Electromagnetic Fields ECE 340
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Electrom agnetics Introduction Coordinates Electrostatics Coulomb s law Gauss law Divergence theorem Point form of Gauss law Potential energy Potential Line integrals Conservative and nonconservative fields Stoke s theorem Poisson s and Laplace s equations Resistance Boundary conditions Inhomogeneous resistors Permittivity Capacitance Inhomogeneous capacitors Force Links to additional applications Actuation and sensing Electromagnetic Fields 24 26 27 30 37 41 43 47 54 59 61 70 74 77 84 86 87 Charge vs mass Cumpare the repmsmh enhe Cemerhb fume e e39 between WEI e1emrunstuthe1r grav1tat1una1 attractmm 1euhe separatmn be 1 m 4 1 m gt F 66731EIquot ms L m lmj kg 5541EIquot N F 1 611ch 1 W N we 8541EquotFm1m 4151u 1 because muS L mater1a1s are e1emr1ca11y a1rhes1 perfemy heutra1 1h fact eharge neutrahty 1s f1e1dsr accuunted fur spee1a1 re1at1v1ty1 Chargedpeope 1mm averagesued peep1e were separated by 1 ms eaeh havmg 1us1 1a11ma1111 hhhau Ves 1a11 a a 1 rh1e7 Ves Greaterthan the Umted States eeWh re a depth er 1n mHes Ves In fact the force would be greater than the weight of the entire eerth 1mm 11 PV them frum nHapsmg eh themseNes Muvmg up frum 1nd1v1dua1 alums hm an em a11ufchem1s uy 151112 mteramun ufurb1ta1e1emruns 1r2 shener range These We farceswurktugethertu pruduce stame hue1e1 waves hhmh eah actua11ytrave1 thruugh empty space Askthat er heat suund er any uthertype ermeehah1ea1ehergy1 E1a mmagnet1l rems Nuclear en 9 er y Censmer a U435 alum undergumg ssmn Fissmn yie ds An examp e of one at the many lvagmenls m reacnans in me uranium235 cess 235 ssion pro unseme assumes What sthe suurce enne Energy Answer arge y ne ectrus tatms What sthe putarma Energy er a Earmm AN 5B alum separated by 1 fmfrum a Kryptun AN SB alum7 PE D SEEjD1EIquot9 c 4KE1D m F39E u 4B1EIquot JZB1D EVZZBDMEV WSWEBIE SFUW Cl 3B5B16 mquot5cv ans Energy 5 m the be park We dwameter used after au sjus t a reasename reugn a H Hux m v r nudear Energy m nssmn s seen m be arge y due m the Cemernb fumeia Cuu umb spnng 9 mm m new them tugether aemneenene rems Electromagnetic waves quote There are two keys to electromagnetic 5 wave propagation The first is that a timevarying magnetic field produces an electric field The second is that a timevarying electric field produces a magnetic field These two effects bootstrap themselves to produce a traveling electromagnetic wave The study of electromagnetics provides the foundation for photonics wireless antennas electrical power microwaves amp RF and highspeed circuits The enormous topic of lumped element circuit analysis lumped elements KVL and KCL is an approximation to the electromagnetic equations the quasistatic approximation The history of electromagnetics is a rich one Following are a few random notes 0 Electron comes from the Greek for amber Greek word for amber nhsmpov lf amber is rubbed with a cloth or with fur it aquires an electrical charge 0 An old story probably apocryphal goes that a Greek shepard Magnus noticed the iron nails in his sandles were attracted by some black stones later called loadstones 0 Early electrical workers learned to store charge in Leyden jars in which the electrical fluidquot was thought to condense 0 To obtain larger stores of charge these Leyden jars were often arranged in rows which Ben Franklin thought looks like batteries batteries of canon that is o Volta used stacks of dissimilar metals separated by a conductiving fluid to produce the first voltaic pilequot what we d today refer to as a battery 0 Early workers notices that upon lightning striking a house which took a course through the cupboards many knives and forks were melted but others were found to be magnetized o In the 1860 s James Clerk Maxwell presented his electromagnetic theory 0 In the 1880 s Heinrich Hertz confirmed the existance of the EM waves The classic history of electromagnetism is a superb work by Edmund Whittaker A history of the theories of Aether amp Electricity which takes the topic to 1926 Selected online sources httpWWW ahrln ar 39 39 39 39 mGhistemndf httn39lhistnrv hvnerieff 39 1 1 httpWWW 39 quot hi Historvhtm Electromagnetic Fields 4 Electromagnetics is fundamental to many areas of science and technology It is a foundational topic in electrical engineering and provides a basis for advanced practice Knowledge of electromagnetics will maintain its utility even as technology constantly changes indeed its importance will likely grow Electromagnetics plays a role in most technologies in electrical engineering and physics such as semiconductor devices LEDs diode lasers transistors diodes etc optics and optoelectronics highspeed electronic systems electrical machines and power antennas and wireless sensors resistive magnetic capacitive optical electromechanical systems MEMS sensors actuators switches NWQH P NT Increasing frequencies used in highspeed design has made it critical that engineers in circuit design and layout develop an understanding of electromagnetics EM At higher frequencies area ll capacitance and connection inductance can no longer be ignored traces become transmission lines conductors become effective antennas At higher frequencies simple lumpedelement models become inadequate and electromagnetics is necessary to simply understand circuit behavior Many of these effects become evident when system dimensions are comparable to signal wavelength Consider the table and graphic below showing wavelengths for an EM wave traveling at a speed of 3108 ms as the frequency is varied frequency vs wavelength 3 Hz a 0 m 300 MHz gt 1 m 3GHz 01m10cm 30 GHz 1 cm At 3 MHz most systems are much smaller than the wavelength At gigahertz frequencies this condition o en no longer holds and one mus resort to electromagnetic fundamentals to understand system behavior Our study will de ne relationships quotquotquot between the eld sources charges and currents and the resulting electric and 239 so 5quot Copyright 2005 Pearsun Prentice Hall Inc Electromagnetic Fields 5 rnagnetrc ners FundarnentaT ouantrtres We energy torce yoTtage and current and tnerr T e reTatron of consequences of Wave oropagatron forumervarymg ners Tne rdea of a electromagnetic field oegan wrtn MrcnaeT Faraday rn tne eany 1800 s Tn ensrty HH nn a assocwated WM he d strength WHTCH together WW1 thew dwrecuon consumes a vector ner Tne eTectnc and rnagnetrcnexds vectonva uedfuncuons of oosrtron and trrne are produced by cnarges and currents Vector notation t r F Fa magmtude F n tne drrectron ortne umtvectorar m mass and VTdEDs vectors wm oe underhned and umtvecturs are denoted by a carrot e 3 Remember to use yector notatron propeny m 5 your Workivym neTo ootn your understandrng and your grade Some necessary rnatnernatrcaT rundarnentaTs rncTude 1 coordrnate systerns rectanguTar cynndncat sonencaT 2 yector operatrons gradrent dryergence cur 3 yector caTcuTus hne rntegraTs urrace rntegraTs ETectrornaonetrc rods 5 Rectangular Cartesian coordinates x distance along x axis or distance from yz plane y distance along yaxis or distance from zx plane 2 distance along zaxis or distance from x y plane position vector rXaXyayZaz differential lengths dx dy dz vector differential length dl dx ax dy ay dz aZ differential areas ds dxdy dydz dzdx ds idxdy a1 ldde ax idde ay differential volume dxdydz variable range 00 to 00 for x 00 to 00 for y 00 to 00 for z z axis ds dxdy a x axis Rectangular coordinates have some attractive properties and are often the standard