r r¢ R2 R μ ¶ 1 2 r = ¡4¼ ±(R) R μ^¶ R 0 E1 = ke qs (r ) R2 Because r is an operator to the field point variable, · μ ^ ¶¸ R 0 r £ E1 = ke qs (r ) r £ R2 E=? ½(r) ²0 r ¢ E for a point charge { For a point charge at the origin, how is r ¢ E at some specific points ? For all charges, N X i=1 Chung‐Ang University N X Ei = r£E =r£ (r £ Ei ) = 0 i=1 Electromagnetics 13 z 1.6 Electric Energy and Potential W ² E for a point charge qs (source) at the origin 1 ³ qs ´ Er = E = ^r Er ; 4¼²0 r2 If a test charge qt is at infinity, there is no energy. r→∞ qt y r x N X V= 1C V qs ² Multiple point charges (not vector sum but scalar sum) Vn ; where Vn = n=1 qn 4¼²0 Rn ² Continuous charge distribution 1 V= 4¼²0 Z v 0 ½(r0 ) dv R We (r = 1) = 0 ² Relation between V and E Required energy to move this charge from 1 to r Z r Z r F ¢ dl = ¡qt E ¢ dl We (r) = ¡ 1 Z Vba = ¡ 1 We (r) = qt ² Electric potential is Electric energy for qt = 1 C. z C4 qt ² Electric potential (V) for a point charge qs , qt x C3 C2 Vb surface C5 qs @V @V @V ¢x + ¢y + ¢z ¢V = @x @y @z ¶ μ @V @V @V ^ ^ ¢y + ^ ^ +y +^ z ¢ (^ x ¢x + y z ¢z) = x @x @y @z qt C6 { From these two expressions y E = ¡r V x qt Va surface b E2 E E1 E 2 Δl E1 Vba=‐E1⋅l r£E=0 ! E » rf ! E = ¡rV ² Since only scalar quantity is associated, it is easier to handle the problem in calculation of E via V . Electromagnetics E E a ² Mathematical validation ( r £ (rf) = 0; (Vector identity) independent of integration path(r £ E = 0) Chung‐Ang University y a 1C C1 a : b { Expression 2 (Gradient) 1.6.1 Electric Potential ² Potential difference Z b E ¢ dl Vba = Vb ¡ Va = ¡ qs E ¢ dl ¢V = ¡E ¢ ¢l ^ ¢y + ^ ¢l = ¢x + y z ¢z) qs 4¼² r qs W(r) = qt 4¼² r b a With dl = ^ r dr V(r) = z { Expression 1 14 { At z = 0, R1 = R2 = R10 = z Example Calculation of E via V L/2 ² Electric potential V for a finite line charge Z L=2 1 ½L 0 dz V= 4¼²0 ¡L=2 R p R = ½2 + (z ¡ z0 )2 Z L=2 1 ½L p V= dz0 4¼²0 ¡L=2 ½2 + (z ¡ z0 )2 dq β2 β ½ E½ = A R10 R A= ρ ρL z § L=2 ½ where where + for i = 1, and ¡ for i = 2 ¶ μ R1 + z + L=2 ½ V = L ln 4¼²0 R2 + z ¡ L=2 { Resulting electric field 8 ¡@ V ½L 1 > > E = = (sin ¯1 ¡ sin ¯2 )) ½ > < @½ 4¼²0 ½ E = ¡rV ! μ ¶ > ¡@ V 1 ½L 1 > > = ¡ :Ez = @z 4¼²0 R2 R1 1 1 ¡ (¡L=2 + R10 ) (L=2 + R10 ) L ½ R10 ¸ ¶ (Same expression) Example V and E for a circular line charge z dV (0,0,z) ² Find V and E for a circular ring charge on the z-axis. { Differential potential dV μ ¶ 1 ½L dl dV = 4¼²0 R p R = b2 + z2 ; { The fact V = 0 when z ! 1 or ½ ! 1 implies that the constant C = 0. In addition tan ¯i = ½2 + (L=2)2 { Along the z-axis (jzj > L=2), we can obtain the same results before. β1 (z ¡ z0 ) = ½ tan ¯ ( dz0 = ¡½ sec2 ¯ d¯ p ! R = ½2 + (z ¡ z0 )2 = ½ sec ¯ Z ¯2 ¡½L V= sec ¯ d¯ 4¼²0 ¯1 ¤¯2 ¡½L £ = ln (sec ¯ + tan ¯) ¯ + C 1 4¼²0 Ri ; ½ · ½L 4¼²0 μ ½ E½ = L 4¼²0 R1 { Introducing a angle ¯ from the ½-axis as sec ¯i = Ez = 0 V = ??? r r′ 0 ‐L/2 R2 p b dl = bdÁ x R 0 y ′ L dq { Electric potential (2 times of the contribution from the half circle) ¶ μ ½L b 1 V= 2²0 R { Similarly we can find V off the axis, and we can know that the derivatives of V along x and y are zero. Therefore the resulting electric field on the z-axis has only the z component as μ ¶ ½L b ¡@ V z = Ez = @z 2²0 R3 ! Same expression. ! Same expressions (See Appendix) Chung‐Ang University Electromagnetics 15 1.6.2 Electrostatic Energy ² Considering all charges ² Situation 1 (only one charge Q1 ) N 1X Qk Vk We = 2 k=1 E-field is formed by source charge Q1 Q2 at r = 1 ! W=0 Vk = 1 4¼²0 ² Situation 2 (moving Q2 to a point R21 ) Required energy Q1 W2 = ¢W = Q2 V21 = Q2 4¼²0 R21 ¡ Subscripts ( 1-st subsript : field point 2-nd subscrip : source point Here the issue may be the contribution due to ½v at the field point where V is evaluated. Because V is known to be finite over the region of interest, and the infinitesimal integration volume ±v ! 0, the contribution of ½v is zero anyway. Therefore W2 = Q1 ² Situation 3 (moving Q3 to R31 ) W3 = W2 + ¢W ¢W = (Q3 V31 + Q3 V32 ) (Q3 V31 + Q1 V13 ) + (Q3 V32 + Q2 V23 ) = 2 (Q2 V21 + Q1 V12 ) W3 = 2 (Q3 V31 + Q1 V13 ) + (Q3 V32 + Q2 V23 ) + 2 1 = [Q1 (V12 + V13 ) + Q2 (V21 + V23 ) + Q3 (V31 + V32 )] 2 1 = (Q1 V1 + Q2 V2 + Q3 V3 ) 2 Chung‐Ang University j=1(j6=k) Qj Rjk ² Generalized to a continuous charge distribution Z 1 ½v (r0 ) V(r) = dv 4¼²0 r0 6=r R Z 1 ½v (r0 ) V(r) dv We = 2 V ¢W : Q2 = Q1 V12 4¼²0 R12 Q2 V21 + Q1 V12 ! W2 = 2 1 ! W2 = (Q1 V1 + Q2 V2 ) 2 Vi : potential due to charges except Qi N X ROI: Region of Interest Q1 ROI r→∞ ² Therefore we can express We with ½v = r ¢ D as Z Z 1 1 ½v V(r) dv = (r ¢ D) V(r) dv We = 2 V 2 V Using vector identity Q1 R12 Q1 R12 R13 r→∞ r ¢ (V D) = V (r ¢ D) + D ¢ (rV) Q2 Q2 R23 Q3 r→∞ Z Z 1 1 We = [r ¢ (V D)] dv ¡ D ¢ (rV) dv 2 V 2 V I Z 1 1 (V D) ¢ dS + (D ¢ E) dv = 2 S 2 V | {z } ! 0 as r ! 1 ¶ Z μ Z D¢E = we dv dv = 2 V V0 we = Electromagnetics D¢E 2 : electric energy density 16 ² Different definition of energy desity 8 ½V > : defined only the source region > <we1 = 2 > > :we = D ¢ E 2 : { Method 2 (with we1 = ½v V=2) The potential V0 at r = a is calculated as Z Z a ¡½v 3 a ¡2 ½v a2 Er dr = a r dr = V0 = ¡ 3²0 3²0 ¡1 ¡1 defined over all problem space and the potential for r < a is Z r ½v ½v (a2 ¡ r2 ) V = V0 ¡ » d» = V0 + 3²0 a 6²0 ¢ ½v ¡ 2 3a ¡ r2 = 6²0 Example Energy required to forming a cloud charge ² Calculate We required to assemble a uniform cloud of charge of radius a and volume charge density ½v . { Method 1 z The total charges inside S and the resulting electric flux density are ¶ 8μ 4¼r3 > > ½v ; r < a < 3 ¶ Qi = μ > 4¼a3 > : ½v ; r > a 3 a Therefore the stored We is calculated as ρv ¢ ½v V ½2 ¡ = v 3a2 ¡ r2 2 12²0 Z 2¼ Z ¼ Z a We = we1 r2 sin μdrdμdÁ we1 = y x 0 0 μ ¶ 4¼½2v a5 4¼½2v a5 5 = a ¡ = 12²0 5 15²0 Therefore, Dr is 8³ ½ ´ v > r; < Qi 3 Dr = = ³ ´ 3 4¼r2 > : ½v a ; 3 r2 0 Same results !!! r<a r>a Now we and We are 8μ 2 ¶ ½v > > r2 ; r < a < 2 18²0 Dr we = = μ 2 ¶ 6 2²0 > a ½v > : ; r>a 18²0 r4 Z 2¼ Z ¼ Z 1 we r2 sin μdrdμdÁ We = 0 = = 4¼ ½2v 0 Z a 4 a5 + a5 5 r dr a 0 μ ¶ 1 ¡2 6 r dr + a 18²0 2¼ ½2v 9²0 0 μZ ¶ = Chung‐Ang University 4¼½2v a5 15²0 Electromagnetics 17 ² Distribution process of charges inserted into the conductor 1) Charge moves to the conductor surface due to repulsive force 2) Finally, ½ = 0 and E = 0 inside the conductor (equilibrium state) I D ¢ dS = 0 Electron energy ² Material in which electrons can move freely, but totally neutralized. 0 just outside a 4) If not, Et makes the charges to move until Et = 0. ² Various electrostatic problems realted to conductors can be solved with 2nd-order differential equations and the boundary-value problems (This will be covered later). E ² What to be noteworthy: { If we can find V for a charge distribution, the E can be found by simple mathematical operation. { Generally it is easier to find V than the direct calculation of E because of scalar operation. Chung‐Ang University Point charge at the center of conducting shell E Conductor E E Inserting positive charges + + +++ + + 4V ‐ ‐ ‐ ‐ E + 4V 2V ‐ ‐ ‐ 0V ² A point charge Q is at the center of neutralized conducting shell. Calculate E and V with inner and outer radius a and b, respectively. { Because the given conductor is neutral, and the electrons in conductor can move freely. { Due to the effect of the point charge +Q at center, the electrons and positive charges move to the inner (by attaction) and outer (by repucsion) surfaces, respectively, as shown in Fig. XXX . { Since the electric field is zero inside conductor in steady state (equilibrium status), the net charge at inner and outer surfaces of the conductor should be ¡Q and +Q respectively. ² Electric field can be found by applying the Gauss' law: Conducting shell b Q a + + + + –– – ––+ +– Q –+ –+ +– b a Er r V V0 r { By the Gauss' law, the electric field becomes E = ^r with 8 < Q ; for r < a and r > b Er = 4¼²0 r2 :0; for a < r < b Electromagnetics E ⊝ ⊕ ⊕ ⊝ ⊕ ⊝ ⊕ ⊝ ⊝ ⊕ Region where a conductor is to be placed E – E After Eext Constant V lines Conductor 5) Finaly, only En can exist on a conductor surface. Before Bandgap Semiconductor Insulator any closed S inside the conductor = E ? Equipotential lines | {z } Band overlap S 3) In addition, Et conductor. { E is always perpendicular to the constant V lines: Conduction band Valance band 1.7.1 Conductor + + + + – –– + – + –– – + 1.7 Conductor and Dielectrics 18 { Now the potential V can be found for r > b at first as Z r Q V=¡ Er dr = ; for r > b 4 ¼² 0r ¡1 The potential V0 at r = a can be found and the resulting We becomes Z a Q V0 = ¡ Er dr = 4 ¼² 0a 1 ¸ · Z a+ 4¼ Q2 Q2 2 ±( r ¡ a ) r dr = We = 2 (4¼)2 ²0 a3 a¡ 8¼²0 a : Same results !!! Inside the conducting shell, the electric field is zero, and therfore the electric potential becomes Q ; 4¼²0 b for a < r < b 6.2 Dielectrics and Electric Dipole Finally the potential for r < a is μ ¶ Z r Q 1 1 V=¡ Er dr + V0 = + V0 ¡ 4¼²0 r a a ² A dielectric material consists of specific atoms, which can be modeled by a lot of electric dipole (pair of + and ¡ charge), which are bounded to some region, and the directions from ¡ to + charge of the pairs are randomly distributed without E. The plots of Er and V are depicted in Figure. Example Energy of a conducting sphere ² Our concern is what happens when the external E is applied. ² A conducting sphere having a radius a is charged with Q. Calculate the stored electric energy. { Method 1: Calculation using Dr , 8 < Q ; r>a Dr = 4¼r2 :0; r<a Q2 Dr Er = ; r>a 2 2 (4¼)2 ²0 r4 ¸Z 1 · Z Q2 2 We = we r sin μdrdμdÁ = 4¼ r¡2 dr 2² 2 (4¼) 0 a v Q2 = 8¼²0 a we = { Method 2: Calculation with ½ and V The stored electric energy is we1 = ½v V=2 with ½v = ½s ±(r ¡ a) Q ½s = 4¼a2 Chung‐Ang University z Q a y x Conducting sphere + ‐ V0 = ² The electron cloud model, (all electron charges are distributed uniformly in a spherical cloud with a radius of its orbit) was suggested and has been used for a simple study the phenomena of an atom under an applied external electric field. E0 = 0 Atom in external electric field { Let us consider an atom whose electron charge is ¡q under an applied external electric field E0 as shown in Fig. By the electron cloud model, calculate the equivalent electron charge density and relative separation d=r0 , where d is the distance between the center of electron cloud and a point positive charge. { The equivalent electron charge density is calculated as ½c = Electromagnetics 3q 4¼r30 – – – q – – – – – E0 ≠ 0 – – – q – – – – – 19 { For an atom with one electron, even a relatively high external field E0 = 106 V/m is applied, d=r0 becomes { When an external electric field E0 is applied, let us assume that the positive point charge is displaced while the electron cloud is fixed. d ' 6:95 £ 10¡6 r0 { Since two kinds of force are exerted on the positive point charge; one by the electron cloud, and the other by the external electric field, and equating these two yields waht we want to calculate. { At first let us consider the force on the positive charge by the electron cloud. The electric field Ec at the point of the positive charge q by the electron cloud can be found by the Gauss' law as μ ¶3 ¡qi d ; q = q Ec = i 4¼²0 d2 r0 { The magnitudes of two electric fields Ec and E0 should be same for equating the two related forces. Therefore qd = E0 4¼²0 r30 { Now d=r0 can be found as μ ¶ d 4¼²0 r20 = E0 r0 q This implies that the separation d is very small comapred to the radius of orbit. ² Many molecules (dipoles) in dielectrics are are aligned if the external electric field is applied. Model of Polarized atom – – – q – – q – – qd d – Chung‐Ang University R1 q – r R2 θ p = qd d 0 + q where d is a vector from negative to positive charges. z { Electric potentail at a far range (d ¿ r) μ ¶ μ ¶ 1 R1 ¡ R2 1 q q ¡ = V= 4¼²0 R1 R2 4¼²0 R1 R2 R1 R2 ' r2 p ¢ ^r ; V= 4¼²0 r2 r θ R2 R1 ¡ R2 ' d ¢ ^r d p 0 x p¢^ r = p cos μ d cosθ z V(r) + ‐ + ‐ + ‐ ^ Eμ E = ¡r V = ^r Er + μ ¡@V p cos μ = @r 2¼²0 r3 ¡@V p sin μ Eμ = = [email protected]μ 4¼²0 r3 Field point r≫d R1 { Electric field at a far range (d ¿ r) Er = Electromagnetics x ‐q ² Electric potential and electric field e = 1:6 £ 10¡19 4¼²0 £ 10¡20 d E0 ' r0 q 6:95 £ 10¡12 ' E0 Ne Field point ² Dipole moment p for a positive charge q and separation d is defined as { With the approximate value r0 ' 10¡10 (m) and Ne the number of electrons q = Ne e; z a pair of charges of ² Electric dipole: same magnitude but opposite sign. r′ + ‐ + ‐ + ‐ + ‐ + ‐ + ‐ + ‐ + ‐ + ‐ R r y x 20 ² Continuous distributed aligned dipoles total dipole E0 pk ¢v !0 k=1 2 (C/m , Same dimension of D) ¢v ( N N¢v : : No. No. of dipoles per unit volume of total dipoles in ¢v { Electric potential for an infinitesimal volume dv μ ¶ μ^¶ ^ P dv (P dv) ¢ R R dV = = ¢ 2 4¼²0 R 4¼²0 R2 · μ ¶¸ 1 1 = P ¢ r0 dv 4¼²0 R · μ ¶ ¸ 1 1 1 0 0 r ¢ P ¡ (r ¢ P) dv = 4¼²0 R R { Polarization surface and volume charge density ( ^0; ½ps = P ¢ n on S0 ½pv = ¡r0 ¢ P; in V0 I Z ½ps ½pv V(r) = dS + dv 4¼² R 4¼² 0 0 0R S V – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + ² Based on this result, a dielectric material under the external electric field can be modeled by an air-filled parallel plates carrying charges Qp with surface S, and the equivalent electric flux density Dp and field Ep are Q ½ps = – + – + { The final expression can be obtained by using the following vector identity: Q – + – + – + Chung‐Ang University Applying external field results in net charge transition Qp – + – + Parallel plate model d – – – – – – Qp ^ =P¢n S The net charge remaining within the volume V (inside S) is the negative of this integral I Z Qv = ¡ P ¢ dS = (¡r ¢ P) dv V Z S Qv = ½pv dv P=Nqd/(Sd) =Nq/S Qp=Nq + + + + + + E0 P Ep – + r ¢ (fA) = rf ¢ A + f(r ¢ A) 1 A = P; f = R μ μ ¶ ¶ 1 1 1 0 0 P ¡ (r0 ¢ P) =r ¢ P¢ r R R R E0=0 P=0 + – N ¢v X P = lim – + S + – + –+ – { Polarization vector P : moment per unit volume Applying the divergence theorem to the first term I Z ^0 P¢n ¡r0 ¢ P V(r) = dS + dv S 4¼²0 R V0 4¼²0 R – + – 1 r3 –+ – – + – –+ – E» – 1 ; r2 – + V» Before applying external field + –+ + –+ – + + – + –+ + –+ – + + – + –+ + –+ – + – { Total electric potential μ ¶ ¸ Z · 1 1 1 0 0 r ¢ V(r) = P ¡ (r ¢ P) dv 4¼²0 V0 R R { Dependency of V and E on r Hence, when r ¢ P 6= 0, the bulk of the polarized dielectric appears to be charged. However, originally since the dielectric body is electrically neutral dielectric body, thc total charge of the body after polarization must remain zero because of the bound charge characteristics. This can be readily verified by noting that Electromagnetics 21 I Total charge = Z ½ps dS + S V I = ½pv dv Z (r ¢ P) dv = 0 P ¢ dS ¡ S V ² In case of free space (no dielectrics), the E and free (not bounded) charge density ½ are related as r¢E= ½ ²0 Now the presence of the dielectric is known to be modeled by polarization charges in free space. Therefore the above should be modified as ½ + ½pv ²0 r ¢ (²0 E + P) = ½ r¢E= As we have studied that the concept of electric flux density comes from ther free charge as ½ = r ¢ D, we can find the following relation: ² Case A (Parallel configuration): We have studied that the E inside the dielectric becomes smaller than the applied E-field. Therefore we may think that E-field in dielectric region is smaller than that in air region. However, this is not the case because the charges on the conductor moves to the left due to the calling from the electric dipoles of dielectrics. Therefore this different charge distribution on the conductor results in P = ²0 Âe E Now D can be written as 8 > ² = ²0 ²r > > > ² < = 1 + Âe ²r = D = ²E ²0 > > > > :Â e D1z = ²r1 D2z ! + + + + + + D0 x Metal electrode – – – Case A + air – – – – – – Case B Dielectric – Dielectric + – – – – – + + + + + – – – – – +Region + + + 2+ – – – – – + + + + + – – – – – + + + + + + + – – – – – – – – – Region 1 (air) – and finally + + z D1z > D2z D = ²0 E + P P describing the properties of dielectrics is known to be a function of E. If we introduce the proportionality condition Â as Metal electrode Ex. Distribution of E and D + + + + – – – + + + – – – – + + + + – + – – + + + – + + + + – + Region 2 – – – – – – – – E1z = E2z This can be verified by the boundary conditions which will be studied later. ² Case B (Series configuration) Physically, the electric flux density becomes same in both regions: (Permittivity) D1z = D2z (Relative permitivity), or (dielectric constant) ! ! (Electric susceptibility) ²r1 E1z = Ez2 E2z E1z = ²r1 For the free space, it is found that ² = ²0 ² ²r = =1 ²0 Chung‐Ang University Electromagnetics + + + Region 1 (air) 22 – – – 7. Boundary Conditions Condition 1 ² Tangential component of electric field Z (r £ E) ¢ dS = 0 r£E= 0 ! S I E ¢ dl = 0 ! Region 1 (1) w Region 2 (2) w!0 ² Normal component of electric field Z r¢D =½ ! (r ¢ D) dv = Q V I D ¢ dS = Q ! Condition 2 h Region 1 (1) D1n dS S s n̂ D2n dS Region 2 (2) S h!0 ¡¡¡! D1n S ¡ D2n S = ½s S ! D1n ¡ D2n = ½s ² Concept of superposition 8 > No source at the boundary: > > < > Source only with ½s : > > : Combining the results D1n ¡ D2n = ½s (D1n )0 = (D2n )0 8 ½ <(D1n )s = s 2 :(D2n ) = ¡½s s 2 x ² A dielectric sheet with a dielectric constant ²r is placed perpendicularly in ^ E0 in free