coordinates used when symmetry considerations do not urge the use of another system 1 The directions of the unit vectors are constant and not functions of position 2 The differential elements are not functions of the coordinates 3 The unit vectors are mutually orthogonal aX ay ay aZ aZ aX 0 Electromagnetic Fields Cylindrical coordinates e dws tance frum 173st ang e frum gtltragtltws m determme sense ahng the thumb ufyuur ngm hand m a ung me zrast mreeuem elf pusmve mtatmn s a ung yuur ngers dws tance a ung 173st Dr dws tance hum gtltry p ane 47 pusmun vemur r a z a W ve ur r4 r4 m we yhndnca cuurdmate syslem mm a and a are functmns elf pesmem uwerenuax Engths e e vemur uwerenuax wean an dp a em a dz a uwerenuax areas a s pdpd 4x dpdz pew d5 e em a men a team a uwerenuax vu ume u d dz vaname range m m mfur p u m 2mm rmtu mfur z z axls ds Dde az x axis a p Lymuneax mammals are user m sys tems mm yhndrma symmetry 1 The umt vemurs m we yhndrma uurdmate sysiem are mutuaHy unhaguna a a a aaa mremem The mremem fur a and a depend un pesmem E simmanent rem Spherical Coordinates r 51 ang e farmed stamng hum pusmve 173st and muvmg m pusmun vectur 6 hand m a ung the zrast mremun uf pusmve rutatmn quuvvs yuur ngers pusmun r r a dxffererma Engths an r swam r as vemur awarenuax Ength dl m a r sme my a r as a dxffererma areas d rsmedwn rzsmeddme rdedr r sme my dr a r sme dipdeah r d6 dr a d5 dwffererma vmume r1 sme dr my d6 vaname range ummmm am More um 2mm 1 axls r as r sino m axis Spherma uurdmates are usem m sysiems thh spherma symmetry 1 curve ufthe umtvemurs have a cunstant mam The mam uf Each s a fummun uf pusmun 2 The umt vemurs are mutuaHy unhaguna a a a a a u E simmanent Fwe ds Electrostatics g In the 18th century CharlesAugustin de Coulomb found the force between two charges acts on a line connecting them that it is proportional to the product of their charges and that it is inversely proportional to the distance between them The force is attractive if the charges are of opposite sign and repulsive for like signs The relation is referred to as Coulomb s law F 0C qtqz aR force on q2 due to q IRIZ In the MKS system F q1q2 2 aR where s is the 47139 SR permittivity of the material between q1 and q2 X The permittivity of vacuum or free space so 8854103912 Fm Coulomb s law is linear with respect to sources so that superposition holds Coulomb s law and superposition an example Calculate the force that n charges q1 through qn at positions r1 through rquot exert on charge q at r F 47321 Ri r ri vector from position of i charge ri to charge q at r Ri lRil distance from im charge to charge q ai RiRi unit vector from im charge to charge q Electrostatic fields g The electrostatic field E is the force per unit charge on a test charge q as the charge of the test charge goes to zero The reason the electric field is defined as a limit is that a finite charge would carry its own field which would affect the very field being characterized by measuring the force on q That is if q were finite its field would affect the F being measured Defining the electric field via an infinitesimal test charge avoids this difficulty Electromagnetic Fields 10 Electric fields and superposition g From the force exerted on q by one charge q q39q F 4778R2 aR The corresponding electric field is E lim5 q 2 aR HO q 4778R R r r39 vector from position q39 at r39 to charge q at r R R distance from q39 to q aR RR unit vectorfrom q39 to q The force exerted on q by a collection of five charges was found above to be quot qrq F a 4778Ri2 I The corresponding electric field is therefore ElimEZ qi 2ari qaoq i147reRi Continuous charge distributions g The results from the preceding example can be extended to include the field from continuous charge distributions rather than from discrete charges dq39 E a J47Z8R2 R q For volume charge distributions dq pvdv PV E 39 m m aR charge dq39 deV39 at r39 infinitesimal source r39 position of source charge dq39source point r position of test chargefied point R r r39 vectorfrom source to field point aR RR unit vector from source to field point Similarly for surface charge distributions E ds39 aR 7Z8 charge And for line charge distributions E I dl39 aR 7Z8 charge Electromagnetic Fields 11 Charge distributions Charge always exists in discreet chunks the magnitude of the charge of an individual electron is 16103919 Coulomb Since charge occurs in discreet quantities charge can never be truly distributed There is always granularity Uniform distributions are useful approximations when the scale of interest is much larger than the scale in which granularity is evident These approximations are used constantly in macroscopic electromagnetics and it is well to discuss them at this time pV Volume charge distributions Cm3 are used for charge distributed over a volume ps Surface charge distributions Cm2 are used to indicate charge is distributed over a given surface This is itself an approximation to what is actually a volume charge density where the charge is distributed in a region very close to some surface Considera charged electrical conductor in which its excess charge is confined to a region very close to the surface of the conductor In this case it is reasonable to describe the charge distribution by a surface charge distribution in Cm2 rather than describe the distribution as a volume charge distribution pL Likewise line charge distributions Cm are used for charge distributed over a length as for example in a charged wire Example infinite line charge g Find the electric field due to a line charge along the z axis Len R chige l ng r z az rpapzaz Rr r papz z az aRRlR pL Pap Z Z39az H Pzt 2392 p2 zz392 E w A pa zz39a dz39 J 2i P 2 2 pL E J 4nsR2daRZL4ng charge Since the limits are from 2 to any finite value ofz does not affect the result E w 3 pap z39azdz39 2 4nsp2 2392E Electromagnetic Fields 12 Since the limits along 2 are symmetric the az component is zero after integration m z M u E papdz p dz 2 47rsp2 23912 L pLPap m dz39 3 4778 2L p2 22 E PLPE p 2 z39 x p zm 4715 lza p2 p field point r 1 La 47139s p 27rsp Example infinite sheet of charge g Consider an infinite plane of charge over the z 0 plane 0 E 5 dsa hf 47r8R2 R carge Forthiscase r x axy ay rxaxyayzaz Rr r x x axy y ay zaz aRRR Useful integrals dz z zdz 1 1 1 z t Iz2 a2 32 22 32 Iz2 a2g Z2 a2 122 612 8 an a E w w psxx39aX yy39ay zz39azdx39dy39 1 i I psz azdx39dy39 7 32 7 32 4 ltxxr WY ltzz39gt2 4 svwxwltxx39gt2 yyr ltzgtzi E pszaz dX39dy39 9 3 X39X39 dy 32 7 4718 mxmxx392 y y392 z 4718 vw v YY 22 gtltgtlt392yy392z2 W m w I w giaz z gt O E pszaz J39 3922 2 dylpszaz dzy 2 pszaz1tan391 8 47128 ywyyv 2 1 27128 ywyyv 2 1 27128 2 z ym pS a 2 lt0 28 z Electromagnetic Fields 13 Gauss law and electric flux To eahy researchers acuonratrardwstance re auons such as Newton s avy of grav tauon ahd Cou omb s avy smacked of mySUC Sm or mag c They sought an exp anauon that h m th sought mechamca modem to deschoe ahd hmk about EM e ds h the 19 quot Century chhaew Faraday proposed that charges and currents create h pamcu ar for e ecmc hewds Faraday pwctured ux hhes erhahathg trorh e ecmc charge Gauss I we q T V V fV MA Where wttsthetotaw 4 j ecjncm A lt7 e ecmc uxcommg out A m I trorhchargeq 5 J 1 K the H Dc39ver Usmgthe ux denSW eauss avy reads H D Dds o sum surface that separates hsrde trorh outswde a boundary wrth ho ho es h ms re auony D s the e ecmc ux dehsrty yector m cm2 The dwfferenua surface exemeht ds rs 5 50 a yector quanmy shoe Gauss avy equates outward ux to charge the drectroh