a uniform electric field E = x space. Determine E, D, and P. { In this case, Et = 0. By the boundary condition on Dn for ½ = 0, C ¡¡¡¡! E2t L ¡ E1t L = 0 ! E2t = E1t Ex. Dielectric sheet in unform E Decomposition Superposition No surface charge density Region 1 (1) D1n0 D2n0 Region 2 (2) Surface charge density only s Region 1 (1) D1ns D2ns Region 2 (2) Dx0 = Dxd ! E0 Ed E0 D0 Dd D0 Air Dielectrics r Ex1 = ²r Exd The subscripts 0 and d stand for air and dielectric regions, respectively. Since Ex0 = E0 , 8 <Ex0 = E0 ! Dx = ²0 E0 :Exd = E0 ²r 8 > in air <0; ¶ μ ! Px = Dx ¡ ²0 Ex = 1 > ²0 Ex ; in dielec. : 1¡ ²r Ex. E, D = ??? for a point charge Q at the center of dielectric shell ² The dielectric shell with inner and outer radius a and b, respectively, has a dielectric constant ²r . { D has only a radial components. of normal D, By the B.C. r Q b a Dielectric shell Dr 0Er Dr0 = Drd { Now Dr and Er can be found as Q 4¼r2 8 Dr Q > > ; < = ²0 4¼ ²0 Er (r) = Dr Q > > : = ; ²r 4¼²0 ²r 0 r Dr (r) = Chung‐Ang University Air Electromagnetics in air in dielelctrics 23 Ex. Boundary conditions at dielectric-air interface ² Case B (Series configuration): becomes same in both regions: z ² A dielectric region having a dielectric constant ²r is placed in air as shown in Fig. XXX, and E in air is assumed to have magnitude E0 and make an angle μ1 with the air-dielectric boundary. Determine the magnitude of E and the angle μ2 . 1 E1 Air D1z = D2z Physically, the electric flux density ! ²1 E1z = ²2 E2z x r 2 ² In a similar way, we can find the related boundary conditions for the following structures (See Appendix). E2 Conducting sphere { In order to apply the boundary conditions, we need to ¯nd the tangential and normal components of electric ¯eld. From the given magnitude and angle information, they are 2 2 1 2 1 E1t = E1x = E0 sin μ1 E1n = E1z = E0 cos μ1 E1=E2 1 1 D1=D2 Er1=Er2 2 Dr1=Dr2 { From the boundary conditions, E2t and E2n become 8. Capacitors and Capacitance E2t = E1t D2n = D1n ! E2x = E1x = E0 sin μ1 E1z E0 cos μ1 ! E2z = = ²r ²r ² Capacitor: Electronic device formed by two conducting plates preserving charges { Therefore the magnitude of E2 and angle μ2 are found as s cos2 μ1 jE2 j = E0 sin2 μ1 + ²2r V=0 V = V1 V = 2V1 ¢¢¢ E2x ²r sin μ1 tan μ2 = = ! μ2 = tan¡1 (²r tan μ1 ) E2z cos μ1 ² Capacitance: Let us assume that the electric field has only the z-component, and there is no fringing field. From the continuity of the tangential component of the electric field, E1z = E2z ! D1z D2z = ²1 ²2 ‐Q Q=0 Q = Q1 Q = 2 Q1 Q ! Q = kV ! C = slope + + Q = CV ² Capacitance C depends ² Case A (Parallel configuration): V Proportional constant of Q w.r.t V Q/V Ex. Parallel conducting plates Q + + 0 V { not on Q and V 1 2 E1z=E2z z x { but on the geometry and medium. ² Capacitance of various configurations of capacitor Circuit symbol C V 1 2 D1z=D2z Chung‐Ang University + Electromagnetics 24 Ex. Parallel plate capacitor ² When a voltage V is applied between two conductors with an assumption of no fringing field, electric field and flux density become ¡V Ez = d Parallel plate capacitor ! ² Let a charge Q is assumed to be uniformly distributed on the surface of the center conductor. Then D and E are found from the Gauss' law as I Q D ¢ dS = Q ! D½ = 2¼½L S Q ! E½ = 2¼²½L z S r d Dz = ²Ez The an induced positive charge on the upper conductor is found from the Gauss' law as I Q = D ¢ dS = Dz S S Therefore the capaciatance becomes Metal Foil Plates Q ²S C= = V d It is found that the capaciatnce increase as dielectric constant and surface area S, but deceases as spacing d, and this is a general trend for any other capacitors. { The other is to use multilayer structure with very high dielectric constant material. Multilayer ceramic capacitor (lower figure), which is widely used in small-scale circuit design. Now the V0 between conductors and the resulting C can be found as μ ¶ Z b b Q V= ln E ¢ dl = 2¼²L a a 2¼²L Q C= = μ ¶ b V ln a Coaxial‐line capacitor L ϵr a b Spherical capacitor Ex. Spherical-shell capacitor Lead Paper Lead Electrolytic capacitors – 100 uF / 10 V ² Based on the previous result, we can man make the capacitor structure yielding large capacitance. Fig. XXX shows two main approaches: { Increasing S with keeping d to be very small Eelectrolithic capacitor (upper figure) Ex. Coaxial capacitor Multi‐layer Ceramic chip capacitor 1005: 1.0 mm 0.5 mm ² Let a charge Q is assumed to be uniformly distributed on the surface of the inner conducting sphere. Then resulting D and C can be found as I Q D ¢ dS = Q ! Dr = 2 4¼r S Q ! Er = 4¼²r2 b a ϵr Now V0 between two conductors C are μ ¶ Z b 1 1 Q ¡ E ¢ dl = V= 4¼² a b a Q 4¼² ¶ C= = μ 1 1 V ¡ a b Multilayer configuration Chung‐Ang University Electromagnetics 25 ² Connection of capacitors Series connected capacitor { Series connection of capacitor Q Q Q V = V1 + V2 + ¢ ¢ ¢ + VN = + + ¢¢¢ + C1 C2 CN ¶ μ 1 1 1 =Q + + ¢¢¢ + C1 C2 CN Q = Cs + VN ‐ CN + V2 ‐ C2 + V1 ‐ C1 +Q ‐Q +Q ‐Q +Q ‐Q +V ‐ Cs +Q ‐Q Parallel connected capacitor 1 1 1 1 = + + ¢¢¢ Cs C1 C2 CN Q = Q1 + Q2 + ¢ ¢ ¢ + QN = C1 V + C2 V + ¢ ¢ ¢ + CN V = (C1 + C2 + ¢ ¢ ¢ + CN )V = Cp V Cp = C1 + C2 + ¢ ¢ ¢ CN S1 ϵ1 S2 ϵ2 { Constant V: battery. Conductors are connected to { Constant Q: Conductors are kept isolated. d d2 d1 ² Coaxial and Spherical capacitors (See Appendix) b1 ϵ2 b b b2 a a ϵ1 ϵ2 a ϵ1 L L Chung‐Ang University dWm = Fdz; ϵ2 ϵ1 b2 b1 XXX) dz V dQ: a ϵ2 F z dWm + dWb = dWe 8 dWm : Additionally supplied mechanical > > > > > energy by the force F > > > <dW : Additionally supplied electric b > energy by the battery > > > > > dWe : Increased stored electric energy > > : in the capacitor S ϵ2 ϵ1 F2=‐F1 F1 z=0 ² Considering two kinds of supplied energy (dependong on the approaches), the above equation can be expressed as ‐Q +Q Parallel connection Series Connection 1 1 1 = C = C1 + C2 + C C1 C2 8 8 Q ²1 S ²1 S1 Q1 > > > = = <C1 = <C1 = V1 d1 V d Q ² > > ²2 S2 Q2 2S : > :C2 = = = C2 = V2 d2 V d dl Constant V + V ‐ Cp ² Parallel plate capacitors (Composite dielectrics) Mechanically supplied energy Wm = –ʃ F1∙dl = ʃ F2∙dl ² It is made by postulating an infinitesimal displacement, and then applying the principle of conservation of energy. Here ² Two approaches (Fig. CN ‐QN +QN ² To calculate the electrostatic forces on the conducting plates, we can use the concept of the principle of virtual work, which is a general and reliable method for calculating forces. ¢Ws = ¢Wi Increased = Increase in supplied energy the internal energy + V ‐ C1 +Q1 ‐Q1 C2 +Q2 ‐Q2 { Parallel connection of capacitor 9. Electrostatic Forces Constant Q dz z FQ S ‐Q z=0 dWb = VdQ extra charge fed into the capacitor by the battery ϵ1 Electromagnetics 26 S Ex. Force on the conductor of parallel-plate capacitor ² Calculate the force exerted on the conductor of parallel-plate capacitor. { Case 1 : With constant V ¡ The variation of the supplied energy by the battery dWb = VdQ = V(VdC) With the following expression of C(z) C(z) = ²0 S z ! ! dC ¡²0 S = 2 dz z ¡²0 S dC = 2 dz z ¡ The resulting dWb can be found with E = V=z: μ ¶ ¡²0 S 2 2 dz dWb = V dC = V z2 = (¡²0 E2 ) S dz Negative sign in dC and dWb implies that the energy flows from the capacitor into the battery. ¡ On the other hand, dWe can be found with E = V=z: μ 2¶ ²0 V2 S ²0 E We = Sz = 2 2z μ ¶ μ ¶ 2 ¡²0 V S ¡²0 E2 dz = Sdz dWe = 2z2 2 ¡ Finally the force on the capacitor can be found as Fdz + dWb = dWe μ ¶ ²0 E2 2 Fdz = ²0 E ¡ Sdz 2 μ 2¶ ²0 E F= S 2 { Case 2: With constant Q (isolated capacitor, dWb = 0) ¡ The related differential energy become dWm = dWe ; dWm = Fdz ¡ In a similar way, dWe can be found with D = Q=S as μ 2¶ D Q2 z Sz = We = 2²0 2²0 S Q2 E2 D2 dWe = Sdz = Sdz dz = 2²0 S 2²0 2 ¡ The resulting F is found to be same as the previous result. μ 2¶ ²0 E F= S 2 Ex. Force on a capacitor with pulling dielectric sheet ² For a dielectric-filled capacitor, calculate the force exerted on the upper conductor wheen a dielectric slab is pull. a w ϵ d x=0 x { With keeping constant Q (isolated capacitor), energy conservation law is stated as Fdx = dWe { It is found that We is a function of C, and C is a function of x. Therefore dWe can be found as 8 ² w <C = C1 + C2 = 0 [x + ²r (a ¡ x)] 2 Q d We = ; :dC = ¡²0 b (²r ¡ 1) dx 2C d ¡Q2 Q2 ³ ²0 w ´ (²r ¡ 1) dx dC = 2 dWe = 2C2 2C d { Therefore F can be found with V = Q=C as F= Chung‐Ang University with Electromagnetics Q2 ³ ²0 w ´ ²0 E2 (²r ¡ 1)wd ¡ 1) = (² r 2C2 d 2 27 Pulling 10. Solutions of Electrostatic Problems 10.1 Uniqueness Theorem ² Up to now, we have studied the following methods for solving the