of ds Shomd be taken as outward trorh the surface f the mega rs to be equa to the tota e ecmc ux corhhg out trorh the charge Surface integrals shoe e ectromagneucs hyohes rhahy surface mtegra s suc we 1 D EIdSr 5m et s take some the to srrhoxy exp ore surface mtegra s h more detah Frst note that a yector dwfferenua area aways has two posswb e drectrohs for us umt horrhax For examp e dxdy a or dxdy ea E ectvumagneuc was 14 To find the flux traveling in the direction indicated in the above diagram the direction for vector differential area is chosen in the same direction For example to find the flux leaving a volume an outward directed differential area leaving the volume would be used Example Cartesian g Given D 2xy aX 3x2 ay Cm2 find 1 passing through the surface y O 0 g x g 1 m 0 Dz 1 m in the ay direction Here since the flux desired is in the ay direction ds is chosen to be dxdz ay 4D ds dxdz ay j ZX ylyoaX 3x2 ay Ddxdz ay 0 20x 2 1 2 1 3 L i 075 C 2 0 2 0 Example spherical surface g Given D 5r2 ar Cm2 find the electric flux passing outward through the surface of a sphere centered at the origin with a radius 15 m ddrarrdeaersined a The differential length is a good place to start r is constant on the surface of a sphere centered at the origin so that ds r d6 r sin 6 d4 ar r2 sin 6 d6d ar 7 27 5 V I I e o o 5 cos Glgo l o 5471 C ar Dr2 r15m sin 6 d6d ar r15m Electromagnetic Fields 15 Example spherical surface g Given D 5r2 sin 6 ar Cm2 find the electric flux passing outward through the surface of a sphere centered at the origin with a radius 15 m 7 27 5 V l l 2 eo o 5 aglj o sin2 ede 52739r jde 90 60 sin 9 ar D r2 sin e d6 d4 ar r15m r15m 5712C Example cylindrical surface g Given D 5p ap Cm2 find the total electric flux passing outward through a closed cylindrical surface with its axis on the zaxis with radius 4 cm extending from 3 cm 2 g 2 cm Step 1 determine differential areas directed outward from surface sides ds plpoo4m dlt1dzap OO4dlt1dzap top ds pd dpaZ bottom ds pd dpaz Step 2 form dot product D ds on each surface sides D Dds E ap DOO4dltldzap 5d dz p p004m top and bottom D Dds 0 Step 3 determine limits and perform the integral 111 Q D gds su rface answer 1 0511 C Numeric Approximation One classic means of estimating the area of a definite integral is to express the integral as a Riemann sum and approximate the area under the curve as the sum of the rectangular areas of constant width and whose heights are determined by the function39s value One possibility in determining rectangle heights would be to use the value of the function at the initial point in the interval as the height of the rectangle This would lead to the formation of a left Riemann sum Another possibility would be to use the value at the end point for a right Riemann sum and another would be to use the value at the midpoint for a middle Riemann sum Here we will use a middle Riemann sum Electromagnetic Fields 16 frsi ine seeeriu usesine eriu pEImL arid the third uses the vaiue at the miupeirii enne iriieivai Taking the vaiue aune miupeirii y uf ari iriieivai resuiis in a rubust V appruximatmn that is mere quotn i accurate furmus i curves x x Numerical approximation nite differences g Given D 2x ax My ax Crn i fin the appruximate eieeirie ux w passing inreugn ine surface 1 mi n smxsi rni U SE ysi rn in me ax dirE iDn Use a i my areas it may be rnere siraigncrewiaru in farm ine surn frum ine iniegiai 2xa3x yanuxuya1 i My dxdy i i2gtlt ax3gtltuyaIJEdgtltdyax iiam 1y a ma a Me n u5u iu i n ii ii Vnx Using a cumputertu eaieuiaie ine surn here Mape is gt ux used sumsum3 WUVEIEIEVSWEIWirEIEISVEH EILiZW iuii in izsuuuu Example electric flux density from charge q anne origin ii in i VhPfiPW miilt7 iiiu u iiu ii i u i we eiemne ux density vemurtu be in me ar dUEEUEIn D D a D not a funnieri uf 1 er s Given this a surface a Gaussian surface is enesen se inai ine Gaussian iniegrai can be evaiuaieu Here we Gaussian surface is a SphErE centered en ine erigin Wigwam a m um I Daurisineuauea um pin r Dr n sin edad Dr Mrcusetq xDr in gt n n Eia mmagnetil rieius Example spherically symmetric distribution g Consider a volume charge density where a charge q is evenly distributed over a sphere of radius a Given this charge the field has spherical symmetry again meaning that the magnitude of the electric flux density vector must be independent of 4 or e and can only depend on r with its direction in the ar direction D D ar 2 q radius a D is not a function of 4 or e v x 39 The Gaussian surface is a sphere centered on the origin The difference here is that there are two cases one for rlta and the other for rgta rlta m DDdsq surface I 2f D ar Dr2 sin 6d6d ar pV dv 2f I j 4 q r392sin639 dr39dG39dcb39 volume 39 eo o oroina3 3 7t 27 Drzj Isin6d6d 4nDr2 4q r3 3 e0 0 na 3 q qr 4an a a 471233 a o i TDar m2 sin eded a mpvdv i4q r392sin639dr39d639d4 eo o volume eoroina3 7t 27 Drzj Isin6d6d 4nDr2 4q a3 3 e0 0 7713 q a 47W2 Notice the only difference in the mathematical procedure between the two cases is in the r limit in the volume integral which changes from a forthe rlta case to r rgta case This one change results in marked differences in field behavior Electromagnetic Fields 18 Electric flux density The electric flux density vector is related to the electric field via permittivity The electric field due to a point charge q at the origin force per unit charge from Coulomb s law q E 4ns al39 The electric flux density due to a point charge q at the origin from Gauss law q a D 4n r Therefore D 8E This relation is general for linear homogenous materials not just for point charges at the origin more about permittivity later Example line of charge g Use Gauss law to determine the field due an infinite line of charge pL on the zaxis Due to the clear symmetry involved cylindrical coordinates will be used The fact that the charge is an infinite line charge along the zaxis permits the form of the field to be determined 1 since the line charge is infinite in extent along 2 the strength of the field cannot depend on 2 and neither can the field have a component in the az direction 2 since the charge producing the field is an infinite line charge symmetry implies that the strength of the field cannot depend on 4 and neither can the field have a component in the a direction and 3 these considerations require the form of the field to be D D ap where D can only depend on p The field s symmetry is the key in using Gauss law to determine the field With this symmetry a Gaussian surface a cylinder with its axis on the zaxis Choose the cylinder to have radius a and length L as shown on the diagram below can be chosen such that the surface integral the Gaussian integral becomes tractable Note the central role of symmetry in this problem 1 The field cannot depend on 2 since the line charge is on the z axis and the line is considered infinite in length The physical environment is not a function ofz so neither is the field 2 The field cannot depend on 4 since the line charge is on the z axis which so that the physics of the situation is also independent of 3 The field can only depend on p and moreover must point in the tap direction depending on whether pL is positive or negative Electromagnetic Fields 19 NDUDE nuvvtne Gausstan surfa e sen251D make 11 tne evamattun uttne Gausstan tntegrat strnpter I t uttcmA 1 1 Tne he d ts perpenmcmartu tne surface nunnat Gausstan I H ntne and uttne yhnder imam i Tne he d ts paraHE tn tne surface nunnat Dn tne 5 cynnuermt van tne new must be a x