electrostatic problems: ² A solution of Poisson and Laplace eqs for the specified B.C. is unique !!! Q3 { Direct calculation (vector operation) Proof: { V ! E { Series Solutions { Let us assume that two different solutions V1 and V2 with the same boundary condition can exist. ¡½ ¡½ r2 V2 = ; r2 V1 = ² ² ( V1 6= V1 ; inside the boundary V1 = V2 ; on the boundary { Numerical Methods With Vd = V1 ¡ V2 { Gauss' law ² Much more geenral methods based on Poisson's and Laplace' equations: { Image Method V=0 on the boundary Q2 V = ??? Q1 ( r2 Vd = 0 Vd = 0; on the boundary ¡ FDM (Finite Difference Method ¡ FEM (Finite Element Method) ¡ MOM (Method of Momentum) { Vector identity with f = Vd and A = rVd ² Poisson Equation { Two governing electrostatic equations in any medium 8 <r £ E = 0 ! E = ¡r V :r ¢ D = ½ ! r ¢ E = ½ ² ! ¡½ r ¢ (r V) = ² ! ¡½ r2 V = ² r ¢ (f A) = f (r ¢ A) + (rf) ¢ A :0 2 »» r ¢ (Vd rVd ) = Vd » (r»¢ » rV d ) + rVd ¢ rVd = jrVd j { With a volume integral and the divergence theorem Z Z 2 [r ¢ (Vd rVd )] dv = jrVd j dv V { Lapalce' equation (with ½ = 0 r2 V = ¡½ ² ½=0 ¡¡¡¡¡¡! S r2 V = 0 2 jrVd j dv V { Since Vd = 0 at the boundary, Z 2 jrVd j dv = 0 ! rVd = 0 V ( Vd = constant Vd = 0on the boundary ! Chung‐Ang University V (Vd rVd ) ¢ dS = ! { Electrostatic boundary-value problems are specified with 1) r2 V = 0 2) Boundary conditions Z I Electromagnetics V1 = V2 ! ! Vd = 0 for all region Contradiction !!! 28 10.2 Simple Boundary-Value Problems Ex. Electric potential for a parallel-plate capacitor ² Calculate a electric potential V for a parallel-plate capacitor, where the plates are assumed to be infinite and the applied potentials are μ ¶ @V ½ =0 @½ @V A1 ! = @½ ½ 1 @ ½ @½ z V0 z=d z=0 V=0 x 1) V(z = 0) = 0 2) V(z = d) = V0 A1 ln a + A2 = V0 From the boundary conditions, it is known that 1) V(½ = a) = V0 2) V(½ = b) = 0 { It is known that V becomes a function of ½ only, and the Lapalce equation becomes simplified as 8 > <A1 = ! ln b > : A2 = ¡A1 ln b z ² Calculate a electric potential V for cylindrical wedge having the following potentials: 1) V(Á = 0) = 0 2) V(Á = Á0 ) = V0 b a { It is known that V becomes a function of Á only, and the Lapalce equation becomes simpli¯ed as ϵr Insulating gap y x V=0 V0 1 @2 @2 = 0 ! =0 ½2 @Á2 @Á2 ! V = A1 Á + A2 V0 V=0 { From the boundary conditions, we can determine the coe±cients A2 and A2 , and the resulting V becomes V = V0 Chung‐Ang University V ³0a ´ Ex. Electric potential for cylindrical wedge V0 A1 = μ ¶ d V0 V= z d A2 = 0; ² Calculate a electric potential V for a coaxial capacitor with the following potentials: V = A1 ln ½ + A2 ³½´ V = A1 ln ½ ¡ A1 ln b = A1 ln b μ ¶ b ln ½ = V0 μ ¶ b ln a @2 V(z) = 0 @ z2 V(z) = A1 z + A2 Ex. Electric potential for a coaxial capacitor ! { Therefore the potential becomes r2 V = 0; ! ½ From the boundary conditions ( A1 ln b + A2 = 0 { The potential V becomes a function of z only, and the related Lapalce equation becomes @V = A1 @½ ! Electromagnetics Á Á0 29 { B1 is given as Ex. Electric potential V for spherical-shell capcitor ² Calculate a electric potential V for spherical-shell capcitor having the following potentials: 1) V(r = a) = V0 2) V(r = b) = 0 { It is known that V becomes a function of r only, and the Lapalce equation becomes simpli¯ed as μ ¶ 1 @ 2 @V r =0 r2 @ r @r { The solution of V(r) can be expressed as 8A 1 > < + A2 ; a < r < b1 r V(r) = > B : 1 +B ; b <r<b 2 1 2 r and considering the given potentials, ¶ 8 μ 1 1 > > <A1 r ¡ a + V0 ; a < r < b1 μ ¶ V(r) = > 1 1 > :B1 ; b1 < r < b2 ¡ r b2 { Therefore Er becomes 8A 1 > < 2 ; a < r < b1 r Er (r) = ¡rV(r) = > : B1 ; b < r < b 1 2 r2 { From the boundary conditions at r = b1 are ( V1 = V2 Dr1 = Dr2 ! ²1 Er1 = ²2 Er2 8 μ ¶ μ ¶ > <V0 + A1 1 ¡ 1 = B1 1 ¡ 1 b1 a b1 b2 ! > :² A = ² B 1 1 2 1 Chung‐Ang University μ b2 a b1 ϵ1 ϵ2 B1 = ²1 ²2 ¶ A1 { Therefore A1 can be found as μ ¶ μ ¶μ ¶ 1 1 1 ²1 1 A1 ¡ = A1 ¡ ¡ V0 b1 a ²2 b1 b2 V0 ¶ μ ¶ A1 = μ ¶ μ ²1 1 1 1 1 ¡ + ¡ ²2 b1 b2 a b1 { Now the potential V(r) can be determined. Ex. Potential distribution of cone structure ² Calculate the potential distribution of cone structure with the following potentials: ( V(μ = ®) = V0 V(μ = 0) = 0 { Considering the structure and assigned potentials, the potential distribution is known to be a function of μ only. Therefore the Lapalace equation and its solution become μ ¶ A1 1 @ @ V(μ) @ V(μ) sin μ =0 ! = sin μ @μ @μ @μ sin μ μ ¶ μ ! V(μ) = A1 ln tan + A2 2 z V0 P(r,,) V=0 { From the boudary conditions, the electric potential V, having the axial symmetry, satisfies the Laplace equation as μ ¶ μ ln tan 2 V(μ) = V0 ³ ®´ ln tan 2 Electromagnetics 30 x ² From the boundary condition (4), the condition of k becomes 10.3 Series Solution 10.3.1 Lapalce Equation k ² For a source-free region, let us consider how to ¯nd the potential V if the boundary conditions are speci¯ed. Ex. Two dimensional boundary-value problem ² One two-dimensional example is shown in Fig. XXX with the following boundary conditions: ( ( 1) X(0) = V0 3) Y(0) = 0 ; 2) X(1) = ¯nite 4) Y(b) = 0 x x→ 0V ² The Laplace equation for this problem becomes r2 V = 0 ! @V @V + 2 =0 2 @x @y x m ² Multiplying sin(kn y) to both parts, and taking integrations them from y = 0 to y = b Z V0 ² The Lapalce equation can be simplified as Ä X ¡Ä Y = = k2 X Y ! 1 X b sin(kn y) dy = 0 V(x; y) = X(x) Y(y) ! m = 1; 2; ¢: Here the unkown coefficient Am can be found from the remaining boundary condition (1): X Am sin(km y) V0 = ² With the seperation of variables, Ä XY + XÄ Y=0 m¼ ; b m 0V V0 0 km = Now the general expression of V becomes X Am e¡km x sin(km y) V(x; y) = y b ! Ä X Ä Y + =0 X Y ( Ä X = k2 X Ä Y = ¡k2 Y Am sin m=1 ³ n¼ ´ ³ m¼ ´ y sin y dy b b Using the integral formula and the orthogonal relation 8 Z b < 2b ; odd m ³ m¼ ´ y dy = m¼ sin : b 0 0; even m μ ¶ Z b ³ n¼ ´ ³ m¼ ´ b y sin y dy = ±mn sin b b 2 0 Now Am can be found as where k(¸ 0) is a constant, and two dots over letter means two times derivative with the related variable. The possible solution sets are ( ( e¡kx cos(ky ) X» ; Y» kx sin(ky) e 8 < 4V0 ; Am = m¼ : 0; Chung‐Ang University even m Therefore V is given as X and their linear combinations become the solution. From the boundary conditions (2) and (3), V(x; y) can be written as V(x; y) » e¡kx sin(ky) odd m V(x; y) = 4V0 exp m¼ m=1;3;5;¢ μ ¶ ³ m¼ ´ ¡m¼ x sin y b b The above expression with only a few terms gives an accurate result because of the decrasing term of 1=m as m increases. Electromagnetics 31 10.3.2 Boundary Value Problem of Poisson's Equation ² AAA Chung‐Ang University Electromagnetics 32 { In a matrix form 10.4 Numerical Solution y 0V 10.4.1 Finite Di®erence Method (FDM) b AV = Vs ² Source free problem ! Laplace equation 2 ¡4 V(x,y) 2 @ V @ V r2 V = 2 + 2 = 0 @x @y A= 0 1 1 0 0 0 0 1 1 { 2D difference approximation V(x + ¢x=2) ¡ V(x ¡ ¢x=2) @V = V0 (x) ! @x ¢x 2 0 0 @ V V (x + ¢x=2) ¡ V (x ¡ ¢x=2) ! @ x2 ¢x V(x + ¢x; y) ¡ 2V(x; y) + V(x ¡ ¢x; y) = ¢x2 2 V(x; y + ¢y) ¡ 2V(x; y) + V(x; y ¡ ¢y) @ V ! 2 @y ¢y2 § V(x,y+Δy) V(x‐Δx,y) V(x,y) 0 Δy Δx V(x+Δx,y) V(x,y‐Δy) V1 . . . V = V6 ; .. . V12 0 0 0 0 0 . . . ¡4 . . . 0 0 0 0 1 0 1 0 0 0 0 V0 1 0 0 1 3 4 6 7 8 9 10 11 12 x V2 2 V5 V6 5 6 V7 7 h V10 10 h { Volution is V = A¡1 Vs § { With ¢x = ¢y = h, x = x § h, y = y § h { Matlab expression of inverse matrix: 4V(x; y) ¡ V(x+ ; y) ¡ V(x¡ ; y) ¡ V(x; y+ ) ¡ V(x; y¡ ) = 0 ! useful formula for programming { V(x; y) is found to be the average of 4 nearby potenials at (x; y). V(x; y) = >> V = inv(A) ¤ VB ; ² For the point of discrete charge with Qs = ½L { Poisson's equation V(x+ ; y) + V(x¡ ; y) + V(x; y+ ) + V(x; y¡ ) 4 Ex. Matlab programming of FDM (See Fig.XXX) Equal grid spacing h = 1 cm, and a = 10 cm, b = 5 cm: { For the node 1 5 + V2 + 0 + V5 ¡ 4V1 = 0 ! ¡4V1 + V2 + 0 + V5 = ¡5 r ¢ (rV) = Ix = [Ex (x + ¢x=2) ¡ Ex (x ¡ ¢x=2)] ¢y¢z £ ¤ Iy = Ey (y + ¢y=2) ¡ Ey (y ¡ ¢y=2) ¢x¢z V5 + V7 + V2 + V10 ¡ 4V6 = 0 ! V2 + V5 ¡ 4V6 + V7 + V10 = 0 Chung‐Ang University ¡½L ±(x)±(y) ² { Taking a volume integral and applying the divergence theorem (extending the region ¢z = 1 m along z-axis with uniform ¯elds) I I Q D ¢ dS = Q ! E ¢ dS = ² I E ¢ dS = Ix + Iy { For the node 6 Q = ½v ¢v = ½L ¢x¢y Electromagnetics 0V a 0V ¡4 ¡5 . . . VB = 0 . . . 