Dnstant Dn tne yhndervvaHs SmEE p ts r Dnstant and t ts Estabhshed tnat D Dp I m D Dds q H D Dds HD Dds H D Dds qme ms mu m t t 2 a 2 a 2 IDaPDpdwdzap 1 IDaPDpda dpax IDaPDpda dp 7a p L n pr n mm t t 2 t2 Dpd d1Dp d dzpt tun fa Dptinpt stemmnagnetm FEMS Example infinite cylinder with its surface charged g Determine the electric field for a cylinder with a surface charge p5 3 pCmz The cylinder has the zaxis as its axis The radius of the cylinder p 015 m The cylinder is infinite in extent lt z lt 3pCm2 p015m 05 o p 015m There is cylindrical symmetry with pL 211 015 m p5 Using the above result one obtains O plt015 m answer E o152n31039 a E pgt015m 2118p p C Example distributed charge g Determine the flux density from an infinitely long cylinder of radius a with a distributed volume charge density find D for p lt a and for p gt a 2p Cms plta V O pgta From the charge distribution it is clear the field has cylindrical symmetry so that the Gaussian surface is a cylinder of length L with its axis along the zaxis ZDL 27 p j 2p pdpd dz plta 22 0p0 DDdSqinswe zDH 27 a zDL 27 p j j 2p pdpd dz I I I0 pdpd dz pgta zzu 0 p0 zzu 0 pa 2p2 T a p p lt a answer D 3 2a a gt a i 0 3p 9 Electromagnetic Fields 2 Example Coulomb s law implies Gauss law For a charge q at the origin the force on a test charge q is F qql 2 47rsr r The electric field the force per unit charge due to the charge q at the origin is q a 47rer r Integrating E over a sphere centered on the origin surrounding the charge q and using the definition of permittivity D 8E 2 m DElds m qzarElds n q ar rzsin6d d 9ar q sphere sphere r sphere4 r meDdsq sphere So that using Gauss law shows the charge to be q as was derived from Coulomb s law The importance of symmetry in analytic solutions In the two preceding techniques using superposition integrals and using Gauss law the ability to find analytic solutions depended upon the charge distribution s symmetry First when using Coulomb s law in the superposition integral the charge s symmetry allowed us to analytically evaluate the resulting integrals to obtain a closedform solution It is true that the integrals could have readily been written without the given symmetry since only a knowledge of the charge s distribution in space is needed for this But for analytic evaluation symmetry was a critical factor Similarly when using Gauss law symmetry often resulted in the existance of a suitable Gaussian surface This allowed the crucial step of evaluating the Gaussian integral to be carried out often not possible without sufficient symmetry With symmetry and the proper choice of Gaussian surface evaluating the Gaussian integral is often trivial Without this symmetry however its evaluation would often be anything but trivial and usually required numeric evaluation How does one approach a more general situation where perhaps the charge is not so conveniently arranged One approach is to use a numeric tool like MATLAB to evaluate the resulting superposition integrals an approach which will be explored via homework Electromagnetic Fields 22 This approach is general forthose cases in which the charge distribution is known Unfortunately in many cases we39re faced with a boundary value problem in which the fields or potential on a boundary surface is known and the charge is an unknown In this case the knowns include the region s geometry material properties and boundary conditions We have therefore two types of problems one in which we know the charge and the other in which we know boundary conditions Both types are important Boundary value problems are solved via differential equations and therefore a mathematical description of electromagnetics in terms of differential equations is important Divergence g Consider a cube of dimension Ax Ay A2 with one corner at xyz The net flux leaving the volume Ax Ay A2 in the aX direction is We xout Dxx AX y Z 39 DxxyvzAy AZ Divide and multiply by Ax We xoui T AX AX Ay AZ In this expression if Ax Ay and A2 a O the result is 0D 0D d ded dz X dv Weoulx ax y ax Similarly 0D 399Dz Wm 3 dv and d ax dv dWe oul dWeoulx d l eouty dWeoulz 0D 0D Edv iydv Edv 4 dV 0X By 02 0X By 02 The quantity within the parentheses occurs often and it called the divergence of D div D can be written as V gD where V is the quotdelquot operator 0 0 0 V Ea an Ea Electromagnetic Fields 23 The divergence is formed by the dot product of the del operator and a vector in this case the current density a a a V gD Raj a ay gal 9DX az Dy ay Dz a2 V 90 g 801 ax By 32 The divergence of D the electric ux density vector is the net electric flux out per unit volume dipwm LDX mv ax By 32 Can this differential relation be used to nd the electric flux out of macroscopic bodies To answer this question consider two adjacent differential cubes The critical question whether the electric ux at the adjacent areas is properly accounted for to allow integration over macroscopic volumes gt It is apparent that when taking the divergence over adjacent differential volumes the common ux across their shared surface areas will be positive for one and negative which will cancel for the composite volume lj This observation for two adjacent differential volumes can be extended to any number of differential volumes This permits the use of integration of an in nite number of differential volumes for the case of macroscopic bodies and allows divergence to be used to nd the electric ux out of macroscopic volumes m v ngv vulume Electromagnetic Fields 24 Example charge within volume g Given that the electric field is E 4 75 ar find the total charge within a sphere of ESJF radius 10 cm which is centered on the origin Do this in two ways with a surface integral and then with a volume integral 1 surface integral by Gauss law q we D D Dds surface 75 N 75 a D a 4739er rC 411W rm2 237 C r01m w 2f 75 a Drzsin6d6d a e oeo47quot5F r r 2 v volume integral Qwe g DDds VDde surface volume 275 ari 1 l4mFc13575c 375C 2 4739 m3 2r324n m3 0127 7 3 2 I VDde J j grslnededzgdr 237C r0 060 volume Divergence Theorem Since the net electric flux out for a macroscopic volume can also be expressed in terms of a surface integral the result is a relation between a volume integral and a surface integral m V 90 dv If D gds volume 5 urface Although this relation has been developed for the case of electric flux the derivation has been purely mathematical and holds for any vector quantity The relation is known as the divergence theorem lfthe volume is reduced to dv it can be seen that the physical meaning of V gD is the net electric flux out per unit volume Electromagnetic Fields 25 What is this quantity the net electric flux out per unit volume Using Gauss law which states the electric flux out of a charge is equal to its charge the flux out per unit volume must be the charge per unit volume pv The result is the point form of Gauss law which is discussed further below Point form of Gauss law g Combining the divergence theorem for electric flux density with Gauss law leout q V ngv pV dv volume volume This relation holds for any volume The only possible way for two volume integrals to be equal for an arbitrary volume is for the integrands themselves to be equal The result is the point form of Gauss law VDDpV From the meaning of divergence of a flux density net flux out per unit volume V D D must be the electric flux out per unit volume Taking this observation together with Gauss law which states that the electric flux coming from a volume is equal to the net charge within the volume the electric flux out per unit volume can be thought of as the charge per unit volume This is the meaning of the point form of Gauss law In terms of the electric field V D SE pV lfthe permittivity is not a function of position it passes unchanged through the del operator VDEamp 8 So far the focus of discussion has been on coordinate systems vectors vector calculus and the relationship between charge and the electric field electric flux and electric flux density Now electromagnetics will be viewed through the lens of energy and potential One advantage in looking at the problem in this way is that one deals with scalar functions energy and potential which can then be used to find fields Electromagnetic Fields 26 Potential Energy g A charge q in an electrostatic field E is acted upon by the field with a force F qE To move charge q against this field an equal and opposite force FapP39ied F must be applied to move charge As the charge moves work is done on the charge by the applied force This work is not lost to thermal energy through friction The work done on the charge acts to increase the charge s potential energy just as pushing against a spring increases the spring s potential orjust as moving a ball uphiII in a gravitational field increases its potential energy The reference point for potential energy the point where the potential energy is considered to be zero is arbitrary Only differences of potential energy are physically meaningful iust as in circuit analysis where the absolute potential of the reference node is arbitrary since only potential differences affect circuit behavior In electromagnetics the reference is typically taken to be at infinity Also by convention the potential at infinity is take to be zero As charge is moved from infinity to a position r a force must be applied to the charge which precisely balances the force exerted on the charge by the electric field In moving the charge this force does work on the charge The differential of this potential energy is dWF applie dDdl qEDd The resulting electric potential energy of the charge q at position r in the field E is W dW 39 EDdl JW 0 Jimq Wr J39rqEDdl Rather than tracking the potential energy of a particular charge q the potential energy per charge the electrostatic potential is often used Vr de Wr q The electrostatic potential is analogous to a node voltage in circuit analysis In nodal analysis a node voltage is defined as having its positive at the node in question and having its negative sign at the reference node Here the electrostatic potential is referenced to infinity Infinity is the reference nodequot for electrostatic potential the place where the negative sign goes Electromagnetic Fields 27 Example potential difference between points a and b g Va fEndl Vb w w q m b q L EDdl vab Va Vb EE dl EE dl fEDdI vab LaEDdl Notice that is the work per unit charge expended in moving charge from b to a against the field E The following formula might be a convenient form in which to remember potential differences vab J39bEDdl v JEDdl Exam pie electrostatic energy and electrostatic potential Coulomb s law gives the force between charges The force F on charge q at r due to charge q at r is q q39 r r F 2 4778 r r39 r 39 r r r39l distance between charges r r unit vector pointing from q39 to q r r Take r 0 q at the origin q q39 L q q 47m lrl2 rl 47m r2 r Electromagnetic Fields 28 Fwd the Wurk ur energy reqmred m muve g at mm an arbmary pusmun r A an aswdey mute the prumern nas spherma symmetry nnren makes the spherma cuurdmate systemthe natura cuurdmate sys LEm m use Tne apphed ruree VF muves agarnsnne enumrnb ruree F 1an rs the dwfferenua ufvvurk dune bythe apphed ruree VF dr q q 1 gvv Fd 39 M Wm 39Arrerz muve g rs transfurmed m m putanua Energy electrostatic potential energy A the apphed e VF mcreases Tne enarge rs bemg ruHed up a putarma energy nm Tne apphed fume VF pusnes e Frust ba anmng the e d s fumey F 5 that Wyeth energy rs nut mcreasmg mg dune agamsi the new wanI Externa fun 0 e spe ks gr Wurk be anamofq rneny the hem Wnren resu ts W F nayrng a ragrax mrectmn the surfaces of es syrn Eqmputentwa are euneentne spner m rnuye g cm tne surfaces uf Eqmputentwa Energy Answer Nun wsevvurkvvumd be g rreg furthemtu rnuye unthe surface mmmh Wemg than nut be surfaces of egmpmenuax energ Freamnaenene Fwe ds Potential energy force potential electric field g From mechanics the differential of energy done by a force F is the dot product of the force and the differential length dW F dl Consider F to be the force exerted by the field on a charge q and that F is the force that must be exerted to move the charge against the field Therefore dW F dl is the differential of the work done by the force F on the charge in moving the charge against the force of the field F The work done on the charge by the applied or external force F is equal to the increase in the potential energy W of the charge q Now considerthe differential of work from purely a mathematical pointofview lfthe scalar function W is a function of position of x y and z in Cartesian coordinates then the total differential must be by the chain rule de Mdy Mdz ax ay 02 This quantity can in turn be expressed as the dot product ofthe gradient of W and the differential length dl dW de Mdy Mdz MaX May MaZ D dx aX dyay dz a2 ax By 02 0x By 02 With this observation one can clearly appreciate the meaning of the first term in the dot product dW F dl aw aw aw aw aw aw F fax 7a iaz a F fax 7 Z ax 6y V 62 E ax 6y V 62 j The expression 6W 6W 6W aX a a2 6y y 02 is the result of the del operator acting on the scalar function W called the gradient of W VW a3 av3 a13 W aX way aZ 6x By 62 6x By 62 The gradient of W is a vector valued function It points directly uphill in the direction of the maximum increase in W lts magnitude is the value of this maximum increase The gradient can be taken of other scalar functions of position and the meaning is analogous The gradient points in the direction of maximum increase in the scalar function and its magnitude is maximum increase Electromagnetic Fields 30 1 The gradient of a scalar function of position is a vector 2 The gradient of a scalarfunction has as its magnitude the maximum rate of increase in the scalarfunction and its direction toward maximum increase Here the scalar function of position is W the electrostatic potential energy What is the meaning of increasing W or decreasing W It simply refers to the fact that when W increases the potential energy of the charge q increases As W decreases the potential energy of the charge decreases What does the gradient of scalar potential energy signify physically The gradient of W would give the magnitude and direction of the maximum increase in potential energy It is analogous to the hiker on the side of a hill The gradient of the gravitational potential energy in this case would indicate uphill The gradient of the electrostatic energy is analogous it also indicates uphill this time the hill being an electrostatic potential hill The force that the field exerts is downhill to lower potential energies Greater energies are only reached if an external force does work on the charge in moving it uphill to greater potential energy The force a charge experiences in an electrostatic field is downhill in the direction of decreasing potential energy The force exerted by the electrostatic field on a positive charge the force F qE points downhill To move the charge uphill to greater potential energy an external force equal to negative the field s