0 2 5 0 0 0 1 33 L W ; i; m = 1 : M; dy = ; j; l = 1 : N M N Source points Field points xsm = (m ¡ 0:5) dx xi = (i ¡ 0:5) dx + ±x ysn = (n ¡ 0:5) dy yj = (j ¡ 0:5) dy + ±y §d d z= z§ s = 2 2 V(x) ¡ V(x + ¢x) ¢x V(x ¡ ¢x) ¡ V(x) Ex (x ¡ ¢x=2) = ¢x V(y) ¡ V(y + ¢y) Ey (y + ¢y=2) = ¢y V(y ¡ ¢y) ¡ V(y) Ey (y ¡ ¢y=2) = ¢y 2V(x; y) ¡ V(x + ¢x; y) ¡ V(x ¡ ¢x; y) Ix + Iy = ¢y ¢x 2V(x; y) ¡ V(x; y + ¢y) ¡ V(x; y ¡ ¢y) ¢x + ¢y ½L ¢x¢y Ix + Iy = ² dx = Ex (x + ¢x=2) = { The charge distribution is expressed by linear combination with basis functions and unknown coefficients: R X l = (m ¡ 1)N + j qsl : point charge at the l-th index 4V(x; y) ¡ V(x+ ; y) ¡ V(x¡ ; y) ¡ V(x; y+ ) ¡ V(x; y¡ ) ½L = h2 ² ½L 2 4V(x; y) ¡ V(x+ ; y) ¡ V(x¡ ; y) ¡ V(x; y+ ) ¡ V(x; y¡ ) = h ² { The charge distribution in the lower plate has the same but different sign. { With an assumed potential difference V0 , the potential on the upper plate becomes V+ = V0 =2. { Calculate the potential Vk at the reasonably chosen points on the upper plate: 10.4.2 Method of Moments ² Here one numerical approach called method of moments is studied. (Integral equation formula z y k = (i ¡ 1)N + j; x 1 2 M μ +Q ‐Q L N 2 W 1 l = (m ¡ 1)N + n ¶ 1 1 ¡ ¡ R+ Rkl kl R 1 X al 4¼²0 l=1 q R+ (xi ¡ xm )2 + (yj ¡ yn )2 kl = q R¡ (xi ¡ xm )2 + (yj ¡ yn )2 + d2 kl = V+ k = VR+£1 = FR£R QR£1 μ ¶ 1 1 1 Fkl = ¡ + 4¼²0 R+ Rkl kl Ex. Capacitance of ¯nite parallel plate capacitor { The conducting plate is divided into small rectanglular section as Chung‐Ang University on the upper plate l=1 { With ¢z = 1, ¢x = ¢y = h ² Up to now we have assumed a uniform charge distribution in conducting plates of a capacitor, and then we can ¯nd electric ¯eld distribution and capacitance. However, the size of a capacitor is ¯nite, and the real distrbution is di®erent to the assumed. For this case, some numerical approaches can give us a solution even though it is not a closed form. qsl ± (x ¡ xsm ; y ¡ ysn ); qs = Q = F¡1 V+ X Qsum = Q ! Electromagnetics C= Qsum V0 34 10.5 Image Theorem ROI = Region of interest ² Same Poisson's equations for a region y ¸ 0 with the same B.C. at y = 0. ² Two problems are the same from the uniqueness theorem. ROI z E Q h V=0 ² Equivalent with each other for y ¸ 0. ρ Planar Conductor ROI z E Q h V=0 ρ h ‐Q Image y ROI d1 Q d2 x z y ROI d1 Q ‐Q d1 d2 z x d2 ‐Q Q ROI Q 30∘ 60∘ x z ‐Q ROI Q Q z 30∘ 60 ∘ x ‐Q ‐Q Q Chung‐Ang University Electromagnetics 35 Appendix Ex. Non-uniform line-charge distribution z ‐5a Ex. Theory of operation of torsion balance appratus ² One conducting sphere (A) connected to the balance bar is charged with Qa . The other conducting spahere (B) charged with Qb is inseted carefully and placed (fixed) as shown in Fig. ² Charge distribution ½L (z) = 1+ Torsion spring L/2 m Z m ρ(z) 1 ½L (z)dz Q=2 0 With a change of variables z = a tan μ ! dz = a sec2 μ dμ Z ¼=2 ¼ Q=2 ½L (z)dz = 2a½0 2 0 = ¼ a½0 { Then the rotation angle μ is measued, and the torque is calculated as L/2 L/2 spring constant 0 L/2 pendulum ² Now an electric force is exerted on the conducting sphere A, resulting in rotation of the balancing bar. · : a Total charge F ¿ = ·μ; ½0 ³ z ´2 θ Fixed Qb F Qa R { The force exerted on the conducting sphere A is calculated 2¿ L ! F= ¿ =F 2 L { The distance of two conducting spheres are also measured. ² E½ along ½-axis ² Charge distribution of Maxwellian-hat function ‐5a a 0 ρ(z) ‐a ½ ½L (z) = p 2 0 2 a ¡z { Repeating this procedure with other conducting sphere B having different amount of charge { The spring constant · can be measure as follows: ¤ A force is applied momentarily. Then the bar connected to the torsion spring oscillates with its center. ¤ The oscillation period T becomes a function of spring constant and length of bar as r μ ¶2 I T T = 2¼ ! ·=I · 2¼ where the Moment of inertia of the balance bar I is given as μ ¶2 μ ¶2 m L2 L L I=m +m = 2 2 2 Chung‐Ang University Electromagnetics 36 ² Condition for zero electric field at the point P ??? ½L2 ¡½L1 = 2 Ex. Chathode-Ray Tube (CRT) = 60o L1 ² An electron ejected from the cathode in vacume tube (chathode-ray tube (CRT)) travels to the right and passes through deflection region, and finally arrivea on the screen. Calculate the distance d0 . P L2 Ex. Circular disk with a uniform line charge distribution ½L =60o ² Electric field formed between two conducting plates and electric force exerted on the electron d ² Ez along the z-axis can be found as 2d dq = ½S dS = ½S ½d½dÁ ¶ Z b Z 2¼ μ ^ R¢^ z Ez = ke ½S ½d½dÁ R2 0 0 The integrand is not a function of Á. Z b ½z Ez = ke ½S (2¼ ) d½ 3 0 R d½ = z sec2 ° d° ½ = z tan °; ¸ Z °0 · 3 z tan ° Ez = ke ½S (2¼ ) sec2 ° d° 3 sec3 ° z Z0 °0 = ke ½S (2¼ ) sin ° d° 0 Circular disk charge (0,0,z) s 0 y t1 = x Vp dp Fe = ¡e E0 = ^ zF0 ; F0 = e Vp dp w ; v0 a= e Vp F0 = me me dp ms = 9:31 £ 10¡31 kg (mass of electron) Therefore d1 becomes Infinite planar charge (0,0,z) Ez d1 = t2 = Circular ring disk charge s b1 vx = v0 ; vz = at1 Vp e w(L ¡ w) dp me v20 Vp e w(L ¡ w=2) d0 = d1 + d2 = dp me v20 d2 = vz t2 = 2 0 L¡w ; v0 Therefore d2 and d0 can be found as Ez (0,0,z) 1 Vp e w2 1 2 a t1 = 2 2dp me v20 ² After the region of two conducting plates, the electrons moves to the right and upward for a travel time t2 with uniform velocities y 0 x Ez = ke ½s (2¼ ) Ex. Circular ring disk with a uniform charge distribution ½s 0 b ρs ² For an infinite plane (b ! 1) E0 = ² Within the region between two conducting plates, electrons move to the right with a uniform velocity v0 and move up simultaneously because of the force exerted on the electrons with acceleration a for a time t1 : Ez dq = ke ½S (2¼ ) (1 ¡ cos °0 ) z cos °0 = p b 2 + z2 Ex. Infinite plane with a uniform charge distribution ½s E = ¡^ z E0 ; b2 y z Vp Vc v0 Cathode Screen Deflection plates d2 dp d1 E 0 w L x Chung‐Ang University Electromagnetics 37 d0 x Ex. Circular Line Charge (N-Polygon Approximation) E1 ² Electric field along the z -axis for an N-polygon uniform line charge Ex. E along the z-axis for rectangular surface charge γ z P(0,0,z) For a single line section, Ez1 can be found from the expression for a finite length line cyrrent as γ R10 y y α L x z dE ¼ N γ (0,0,z) γ L = p sin ®; 2 q = p cos ® μ ¶μ ¶ L=2 z Ez = N (2ke ½L ) R10 R2 μ ¶μ ¶ c0 CN z = ke ½L R10 R2c0 CN = NL dq = ½s dxdy = ½L dx ! ½L = ½s dy p ½ ! Rc0 = z2 + y2 p R10 = (L=2)2 + z2 + y2 μ ¶ sin ¯10 Ez1 = 2ke ½L cos ° Rc0 L=2 z sin ¯10 = ; cos ° = R Rc0 μ 10¶ μ ¶ L=2 z Ez1 = 2ke ½L R10 R2c0 L ρL Thus the total electric field can be found as ®= 0 x R 0 y ϕ′ r′ x ρL Chung‐Ang University L/2 R ρs r’ ‐W/2 y dq ‐L/2 x W/2 z Ez R10 L/2 Rc0 ρs L ‐W/2 x y ‐L/2 W/2 ² Thus the total electric field can be found as ¶μ ¶ Z W= 2 μ L=2 z Ez = 2ke ½L dy = 2ke ½L Lz I(z) R10 R2c0 ¡W=2 Z W=2 Z W=2 dy dy p I(z) = 2 = 2 R R (L=2) + z2 + y2 (z2 + y2 ) 10 c0 0 0 dq From cos ° = z=Rc0 Rc0 ydy ! sec ° tan ° d° = z zRc0 μ ¶ dy y sin ° sin ° = =! d° = sin ° 2 Rc0 cos ° z z dy = cos2 ° sec ° = (Total length of polygon) ² For a circular line charge distribution p C1 = 2¼ p; Rc0 = R10 = p2 + z2 " # μ ¶ C1 z 2¼ p z Ez = ke ½L = ke ½L R310 (p2 + z2 )3=2 Ez ² E=^ z Ez due to the structural symmetry. Rc0 Due to the axial symmetry, E = ^ zEz . For a single line section, Ez1 can be found from the expression for a finite length line cyrrent as p p ½ ! Rc0 = q2 + z2 ; R10 = p2 + z2 μ ¶ sin ¯10 Ez1 = 2ke ½L cos ° Rc0 L=2 z sin ¯10 = ; cos ° = R10 Rc0 μ ¶μ ¶ L =2 z Ez1 = 2ke ½L R10 R2c0 z ² For a circular line charge distribution p C1 = 2¼ p; Rc0 = R10 = p2 + z2 " # μ ¶ C1 z 2¼ p z Ez = ke ½L = ke ½L 3= 2 R310 (p2 + z2 ) Electromagnetics 38 Ex. Circular Disk Charge and Its Variants { Therefore dEz becomes ² Circular disk with a radius a carrying a uniformly distributed charge with a surface charge density ½s . ² Due to the axial symmetry. it is su±cient to ¯nd Ex and Ez on the xz-plane as before. With r = (x; 0; z) r0 = (½0 cos Á0 ; ½0 sin Á0 ; 0) R = r ¡ r0 = (x ¡ ½0 cos Á0 ; ¡½0 sin Á0 ; z) dEx and dEz for a di®erential surface dS = ½0 d½0 dÁ0 becomes μ ¶ ^ ½s R¢x dS dEx = 4¼²0 R3 μ ¶ ½s R¢^ z dS dEz = 4¼²0 R3 where R3 = (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2 ^ = x ¡ ½0 cos Á0 R¢x R¢^ z=z ² At ¯rst, let us calculate the electric ¯eld on the z-axis. Due to the axial symmetry. we know that the electric ¯eld has only the z -component, and dEz becomes dEz = ½s dS z 0 0 0 ½ d½ dÁ 4¼²0 R3 where R3 (z2 + ½02 )3=2 dEz = ½s ½0 z 02 d½0 dÁ0 4¼²0 (½ + z2 )3=2 { Considering the integral along Á0 becomes 2¼ , the total Ez becomes Z a ½s ½0 z d½0 Ez = 02 + z2 )3=2 2²0 ( ½ 0 0 With a change p of variables ½ = z tan ® and introduce R0 = z2 + a2 , μ ¶ ½s z Ez = 1¡ 2²0 R0 This is valid for z > 0, and Ez including the region z < 0 becomes μ ¶ jzj ½s Ez = § 1¡ 2²0 R0 ( + for z > 0 ¡ for z < 0 { One special case : circular ring disk with inner and outer radius a and b ·μ ¶¸ ½s jzj jzj ¡ Ez = § 2²0 Ra Rb p Ra = z2 + a2 p Rb = z2 + b2 { Second one : in¯nitely extended surface charge