force F qE must be applied to exactly balance the field s force and allow the charge to be moved dW F Ddl VW Ddl F VW where F is the field force F VW where VW is the gradient of the potential field W W is a scalar function of position a scalar field and VW is a vector field V is the del operator kind of a vector derivative operator It has different forms in the coordinate systems we use 0 0 0 ln rectangular coordInates V ax ay az ax By 02 In cylindrical coordinates V ap3 13 a23 0p p M OZ ln spherical coordinates V are3 a 13 a 3 0r r 06 r sine 0 FLVWL axM ayM aZM ax 0y 02 Electromagnetic Fields 3 Dividing the electrostatic energy by charge gives the potential work per unit charge Taking the gradient would then result in the force per unit charge or electric field V q EVV ax ay az 0x E q 0y Oz Restating an alternate and equivalent way of thinking about the electric field is that it is the negative gradient of the electrostatic potential which is the electrostatic potential energy per unit charge E VWLW q E39VWE q q Electrostatic potential is the electrostatic potential energy divided by charge The SI unit for energy per unit charge is volts V JC Since the electric field is the negative gradient of the electrostatic potential also referred to as just the potential equivalent units for the electric field are volts per meter the del operator has units of m391 The electric field can therefore be specified with two equivalent sets of units JC or Vm Two ways in which to think of the static electric field 1 The electric field is the force per unit charge From the example above q39 r r39 47m Ir r39I2 Ir 39 I39ll EE q For q at the origin r O 221 SI units of E are WC 4778 r 2 The electric field is potential difference per unit length From the example above q 47139s r W V SI units of E are also Vm Electromagnetic Fields 32 Coulomb s law and superposition an example If the charge distribution is known the superposition integral can be simplified by first finding the electrostatic potenital and then finding the electrostatic field by talking the negative gradient of the potential function dq39 V r J47Z8R q As was true when with using superposition integral with Coulomb39s law the integration will be a single integral if the charge is distributed along a path will be a double integral if the charge is distributed on a surface will be a triple integral if the charge is distributed in a volume and will be a sum if the distribution is a collection of discrete charges Example E in spherical and rectangular coordinates This Consider a point charge q39 at the origin Use E VV to find E givenV 4 q will be done first using rectangular coordinates and then using spherical coordinates i Using rectangular coordinates V i 4y 47rsr 4 g X2y2222 EVV ax ay 32 0x By 02 q 0 2 2 2 7 0 2 2 2 7 0 2 2 2 7 4 gaxax y 2 aygx y 2 azEx y 2 q39 x y z 7 ax a a2 47m x2 y2 Z2 y x2 y2 Z2 x2 y2 Z2 ii Using spherical coordinatesVV gar 1 a a ar r 06 r Sln6 a E i areilj 3 3 ia r iza 475 0r r r 36 r r Sln6 M r 475 r 47rgr The electric field derived is the same regardless of the coordinate system used Can you show that the two solutions are the same Electromagnetic Fields 33 Example electric field from the potential The electric field E is the negative gradient of the electrostatic potential E W Find the electric field given the potential Vxyz x2y 32 4 volts EVv ax0x2y3z4 aly0x2y3z4 alz 9x2y3z4 2 0X 0y 2xyaX xzay 3az Vm Exam pie electric field from the potential Find the electric field given the potential below Vr 2 sin7rx 3yz V ax 02 sinr 3yz ay 02 sinrl 3yz a 02 sin7rx 3yz EW 02 E 27139 cos7rxaX 32 ay 3y a2 Vm Example directional derivatives The rate of change of a scalarfunction in a certain direction can be found by taking the dot product of the unit vector in that direction with its gradient For the electrostatic potential W Dal dl Find the potentials directional derivative in the a 3 325 direction if the potential field is given by Vxyz 2x 5y volts 9 lt 02x5y a 02x5y a 02x5y DEi ax y 0y 2 02 J5 are I VVDa 2 3llt ll Electromagnetic Fields 34 Potential Surfaces g If the directional derivative of the electric field is zero for a particular direction this indicates that the potential does not change in that direction Since the electric field points in some direction any direction that is perpendicular to the electric field give a directional derivative of zero Taken overall all space these form a surface an equipotential surface which is perpendicular to the electric field or which is the same thing to the negative gradient of the potential The equipotential surface is perpendicular to the gradient or the negative of the gradient so that the electric field is always perpendicular to equipotential surfaces The gradient of the potential the negative of the electric field is perpendicular to these equipotential surfaces which are analogous to contour maps which show lines of constant elevations In the case of contour maps the gradient of the gravitational potential would be perpendicularto these elevation contours Using the physical definition of potential and the gradient operation consider the diagram below which shows a potential V which depends only on x and y This plot provides information regarding energy electric eld and the charge distribution required to produce it Graphically characterize the potential gradient and the field f at points a b and c k 3 Sketch some representative a equipotential contours I Electromagnetic Fields 35 Examplkenergy potential and polarity u 5 y yum 1euu1urnb elf y enarge frum ElElEltu 11m meters At Dun the charge s pmenna1 energy 1 n J At 1 1m 11 pmenna1 energy 151 J Therefurey 11 takes 1 J m muve the enarge frum ElElEltu 11m meter vvna 1 the pmenna1 mfference between Dun and 11D 11s1J1c1v vvna 111 pman 7 magmtude mne yunage d1 erence151vyvmye are the pusmve and neganye s1gns7 eqmp enm umas A the 1c gr enarge 1s bemg pusneg by an Externa fume frum nun m 1 1n Wurk 1s dune an the enarge and1stransfurmed1ntuthe charge s pmenna1 energy Potential energy and electrostatic potential the path taken sthe energy reqmred m muve the enarge amng ax frum nun m 1mm n rn 1 straghthnetu 11cm inhegrals amng the WEI paths Tne mtegrand 1n n1 ease1s gvv VF dl rqE dl W new aw 1 dl seemnaenene regs 36 Line integrals potential energy and voltage drops The differential of energy required to move the charge is a given direction is dWi qE Ddli Integration is used to find the total energy required to move along a given path Since the differential may be a function of position the integration must be performed in a manner which incorporates the dependence The result is a line integral in which the effect of the path on the integrand is taken into account In terms of potential the potential rise in a given direction is simply the energy divided by the charge dVi E Ddli A path integral would then be used to determine the total voltage rise along the path a N Vab J E Ddl which is numerically approximated as Vab 2 2E DaiM i1 Line Integrals g Line Integrals are evaluated using these three steps Step 1 Form the dot product of the integrand and the differential length such as dWi qE Ddli or di E Ddli Step 2 Incorporate the effects of the path on the integrand and on the path Are any of the variables constant over the path If so this would allow them to be treated as constants and their differentials would be zero Are any variables zero over the path Step 3 Integrate Exam ple line integral rectangular coordinates nonconservative field Given the electric field