distribution Taking a limit on a to in¯nity, then Ez = § Chung‐Ang University Electromagnetics ½s 2²0 39 Ex. Conical surface (like an empty ice-cream cone) { Using the following integral Z p x dx p = x2 + a2 x 2 + a2 Z ³ ´ p dx p = ln x + x2 + a2 x 2 + a2 z ² The cone surace with a height h carrys a uniform surface charge ½S . Find the potential difference between points A (the vertex) and B (the center of the top). γ B h h‐z′ z′ { We can use the previous result with b = z0 p p R = b2 + (z ¡ z0 )2 = z02 + (z ¡ z0 )2 h Iz = Iz1 + Iz2 Z h¡z=2 p Iz1 = z′ ρs A z=0 ¡z=2 { VA and VB μ VA = V(0) = V(z) = ¶Z hμ 0¶ 0 z R μ dz0 = ½ ps 2²0 ¶Z 0 h " z0 p z02 + (z ¡ z0 )2 # dz0 { With ´ = z0 ¡ z=2, the argument of square root can be written as · ³ ³ z ´2 ¸ z ´2 z2 2z02 ¡ 2zz0 + z2 = 2 z0 ¡ + = 2 ´2 + 2 2 2 d´ = dz0 μ ¶ ½s V(z) = Iz 2²0 Z h¡z=2 ´ + (z=2) p Iz = d´ ´ 2 + (z=2)2 ¡z=2 Chung‐Ang University ½s h 2²0 ¶ p ¸ · h 1=2 + 1= 2 p VB = V(h) = ln 2 ¡1=2 + 1= 2 μ ¶ μp ¶ ½s h 1 2+1 p = ln 2²0 2 2¡1 μ ¶ μq ¶ p ½s h = ln 3+2 2 2²0 μ ¶ ³ p ´ ½s h = ln 1 + 2 2²0 μ { The total potential on the z-axis ½ ps 2²0 d´ p p (h ¡ z=2)2 + (z=2)2 ¡ z2 =2 Z h¡z=2 z=2 p Iz2 = d´ 2 ´ + (z=2)2 ¡z=2 " # p h ¡ z=2 + (h ¡ z=2)2 + (z=2)2 z p = ln 2 ¡z=2 + z2 =2 ½L = ½s dr0 ! dr0 z0 = r0 cos μ ! dz0 = p 2 μ ¶ ½L b 1 V(z) = 2²0 R μ ¶ μ ¶ 0 ½s z0 1 p ½s z 0 2 dz = p dz0 dV(z) = 2²0 R 2²0 R μ + (z=2)2 = { In addition, the uniform line charge density ½L can be found as dq = ½s dr0 bdÁ0 = ½L bdÁ0 ´ ´2 ½s 2²0 ¶ { Resulting electric field has only the z -component. μ ¶μ ¶ ¡½s @ Iz1 @ Iz1 ¡@ V(z) Ez = = + @z 2²0 @z @z p p 2 2 2 Iz1 = (h ¡ z=2) + (z=2) ¡ z =2 " # p h ¡ z=2 + (h ¡ z=2)2 + (z=2)2 z p Iz2 = ln 2 ¡z=2 + z2 =2 Electromagnetics 40 Ex. Hemispherical Charge ² Uniformly distributed spherical charge. ² Because of a spherical symmetry, the resulting electric filed is known to have the r-component only and it is a function of r only. It is sufficient to calculate Ez at (0; 0; z). ² Second way : The result for a circular disck charge can be applied extended with the following surface charge density ½s as dq = ½v dv = ½s dS ² What is noteworthy is that this expression can be rewritten in terms of the total charge of the sphere Q as z ! ½s = ½v dz0 E P(0,0,z) Q 4¼²0 z2 4¼ a3 ½v Q= 3 Ez = a ρv 0 y a x ² Electric field inside the sphere, that is z < a ½v (I3 + I4 ) 2²0 ¶ Z zμ z ¡ z0 1¡ p dz0 I3 = z2 + a2 ¡ 2zz0 ¡a ¶ Z aμ z0 ¡ z I4 = ¡ 1¡ p dz0 2 2 0 z + a ¡ 2zz z Ez = z dE P(0,0,z) R0 b ρs z′ a Therefore dEz for z > a (outside the sphere) becomes μ ¶ ½v dz0 z ¡ z0 1¡ dEz = 2²0 R0 p R0 = z2 + a2 ¡ 2zz0 0 ² Taking an integration from z0 = ¡a to a, Ez can be found as ¶ Z aμ z ¡ z0 ½v ½v Ez = 1¡ p dz0 = (a + Iz1 + Iz2 ) 2 2 0 2²0 0 2 ²0 z + a ¡ 2zz Z a 1 p dz0 Iz1 = ¡z 2 z + a2 ¡ 2zz0 0 Z a z0 p Iz2 = Iz2 = dz0 2 2 0 z + a ¡ 2zz 0 Using the integral formula with Z 1 dx = p px + q Z x dx = p px + q y x p = ¡2z and q = z2 + a2 : p 2 px + q p 2(px ¡ 2q) p px + q 3p2 Following the similar procedures and algebraic manipulations, μ 3¶ ½v 2z Ez = 2²0 3z2 This expression can be rewritten in terms of the charge Qin , which is total charge inside the sphere with a radius z as Qin 4¼²0 z2 4¼ z3 ½v = 3 Ez = Qin ² Conclusions: { Only the charge inside the sphere with a radius z can make a contribution to E. This can be easily proved by the Gauss' law which we will study later. { One more interesting problem is E from a uniformly distributed spherical shell charge. This problem can be solved by extending the above results as and taking some algebraic manipulations Ez =??? Chung‐Ang University Electromagnetics 41 Qin 4¼²0 z2 8 0; z·a > > > < 4¼(z3 ¡ a3 )½v ; a·z·b = 3 3 > > 4¼(b3 ¡ > a ) ½ v : ; z¸b 3 { z-point for maximum Ez Ez = Qin dEz =0 dz Ex. Circular Line Charge (O® the Axis) ² Due to an axial symmetry, EÁ = 0. ² On the xz-plane dEx and dEz become μ ¶ ½L R¢^ x dEx = a dÁ0 4¼²0 R3 μ ¶ ½L R¢^ z dEz = a dÁ0 4¼²0 R3 8 r = (x; 0; z) > > > > > r0 = (a cos Á0 ; a sin Á0 ; 0) > > > <R = r ¡ r0 = (x ¡ a cos Á0 ; ¡a sin Á0 ; z) > R3 = (a2 + x2 + z2 ¡ 2ax cos Á0 )3=2 > > > > > R¢^ x = x ¡ a cos Á0 > > : R¢^ z=z · ¸ ½ a x ¡ a cos Á0 dEx = L dÁ0 4¼²0 (a2 + x2 + z2 ¡ 2ax cos Á0 )3=2 · ¸ ½L a z dEz = dÁ0 4¼²0 (a2 + x2 + z2 ¡ 2ax cos Á0 )3=2 z dE P R r b 0 r’ x ρ ϕ’ dq y ρL · ¸ Z ½L a 2¼ z Ez = dÁ0 4¼²0 0 (a2 + z2 )3=2 μ ¶ ½L a z = 2²0 (a2 + z2 )3=2 ! ! a2 ¡ 2z2 = 0 · ² Ex and Ez off the axis · ¸ Z ½L a 2¼ x ¡ a cos Á0 ½ a Ex = dÁ0 = L (I1 + wI2 ) 4¼²0 0 4¼²0 x (a2 + x2 + z 2 ¡ 2ax cos Á0 )3=2 Z ¼ 1 p I1 = dÁ0 2 2 2 0 a + x + z ¡ 2ax cos Á 0 Z ¼ 1 I2 = dÁ0 2 + x2 + z2 ¡ 2ax cos Á0 )3=2 (a 0 w = x2 ¡ a 2 ¡ z2 Z 2¼ ½ a z Ez = L dÁ0 2 2 2 4¼²0 0 (a + x + z ¡ 2ax cos Á0 )3=2 = ½L a (z I2 ) 2¼²0 Ex. Circular disk charge (Off the axis) ² Ex and Ez off the z-axis dEx = ½s 4¼²0 ½s dEz = 4¼²0 ² On the z-axis (x = 0), Ex = 0 ¸ 2z2 =0 (a2 + z2 )5=2 a ! zmax = § p 2 1 3 ¡ 2 2 3 = 2 2 (a + z ) · · (x ¡ ½0 cos Á0 ) ½0 d½0 dÁ0 (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2 z ½0 d½0 dÁ0 (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2 ¸ ¸ Considering the integral along Á0 at first Z aZ ¼ ½s (x ¡ ½0 cos Á0 ) dÁ0 ½0 d½0 Ex = 2¼²0 0 0 (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2 Z aZ ¼ ½s z dÁ0 ½0 d½0 Ez = 4¼²0 0 0 (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2 ² Numerical calculation is needed ??? Chung‐Ang University Electromagnetics 42 Ex. Find V and then E for a circular ring charge ² Due to an axial symmetry, we can choose a field point P(x; 0; z) on the xz-plane (Á = 0). ² Differential potential dV ½ dV = L 4¼²0 8 r = (x; 0; z) > > > <r0 = (a cos Á0 ; a sin Á0 ; 0) >R = r ¡ r0 = (x ¡ a cos Á0 ; ¡a sin Á0 ; z) > > p : R = a2 + x2 + z2 ¡ 2ax cos Á0 μ ¶ 1 a dÁ0 R ² Electric potential (2 times of the contribution from the half circle) Z ½L a ¼ dÁ0 p V= 2¼²0 0 a2 + x2 + z2 ¡ 2ax cos Á0 ² With a change of variables ¼ ¡ Á0 ! dÁ0 = ¡2d® 2 cos Á0 = cos(¼ ¡ 2®) = 2 sin2 ® ¡ 1 Z ½L a ¼=2 d® ½ a 1 p V= = L p IV 2 ¼²0 0 ¼²0 (a + x)2 + z2 (a + x)2 + z2 ¡ 4ax sin ® Z ¼=2 1 p IV = d® = K(k) 0 1 ¡ k sin2 ® 4ax k= (< 1) (a + x)2 + z2 : Complete elliptic integral ®= ² Generalization (x to ½) V= ½L a 1 p K(k) ¼²0 (a + ½)2 + z2 4a½ k= (a + ½)2 + z2 Chung‐Ang University ² Induced charges on metallic conductors { When the point charge Q is moved to a point off from the center of the spherical conducting shell, it is difficult to find the analytical solution of induced charge distribution. + { E inside the shell is now distorted to ensure the normal component of E only on the inner surface of the shell. + + ‐ ‐ ‐ + ‐ +Q ‐ ‐ + ‐ + ‐ + + { In the inner surface of the shell, the induced (¡) charge is = ½s , being non-unifomly distrbuted, in accordance with Dn denser at points of the surface closer to the charge Q, where the electric field magnitude is higher. { On the other hand, the (+) charge layer on the outer surface of the shell remains uniformly distributed, resulting in the same magnitude at all points as if the exterior field originated from a point charge located at the center of the shell. Ex. Connection of two conducting spheres Q1 q1 ² Different charges Q1 and Q2 are put on two conducting spheres with radii R1 and R2 (R1 < R2 ), respectively. These conducting spheres are places a long distance D apart as shown in Fig. After connecting these conduction spheres, calculate the normal component of E on the two spheres. Q2 q2 R1 R2 D { Because D À R1 + R2 , each sphere can be treated as isolated. potential on each sphere is then V1 = Q1 ; 4¼²0 R1 V2 = The Q2 4¼²0 R2 { If a wire is connected between the spheres, they are forced to be at the same potential V0 with redistribution of charges q1 and q2 : q1 q2 = 4¼²0 R1 4¼²0 R2 q1 + q2 = Q1 + Q2 V0 = Electromagnetics 43 { The lower equation comes from the charge conservation. Now we can find new charges q1 and q2 as q1 = R1 (Q1 + Q2 ) ; R1 + R2 q2 = Ex. Coaxial and Spherical shell capacitors ² Coaxial capacitor { Parallel connection R2 (Q1 + Q2 ) R1 + R2 C1 , C2 , and C can be found from C0 , the capacitance of air-filled coaxial capacitor of length L (already been studied): { The resulting system potential is found to be V0 = C1 = ²r1 C0 ; 2 C2 = ²r2 C0 2 2¼²0 L μ ¶ b ln a μ ¶ ²r1 + ²r2 C= C0 2 En = ½s =²0 at the surface implies that the surface electric field is stronger as the surface charge density is larger. a ϵ1 L b2 b1 aϵ 1 ϵ2 C0 = { Therefore even though the smaller sphere carries less total charge, it is found that E1 > E2 . L { Series connection q1 V0 E1 (r = R1 ) = = 4¼²0 R21 R1 q2 V0 E2 (r = R2 ) = = 4¼²0 R21 R2 1 1 1 = + C C1 C2 2¼²1 L C1 = μ ¶ ; b1 ln a { For this reason, E is always largest near corners and edges of equipotential surfaces, which is why sharp points must be avoided in high-voltage equipment. C2 = 2¼²2 L μ ¶ b2 ln b1 ² Spherical shell capacitor z PB { Parallel connection Circle ² When the electric field exceeds a critical amount Eb , called the breakdown strength, spark discharges occur as electrons are pulled out of the surrounding medium. Chung‐Ang University b C = C1 + C2 Q1 + Q2 4¼²0 (R1 + R2 ) ² Air has a breakdown strength of Eb ' 3x106 V/m. If the two spheres had the same radius of 1 cm, the breakdown strength is reached when V0 ' 30; 000 V. This corresponds to a total system charge of Q1 + Q2 = 6:7x10¡8 Coulombs. ϵ2 C PO μ C = C1 + C2 = ‐Q PA d C0 = μ s Infinite Conducting plate y x ²r1 + ²r2 2 ¶ b ϵ1 { Series connection Electromagnetics a C0 4¼²0 ¶ 1 1 ¡ a b 1 1 1 = + C C1 C2 4¼²1 ¶; C1 = μ 1 1 ¡ a b1 ϵ2 b2 a C2 = μ 4¼²2 ¶ 1 1 ¡ b1 b2 b1 ϵ1 44 ϵ2 Ex. Conducting cylinder in uniform E ² Initially exists in uncharged placed in x E0 a uniform electric field E0 = ^ an unbounded free space. If an conducting cylinder of radius a is this field, determine V and E. { Induced charge on the conducting cylinder due to the uniform electric field, generates scattered field Es . Therefore the total electric field can be expressed as ^ E0Á ^ E0½ + Á E0 = ½ Conducting cylinder E0½ = E0 cos Á; E0Á = ¡E0 sin Á Z x V0 = ¡ E ¢ dl = ¡E0 x = ¡E0 ½ cos Á a 0 y { The scattered field should be zero as ½ ! 1, therefore the ¡1 should be chosen for our solution as Ei = E0 0 a z Etotal = E0 + Es Vtotal = V0 + Vs Vi (r = a) + Vs (r = a) = 0 C1 ! ¡E0 a + =0 ! a Equipotential surface { Furthermore ©(Á) is known to be 2¼ -periodic and also even in Á, theerfore ©(Á) becomes m=0 m 6= 0 { Now the equation of R(½) becomes μ ¶ @ @2 ½2 2 + ½ ¡ m2 R(½) = 0 @½ @½ ( ln ½; m=0 R(½) = m ¡m ½ ; ½ ; m 6= 0 { We have to consider the expression of the original potential in the cylindrical coordinates. With a zero reference potential at x = 0, Chung‐Ang University cos Á ½ { Applying the boundary condition V(r = a) = 0 Electric field { Considering structure, Vs = R(½)©(Á), and the Laplace equation of scattered potential Vs in cylindrical coordinates, becomes · μ ¶ ¸ @ 1 @ 1 @2 ½ + 2 2 R(½)©(Á) = 0 ½ @½ @½ ½ @Á μ ¶ Ä ©(Á) ½ d @ ½ R(½) + =0 R(½) d½ @½ ©(Á) Ä ©(Á) = ¡m2 ©(Á) ( Á, or constant; ©(Á) = cos(mÁ); Vs (½; Á) = C1 x C1 = E0 a2 { Therefore Vs and the total V are found as E0 a2 cos Á ½ μ ¶ ½ a V = ¡E0 a cos Á ¡ a ½ Vs = { The electric fields can be found as μ ¶ @V ^ @V +Á E = ¡rV = ¡ ½^ @½ ½@Á μ ¶ μ ¶ 2 a a2 ^ = ½^E0 1 + 2 cos Á ¡ ÁE0 1 ¡ 2 sin Á ½ ½ As shown in Fig. XXX displaying the electric field and potential distributions, field values increases as Á moves to 0 and ¼ , and becomes maximum as (2E0 ) at ½ = a, two times of original field. { Now the surface induced charge on the cylindrical conductor can be found from the Gauss' law as ½s = D½ = ²0 E½ = 2²0 E0 cos Á Therefore the total charge induced can be found by surface integration, and is found to be zero. Electromagnetics 45 Ex. Conducting sphere in uniform E ² Initially exists in uncharged placed in z E0 a uniform electric field E0 = ^ an unbounded free space. If an conducting sphere of radius a is this field, determine V and E. { Consodering the scattered electric filed due to the induced charge on the conducting sphere, the total electric field and potential can be expressed as Conducting sphere { Two independent differential equations with a constant k · ¸ @ R(r) 1 @ r2 = k2 R(r) @ r @r · ¸ @ £(μ ) 1 1 @ sin μ = ¡k2 £(μ ) sin μ @μ @μ a x Ei = E0 { R(r), satisfying the following differential equation, becomes 0 dR(r) d2 R(r) + 2r ¡ k2 R(r) = 0 2 dr dr n(n ¡ 1) = k2 R(r) » r¡(n+1) ; n = 0; 1; 2; ¢ ¢ ¢ : z r2 Etotal = E0 + Es Vtotal = V0 + Vs { The expressions of the original electric field and potential in the spherical coordinates, which are needed later, are found as ( E0r = E0 cos μ ^ E0 = ^r E0r + μ E0μ ; E0μ = ¡E0 sin μ Z z V0 = ¡ E0 ¢ dl = ¡E0 z = ¡E0 r cos μ Electric field Equipotential surface x 0 z { £(μ) is known to satisfy the Legendre differential equation, · ¸ d d£(μ) sin μ + n(n + 1)£(μ) sin μ = 0 dμ dμ k2 = n(n + 1) and the solution of candidate, which is called the Lengendre function, is 0 n £(μ ) = Pn (cos μ) 0 1 1 cos μ 2 (3 cos2 μ ¡ 1)=2 3 (5 cos3 μ ¡ 3 cos μ)=2 Here, the potential at z = 0 is chosen as a reference zero potential. { Considering the axial symmetry of Vs , the related Laplace equation in the spherical coordinaes becomes μ ¶ μ ¶ @ 1 @ @ Vs 2 @ Vs r + sin μ =0 @r @r sin μ @μ @μ { Applying the separation of variables Vs (r; μ) = R(r) £(μ) · ¸ · ¸ @ £(μ) @ 1 @ 2 @ R(r) r £(μ) + sin μ R(r) = 0 @r @r sin μ @μ @μ · ¸ · ¸ 1 1 @ @ £(μ) 1 @ 2 @ R(r) r + sin μ =0 R(r) @ r @r £(μ) sin μ @μ @μ Chung‐Ang University { If n is a valid index, then a new n0 = ¡(n + 1) is alos valid as n0 (n0 + 1) = ¡(n + 1)(¡n) = n(n + 1) = k2 This implies that functions with index n and ¡(n valid. Therefore + 1) become Vs (r; μ ) = R(r) £(μ) 1 h i X = An r¡(n+1) + Bn rn Pn (cos μ ) n=0 Electromagnetics 46 { The scattered field, formed according to original, is known to be dependent of only the cos μ (orthogonality), that is, n = 1. Therefore the scattered potential and field components are £ ¤ Vs = A1 r + B1 r¡2 cos μ ! Vs = B1 r¡2 cos μ ¡@ Vs Esr = = 2B1 r¡3 cos μ @r ¡@ V s Esμ = = B1 r¡3 sin μ [email protected]μ { From the boundary condition V(r = a) = 0, V0 (r = a) + Vs (r = a) = 0 ¡2 E0 a cos μ = B1 a cos μ ! B1 = E0 a 3 { Therefore total electric potential and field components are · ³ a ´3 ¸ V = V0 + Vs = ¡E0 r 1 ¡ cos μ r · ³ a ´3 ¸ Er = E0r + Esr = E0 1 + 2 cos μ r · ³ a ´3 ¸ Eμ = E0μ + Esμ = ¡E0 1 ¡ sin μ r { Surface charge density induced on the conductor surface ½s = Dr (r = a) = ²0 Er (r = a) = 3²0 E0 cos μ { Total induced charge is obviously zero. ZZ Q= ½s a2 sin μ dμdÁ S Z ¼ 2 = 6¼²0 a E0 sin μ cos μ dμ 0 Z ¼ = 3¼²0 a2 E0 sin(2μ) dμ0 = 0 0 Chung‐Ang University Ex. Dielectric sphere in uniform E Dielectric sphere ² A dielectric sphere of radius a is placed in an initially formed uniform electric field E0 = ^ z E0 . Determine V and E. { Similar to the case of conducting sphere, the scattered electric potential and field components are ( Vso (r; μ) = B1 r¡2 cos μ Vsi (r; μ) = C1 r cos μ ( (Eso )r = 2B1 r¡3 cos μ (Esi )r = ¡C1 cos μ ( (Eso )μ = B1 r¡3 sin μ (Esi )μ = C1 sin μ clear all, close all, clc a=1; E0=1; dx=0.05; dz=0.05; x=‐2.4:dx:2.4; z=‐2:dz:2; [xx,zz]= meshgrid(x,z); rr=sqrt(xx.^2+zz.^2); th=acos(zz./rr); with the boundary conditions at r = a: a x Ei = E0 0 z Electric field Equipotential surface x ??? z 0 1) Vo = Vi 2) Do r = (Di )r V=‐E0*(1‐(a./rr).^3).*rr.*cos(th); [Ax,Az]=gradient(V,dx,dz); Ex=‐Ax; Ez=‐Az; (Eo )r = ²r (Ei )r ! { Therefore ii0=find(rr<=1); V(ii0)=0; Ex(ii0)=0; Ez(ii0)=0; 1) B1 = C1 a a2 ! C1 = B1 a3 8 ¡3 > <E0 + 2B1 a = ²r (E0 ¡ C1 ) μ ¶ 2) (E0 )r + (Eso )r = ²r [(E0 )r + (Esi )r ] ! 2 ²r > + 3 :! E0 (²r ¡ 1) = B1 a3 a · 3 ¸ 8 a (²r ¡ 1) > > <B1 = E0 2 + ²r μ ¶ ! > ² ¡1 > :C1 = E0 r 2 + ²r xth=(0:1:360)*pi/180; figure(1) pcolor(zz,xx,V), axis equal, shading interp, colorbar, hold on plot(cos(xth),sin(xth)) figure(2) quiver(zz,xx,Ez,Ex) axis equal, hold on plot(cos(xth),sin(xth)) { Final expression of electric potential is · μ ¶¸ 8 ²r ¡ 1 a3 > > r cos μ <Vo = ¡E0 1 ¡ 2 + ² r3 r μ ¶ > 3 > :Vi = ¡E0 r cos μ 2 + ²r Electromagnetics 47 { Electric field can be found from the obtained coeffcients and combing the components. { However, it can also be found as E = ¡rV Applications of Image Theory y ² Capacitance of two conducting cylinders with ( a = radius d = center to center distance ‐V0 d a 0 x V0 Outside the sphere μ ¶ @ Vo ^ @ Vo +μ Eo = ¡ ^r @r [email protected]μ · μ ¶¸ 2(²r ¡ 1) a3 rE0 1 + =^ cos μ 2 + ²r r3 · μ ¶¸ ² r ¡ 1 a3 ^ + μ E0 1 ¡ sin μ 2 + ² r r3 Inside the sphere μ ¶ @ Vi ^ @ Vi Ei = ¡ ^r +μ @r [email protected]μ μ ¶ 3 = ^rE0 cos μ 2 + ²r μ ¶ 3 ^ E0 +μ sin μ 2 + ²r Here we can find the Ei has only the z-component and uniform as μ ¶ 3 Ei = ^ z E0 2 + ²r { Since this is equivalent to a conducting cylinder above a grounded conducting surface. Let V0 be the potential on the conductinf cylinder y d/2 V0 { Our concern is if there is an equivalent line charge. If it exists, then the problem is solved by considering two oposite line charges in free space (having line charge densities ½L and ¡½L ). y E½ = ½L 2¼²0 ½ V1 (½) = ¡ x0 ρL V0 { E and V1 of a single line charge at ½ = 0 ½L ln(½) + Vref 2¼²0 ‐ρL x0 0 ρL V(x = 0) = 0 (* ½1 = ½2 ) { Now let us check if these line charges can generate the potential V0 on the conducting cylinder surface. This