below find Vab between 2 a 110 and b 000 along the path b Vab y 000 i 100 a 110 000 x a 110 Exax 22ay 3xaZ Vm Step 1 Form the integrand by taking the dot or scalar product of E and dl dVE Ddl xaX 22ay 3xaz DdxaX dyay dzaz dVxdx22dy3xdz Step 2 Incorporate the effect of the path on the integrand Electromagnetic Fields 37 1 Vab Ixdx22dy3xdz x0 1 Ixdx22dy3xdz 5210 v W xdz0 1 1 vab jxdx jo x0 y0 Step 3 1 1 x2 1 1 Integrate Vab I x dx I O 7 7 V x0 Y0 2 0 2 It would require 1C Vab O5 J to move 1 C from b to a Does the potential depend on the path taken V ab 000 x a 110 Consider taking the path below in the same field b 000oo1 a 101 a 111 010 Exax 22ay 3xaZ Vm Ste 1 pForm the integrand by taking the dot or scalar product of E and dl dVE Ddl xaX 22ay 3xaz DdxaX dyay dzaz dVxdx22dy3xdz Step 2 Incorporate the effect of the path on the integrand 1 Vab Ix dx22 dy 3x dz 2 0 1 x dx 22 dy 3x dz 0 xy0 X 1 Ixdx22dy3xdz y0 Vab j39o J139xdx 121dy j3931dz z0 x0 y0 z1 Step 3 lntegrate 21 vab 3L 211 3zlf 1 4 3 v 51 v 2 0 2 2 Forthis field the energy required to move charge between points depends on the path taken That is the potential difference is path dependent Wab anb 1C 55 V 55 J Electromagnetic Fields 38 Conservative Fields What is implied if the energy required to move charge against the field depends on the path taken What are the causes and consequences Physically there must be a changing magnetic field present If a changing magnetic field is enclosed by the loop Faraday s law states that a net electromotive force is induced in the loop NE gdl loop ln circuits Kirchoff39s voltage law states the sum of voltage drops about any closed loop is zero This is assuming a special case exists The assumption underlying KVL is that the loop does not enclose changing magnetic fields Assuming there are no changing magnetic fields present the sum of the voltages about any loop is zero The analogous statement in electromagnetics is that the path integral of the electric field about any closed path is zero NE gd0 loop lfthis condition holds the field is said to be conservative No net work is required to move the charge about any closed loop which implies a singlevalued electrostatic potential function energy per unit charge can be defined If the field is derived from an electrostatic potential the field must be conservative Example cylindrical coordinates conservative field g Vr 05 x2 05 y2 05 p2 EVV a l a az 0p 9 p 59 l 02 E p ap Vm Determine the work required to move 1 C from b p 2 1 114 4 to a 2 112 1 via the two paths shown Electromagnetic Fields 39 x axis ath 1 Wm reqmred m muve q frum b m a s W qvn b v IVE gdl E gdl b a n vib Irp qdlafpdw dp l maqdunm axde 2 m rpagdzaipda dpa n 2n 2 V agdpa agdl 1 pa gum r 1 zrusw 2 m 2 2 2 n vvqvb 7 cm 5V 4 SJ Pathz o m V rpapqdlaxpdmdpapl I rpapqdzaxpd dpaPJ jrpapquzppuipgmpap km W m rpapgizax rpapguddm Irpapgjpap mam v 9 m 2 m Wqvab 71 c15v45J 1 V I E simmanent Fwe ds Conservative and nonconservative electric fields g What causes the potential difference to be pathdependent or pathindependent What is implied What are the implications and consequences Terminology If the potential difference between two points is independent of the path taken the answer is the same no matter what path is taken and the electric field is said to be conservative Otherwise if the potential difference is path dependent the electric field is said to be nonconservative Four implications of conservative fields b 1 For a conservative field the path integral J39E gdl is independent of path taken between a and b 2 For a conservative field the path integral about any closed loop E gdl is zero 3 Electric fields are conservative when no changing magnetic fields are present The d4 Inte ral form of Farada 5 law states dl 9 Y NE 9 dt 4 Any electric field derived from an electrostatic potential E VV is a conservative field For a conservative field the following path integrals would be equal J39EDdl J39EDdl J39EDdl J39EDdl path1 path 2 path 3 path 4 path 1 b on Electromagnetic Fields 4 lfthe path integral between any two points is path independent the path integral about any closed loop is zero mEDdIO Consider a few closed loops associated with the diagram above mEDdI IEDdI31EDdI iEDdIjEDdIO gtpath1 h a h h e path 2 path1 pach path1 pach mEDdI iEDdl3EDdl iEDdljEDdIO gt path 2 h a h h e path 4 path 2 path 4 path 2 path 4 As stated above the electric field is nonconservative whenever there is a changing magnetic field present in which by Faraday s law d EDdI dt where pm is the magnetic flux The magnetic flux is related to the magnetic flux density vector via a surface integral 4 BDds surface A consequence of having no changing magnetic fields present is that a singlevalued electrostatic potential function can be defined such that E VV as stated above Given the discussion to this point can one say with certainty whether the fields below are conservative or not Why or why not If a closed path is found for which mE Ddl 7t 0 then it can be stated with certainty that the field is nonconservative One need look no further One path for which mE Ddl 7t 0 is sufficient to show once and for all that the field is nonconservative If one must use mE Ddl to test whether the field is conservative and path after path gives mE Ddl O one can never say with certainty that the field is conservative one can only say that no path has been found yet for which mE Ddl 7t 0 and that the field may be conservative The only certainty to be gained here would be if the electric field is given in analytic form and an electrostatic potential is found for which E VV For this case one can say that the field is conservative A better test is needed Electromagnetic Fields 42 Conservative and nonconservative elds um and Stoke39s theorem uhee and fur aH The curl sthe thhe vemur uperatmn mvu vmg the de uperatur mseussee Sn far the rsttvvu Ware the gradteht and the dwergence Termhptpg the eheutatph uf F S the hhe htegrat at a vemur F abuut a dusted path ff gdl dl rm eheutatmh S a term fmm md e a t tta me svetp ty 7 Er mmu atmm the m mee be eheutathg Mk whmpppt Extehmhgthtsmea bEyDHd mdS VWEH ur has a HEIHVZEVEI pat thtegrah t 5 sad tn ave a HEIHVZErEI etreutatmh abuut that path mm 1 and mm 2 petpw mee eaheet and phty the putstee Eunmbutmns remam FXYanrm Hm hght heft XV d dy dy tpp FVgtlt yydgtlt d puttpm Rayegmx dx thhhtestmat path E a mmagnetu Fte ds 43 The circulation about this infinitesimal loop is 2F gdl x d7X4 dy Fxxy dZ y dx Fyx d7Xydy Fxx y dydx loop Rearranging the terms and expressing as a product of derivatives and differential surface areas Fx 3 dy Fx dx dy Fxxy dim Fxxy dydx 239 dl 2 2 A 2 2 4 9 v v loop dx dy dx dx d d Fx 13 Fx av Fxxy 1 Fxxy 1 Zngl 2 2 ddy 2 2 ddy loop dx dy dF Zngl V dxdy V x F dxdy dx dy Z loop This quantity V x F is the zcomponent of the curl of F in rectangular coordinates 0F 0F V By 02 02 0x y 0x By For a loop of arbitrary orientation the circulation of F about a closed loop is not just the surface integral of the zcomponent of the curl but simply the surface integral of V x F Nng V x F gds This relation is Stoke39s theorem and can be seen to provide a physical meaning of the curl of F Considering some infinitesimal path the normal component of the curl of F is the ratio of the circulation of F about the infinitesimal loop divided by the area of the loop Conservative fields and V x E The curl of E can be understood qualitatively as quite literally the curliness of the vector E lfthe vector E has a nonzero curl then the line integral