can be written with introducing a variable R as s ½2 (x + x0 )2 + y2 = (> 1 ) R= ½1 (x ¡ x0 )2 + y2 μ ¶ ½L 2¼²0 V0 ln R = V0 ! R = exp 2¼²0 ½L Chung‐Ang University Electromagnetics x y x0 { Potential for two opposite line charges at x = x0 and ¡x0 μ ¶ ½L ½2 V(½) = ln 2¼²0 ½1 ( p ½1 = (x ¡ x0 )2 + y2 p ½2 = (x + x0 )2 + y2 This can be verified from the electric field and potential distributions in Fig. XXX. x 48 x ² Capacitance of two conducting spheres Two identical conducting spheres havong radius a are separated with a center to center spacing 2c. Calculate the capacitance. { Therefore μ 2 R +1 R2 ¡ 1 x2 ¡ 2 ¶ x0 x + +y2 + x20 = 0 2 (x ¡ xc ) + y2 = a2c : equation of circle μ 2 ¶ 8 R + 1 > > x0 <xc = 2 μR ¡ 1¶ 2R > > :ac = x0 2 R ¡1 8 2 2 2 > <xc ¡ ac = x0 ¢ ¢ ¢ (1) ! x R2 + 1 > : c = ¢ ¢ ¢ (2) ac 2R ¡ From this, we can determine 8 d > > <xc = 2 ac = a > > : V0 $ C= : ( x0 ½L Charge per unit length Q ½ ¼²0 = L = = 2V0 2V0 ln R ¼²0 μ cosh¡1 d 2a ¶ s R= ² At first, let us consider a positive point charge Q located at x = d and a conducting sphere of a radius a centered at the origin (a < d) having a potential ¡V0 . { The image charge Qi (negative point charge) is expected to exist inside the sphere at a distance x = di with: { Potential on the surface of the sphere caused by these two chages μ ¶ 1 Q Qi + =0 V= 4¼²0 ro ri ! sμ ¶ ¶ 2 d d ¡1 ¡ 1 ; x0 = a x=a 2a 2a sμ ¶ 2 d d R= + ¡1 2a 2a μ Chung‐Ang University the conductor z 0 a Q d z c { Thetrefore there are two unknowns di and Qi (This is different to the case of two opposite line charges in cylindrical coordinates). With (En )on ‐V0 ro a ri Qi ‐z0 di d a ro ‐V0 Qi 0 Qa Q z Q z d Qi ri = ¡ = R (constant); 0 < R < 1 ro Q ( p ro = x2 + y2 + (z ¡ d)2 p ri = x2 + y2 + (z ¡ di )2 { Equation of zero-potential surface s x2 + y2 + (z ¡ di )2 ri = =R r0 x2 + y2 + (z ¡ d)2 = Electromagnetics z c V0 c Qi 6= ¡Q (x + x0 )2 (x ¡ x0 )2 ¡ Charge density on the conducting cylinder a V0 a a c 0 because ¡Q and the original Q can not make the spherical surface with a radius a of a zero-potential (but a planar surface). ¡ Capacitance per unit length of two conducting cylinders Q = ½L ‐V0 49 { Step 3: In order to restore VS charge of ¡Q0 , is put at z = z1 . x2 + y2 + (z ¡ z0 )2 = a2 : equation of sphere μ ¶2 di ¡ R2 d R(d ¡ di ) 2 z0 = ; a = 1 ¡ R2 1 ¡ R2 di z0 = 0 ! R2 = a 2 = di d ! d Q1 = Q0 °; z1 = c ¡ di1 ; di1 = { Now the image charge can be determined as ¡Qi = R ! Qi = ¡RQ Q di di d a2 R2 = = 2 = 2 ! d d d ³a´ ! Qi == ¡ Q d VS = V0 ; { Step 4: a R= d at z = 0 z2 = c ¡ di2 ; Electromagnetic Fields and Waves, Lorraine, Cordon, 2nd ed. di2 = d0 { Step 1: A charge Q0 is set at z = z0 = c (center of the sphere) to provide a potential V0 on the surface S. Q0 = 4¼²0 a V0 ; VS = V0 ; @ z = z0 = c but non-euipotential VS on S VS = V0 ; { Step 6: VSg 6= 0 μ but ¶2 z2 d1 = 1 ¡ °2 d0 di2 = a d0 = di1 = °2 d0 d1 di1 =1¡ ! d0 d0 μ ¶ 1 2 Q 2 = Q0 ° 1 ¡ °2 a2 d ² Next a Conducting Sphere above a Ground Plane is considered. VSg 6= 0 d1 = d0 ¡ di1 ; at z = d at di = di1 = °2 d0 { Step 5: Putting a charge Q2 (image charge of ¡Q1 ) at z restores the equipotemtial on S. μ ¶ μ ¶ a a d0 d0 = Q1 = Q0 ° 2 Q2 = Q1 d1 d0 d1 d1 ² Is it possible to make non-zero potential ¡V0 on the conducting sphere ? Sure, it can be made by putting an additional point charge Qa at the center of conducting sphere. 8 Q > > > ³a´ < Qi = ¡Q d > > > : Qa = ¡4¼²0 a2 V0 a2 ; d0 d0 = 2c Putting a charge ¡Q1 at z = ¡z1 yields VSg = 0; { Now the V and E external to the grounded sphere can be calculated by the two point charges Q and ¡(a=d) Q in free space. but a < 1; d0 °= where V0 , a charge Q1 , the image = d0 d1 a2 d1 ! di2 °2 = d0 1 ¡ °2 QSg 6= 0 Putting a charge ¡Q2 at z = ¡z2 VSg = 0; but non-equipotention VS on S { Step 2: Putting an image charge ¡Q0 at z = ¡z0 yields VSg = 0; but non-euipotential VS on S Chung‐Ang University Electromagnetics 50 { Step 7: Putting a charge Q3 (image charge of ¡Q2 ) at z = z3 μ ¶ μ ¶μ ¶ μ ¶μ ¶ a a d0 °3 d0 Q3 = Q2 = Q2 = Q0 2 d2 d0 d2 1¡° d2 d2 = d0 ¡ di2 ‐V0 di4 = d0 VS = V0 ; { Step 8: but { Step 9: 0 z c ¶ ‐V0 S ‐Q0 ‐Q1 QSg 6= 0 Sg 0 di1 Q2Q1Q0 z z2 z1 di2 c μ Q5 = Q4 di3 d3 1 ¡ °2 =1¡ = 1 ¡ °2 d0 d0 1 ¡ 2° 2 1 ¡ 3° 2 + ° 4 d3 ! = d 1 ¡ 2° 2 μ0 ¶ 1 4 Q4 = Q0 ° 1 ¡ 3° 2 + ° 4 a2 z4 = c ¡ di4 ; di4 = d3 a d4 ¶ μ = Q4 a d0 ¶μ d0 d4 ¶ μ = Q0 ° 5 1 1 ¡ 3° 2 + ° 4 ¶μ d0 d4 ¶ d4 = d0 ¡ di4 Putting a charge Q4 (image charge of ¡Q3 ) at z = z4 μ ¶ μ ¶μ ¶ μ ¶μ ¶ a a d0 1 d0 Q4 = Q3 = Q3 = Q0 ° 4 d3 d0 d3 1 ¡ 2° 2 d3 μ but non-equipotention VS on S { Step 11: Putting a charge Q5 (image charge of ¡Q4 ) at z = z5 μ ¶ di4 d4 1 ¡ 2° 2 =1¡ = 1 ¡ °2 d0 d0 1 ¡ 3° 2 + ° 4 1 ¡ 4° 2 + 3° 4 d4 ! = d0 1 ¡ 3° 2 + ° 4 μ ¶ 1 5 Q5 = Q0 ° 1 ¡ 4° 2 + 3° 4 but non-equipotention VS on S Chung‐Ang University c 1 ¡ 2° 2 1 ¡ 3° 2 + ° 4 VSg = 0; QSg 6= 0 d3 = d0 ¡ di3 a V0 a { Step 10: Putting a charge ¡ Q4 at z = ¡z4 Putting a charge ¡Q3 at z = ¡z3 VSg = 0; but d0 d3 μ ¶ c a2 z3 = c ¡ di3 ; di3 = d2 μ ¶2 μ ¶ di3 a d0 = d0 d0 d2 μ ¶ di3 1 ¡ °2 2 =° ! d0 1 ¡ 2° 2 VS = V0 ; a d0 ¶2 μ di4 = °2 d0 ! di2 d2 °2 = 1¡ =1¡ d0 d0 1 ¡ °2 1 ¡ 2° 2 d2 ! = d0 1 ¡ °2 μ ¶ 1 3 Q3 = Q0 ° 1 ¡ 2° 2 μ ² Repeating this process to get the limit of potential on S and Sg . ² Total induced charge on S = the total point charges Q = Q0 + Q1 + Q2 + Q3 + Q4 + ¢ ¢ ¢ = Q0 G · ¸ °2 °3 °4 + + + ¢¢¢ G= 1+°+ 1 ¡ °2 1 ¡ 2° 2 1 ¡ 3° 2 + ° 4 ¶ ² Capacitances of two cases Q Q0 G = S = 4¼²0 aG V0 V0 Q C0 = C= 2V0 2 Conducting sphere and ground plane: C0 = Two conducting spheres: Electromagnetics 51 Convolution Integral in Time Domain ² Impulse function (or Dirac delta function) ( 0; for t = 6 0 ± (t) = 1; for t = 0 Z 1 ± (t) dt = 1 Convolution Integral in Space Domain (t) h((t) (t) Integral = 1 ² Impulse function in the space domain 1 ± (r) = ± (x) ± (y) ± (z) Z ± (r) dv = 1 h(t) 0 V t ¡1 ² Approximate impulse function ( 0; for jtj > ¢=2 ±¢ (t) = 1=¢; for jtj < ¢=2 (t) 1/ Area = 1 (Integral = 1) ‐/2 /2 ² Output y(t) for an arbitrary input x(t) in an LTI (Linear Time Invariant) ststem Input ± (t) ! Output h(t) Input x(t) ! Output y (t) = ??? where h(t) is called the impulse response. Then y(t) becomes a convolution: Z 1 x(t0 ) h(t ¡ t0 ) dt0 y(t) = x(t) ¤ h(t) = ¡1 (t) h(t) 1/ t ² If we introduce Gv (r) to denote an impulse response of electyric potential (Green's function) then total electric potential for the source charge distribution can be expressed by Space-domain convolution representation as Z V(r) = Gv (r) ¤ ½(r) = Gv (jr ¡ r0 j) ½(r0 ) dv0 v0 1 Gv (r) = 4¼²0 jrj h(t) ‐/2 /2 x’=0 |X|=|x‐x’|=|x| E0 qs 0 Source point x Field point x’≠0 |X|=|x‐x′| qs 0 x Field point x’ Source point Discrete charge distribution |Xn|=|x‐xn’| qs1 qs2 t E1 ² Appropriate expression of a point source qs at the origin qsn 0 En qsN x xn’ qs ! ½(r) = qs ± (r) x(t) x(t) x(n) (t‐n) Continuous charge distribution ρL(x) |X|=|x‐x’| Proof dE { Approximate expression of x(t) and y(t) X x¢ (t) = x(n¢) ±¢ (t ¡ n¢) ¢ 0 0 n t x’ Source point x Field point n X y¢ (t) = x(n¢) h¢ (t ¡ n¢) ¢ n { Taking a limit with ¢ ! 0 with the corresponding variable parameters: n¢ = t0 ; ¢ = (n + 1)¢ ¡ n¢ ! dt0 Z y(t) = lim y¢ (t) = x(t0 )h(t ¡ t0 )dt0 Time domain Input Output h(t) δ(t) LTI system h(t) x(t) y(t) y(t)=x(t)∗h(t) Space domain Input δ(r) ρ(r) LSI system Gv(r) Output GV(r) V(r) V(r)=GV(r)∗ρ(r) ¢!0 Chung‐Ang University Electromagnetics 52 Integral of sec(x) Expressions of Some Derivatives ² For a sec x ² For ln P with Z Z 1 dx cos x Z Z cos x cos x dx = dx = 2 cos x 1 ¡ sin2 x Z cos x = dx (1 ¡ sin x)(1 + sin x) N D N = R1 + z + L=2; P= sec x dx = I= Derivatives of ln P with respect to ½ and z N0½ D0½ @ ln P 1 @P = = ¡ @½ P @½ N D 0 @ ln P 1 @P N D0 = = z¡ z @z P @z N D With u = sin x, du = cos x dx Z 1 I= du (1 + u)(1 ¡ u) ¶ Z μ 1 1 1 = + du 2 1+u 1¡u 1 = [ln(1 + u) ¡ ln(1 ¡ u)] + C 2 · μ ¶¸ 1 1+u = ln +C 2 1¡u where the subscripts ½ and z stand for taking a derivative with respect to ½ and z, respectively. The related derivatives are ½ ½ ; D0½ = R1 R2 R + z + L = 2 R 1 2 + z ¡ L=2 N0z = ; D0z = R1 R2 · @ ln Q 1 =½ @½ R1 (z + L=2 + R1 ) ¸ 1 ¡ R2 (z ¡ L=2 + R2 ) ½ 1 ½2 = ½ R21 [1 + (z + L=2)=R1 ] ¾ ½2 ¡ 2 R [1 + (z ¡ L=2)=R2 ] μ2 ¶ 1 cos2 ¯1 cos2 ¯2 = ¡ ½ 1 + sin ¯1 1 + sin ¯2 1 = (sin ¯2 ¡ sin ¯1 )) ½ 1 @ ln Q 1 = ¡ @z R1 R2 N0½ = Replacing u with tt sin x (1 + sin x)2 1+u 1 + sin x = = 1¡u 1 ¡ sin x (1 ¡ sin x)(1 + sin x) μ ¶2 1 + sin x 2 = = (sec x + tan x) cos x Therefore 1 2 ln (sec x + tan x) + C 2 = ln (sec x + tan x) + C I= Summarizing Z sec x dx = ln (sec x + tan x) + C Chung‐Ang University D = R2 + z ¡ L=2 Electromagnetics 53