of E about a closed path can be nonzero On the other hand if V x E is zero then by Stoke s theorem the circulation of E about any closed path must be zero and the field is therefore conservative NEng V x E gds Surface Electromagnetic Fields 44 If V x E 0 it then follows that 15ng J V x Egds Ogds Surface Surface 15ng 0 fV x E Othen 1 2 3 4 5 vv NEgdlO E is a conservative field b J39E Ddl is independent of the path taken between a and b no timevarying magnetic field is present the electric potential is singlevalued Line integrals along prescribed paths g The line integrals considered so far have always been along paths which have been split into segments where only one variable changes at a time What is done in cases where this does not hold What if the path is chosen in which more than one variable varies The answer is that constraint equationsquot defining the path must be incorporated into the integrand To illustrate consider finding the line integral 31E Ddl b between xy 00 and 12 for E xy ax y2 ay For rectangular coordinates d dx ax dy ay dz aZ Edl xydx y2dy Electromagnetic Fields 45 Evaluation In line integrals there can be only one independent variable If an integrand has more than one independent variable constraint equations relating the variables are needed to so that a single remaining independent variable in the integrand Path 1 A straight line from 00 to 12 The constraint equation describing this path is y 2x The differential relation is dy 2 dx E dl x2xdx2x22dx 6x2dx or Edl 2yy 2dy y2dy y2dy E gdl 6x2 dx 2 or 2 3 Ed 72d 9 4yy 1 l i 0 Path 2 The parabola y 2x2 from 00 to 12 The constraint equation is the equation of the parabola The differential relation is dy 4x dx E dl x2x2dx 2x224x dx 2x3 16x5d or E dl 2y39Wdy 2y391y2dy yy2dy E gdl 2x3 16x5 dx 2 or 0 2 E gdl gyyzdy 2 0 Line integrals in 3space For a path integral in 3dimensional space two constraint equations would be required to define the path Each of the constraint equations in three space would define a surface and their intersection obtained through their simultaneous solution would define the path of integration The integration would be with respect to whichever of the three variables is not eliminated with the two constraint equations Electromagnetic Fields 46 In fact we could consider the path integrals above as being performed in 3space where one of the constraint equations is z O Poisson s and Laplace s equations g The electrostatic field is the negative gradient of the electrostatic potential E VV For a region bounded by surfaces of known potential casting the problem in the form of a boundary value problem using potential can be the most natural solution path Starting with Gauss law the use of E VV gives the differential equation in V known as Poisson s equation Solving the boundary value problem for the potential avoids having to solve the vector differential equation directly It offers an end runquot by first solving for the potential the negative gradient of which is electric field V D VV pi using E VV in the point form of Gauss39 law V DE pl 8 S vmvv ii 8 V D VV is the Laplacian operator operating on V and is written V D VV V2V The result is Poisson s equation VZV Piv S In rectangular coordinates this implies 0 0 0 VDVVV axi ayi azi D axia 7 azi 7 ax By 02 02v 02v 02v amp 7 7 0x2 ay2 022 s For pV O the result is Laplace s equation VZV 0 Poisson s and Laplace s equation and the subsequent gradient operation give the electric field only for electrostatics when the source of the fields charge is at rest When timevarying currents are present Faraday s law shows that the line integral of the electric field about a closed path is no longer zero the electric field is no longer a conservative one A nonconservative field cannot be expressed as the negative gradient of a scalar potential In this case the concept of a vector potential is often introduced Here the static case is treated Electromagnetic Fields 47 Laplace39s Equation g Laplace s equation is present is many branches of science and engineering VZV 0 ln electromagnetics Laplace s equation gives the electrostatic potential from which the electric field can be found E W given no changing magnetic fields are present In rectangular coordinates Laplace s equation reads 02 02 02 2 2 2 x by 02 Laplace39s Equation will be solved analytically for the case of variations with involving one variable When more than one variable varies the techniques of partial differential equations must be used Consider a potential for which the V potential is a function ofz and which 5 a 0 z 0 is independent of x and y 02V d2V 0 0x2 dy2 solving 2 gi o Id IOdZA dz dz dz dz dV E s J39dV j Adz Vz A2 B applying the bc39s of VO O and VL VS VOBO V VLVSAL ATS V2 V2 S L Using the gradient Electromagnetic Fields 48 Laplace s equation in cylindrical coordinates g Consider two concentric cylinders that are PECs inner radius a outer radius b Let the voltage of the outer cylinder be 0 and the voltage of the inner cylinder be V0 Vb o Va vo Laplace39s Equation in cylindrical coordinates VZV 7 1 0 av 1 02v 02v 0 77 7 7 p 0p 0p p2 of 022 For this situation there are no variations in e or and Laplace39s Equation reduces to 1 0 av 1 d dV we a pap 0p pdp dp Solving this dV dV d i 0 d 7 7 J pdp J p p pdp A jdvj7dp 7 VAnpB p Applying the boundary conditions bc s VbAnb BO BAnb Ve1AnaAlnbVo V n j Electromagnetic Fields 49 Finding E W 0V 1 0V 0V VV p aZ 0p p M 02 0V 0 V V E 7a 7 np nb a p p p 0p rig rig a a E V l a Numeric Techniques in solving Laplace39s Equation it In rectangular coordinates Laplace39s equation reads 62V 62V 62V ax2 ay2 622 Defining potentials at points on a grid Evaluating the first derivative with respect to x at A 5 vo va 0x A Ax Az A and at C L 0 VF 39 Vo Ax A F ax c Ax c y F 0 Evaluate the 2quotd derivative with respect to x at O D Electromagnetic Fields 50 0V 0X c ax A 0x2 0 Ax substituting 62V VF VB 2Vo We Axf Similar expressions can be found for the second derivatives with respect to y and z iv vR vL 2vo ay2 0 Ayf iv vU vD 2vo 022 o A22 Take Ax Ay A2 A The numerical approximation for Laplace39s equation reads 6v 6v 6v 7 fix2 0y2 022 VZV VFVBVRVLVUVD6Vo A2 This gives the reasonable result that the voltage at O is just the average of the surrounding voltages Why is this reasonable vFvEvRvLvUvD 6 V For the 2D case VRVLVUVD 4 V Electromagnetic Fields 5 Linear equations 100v ov 50V 50v Develop nine equations and nine unknowns V1 through V9 1 VRVLVUVD4Vo 3 V2OV1OOVV44V1 2 3 4 5 6 7 8 9 Once the potential voltage at the nodes is know linear interpolation can be used to find the potential at any point Electromagnetic Fields The general guideline that can be relied upon in homogenous resistors is that the potential at any point or is the average of the potential at surrounding points arrayed symmetrically about the point 0 Using MS Excel For 2D problems iteration can be implemented on a spreadsheet For example in the problem above twentyone 21 cells would be required 9 for V1 through V9 and 12 for the boundaries The potential of each of the cells corresponding to V1 through V9 is calculated as the average of the four surrounding cells To begin give the boundary cells their voltages and the interior cells zero volts Begin iterating V1 through V9 will usually stabilize quickly Table as entered After 15 iteration circular After 100 h iteration reference enabled by clicking iteration under tooIsoptionscalculation Electromagnetic Fields 53 ohm39s law and resistance rtutne where 5 ts Swemens the st umt tcr ccnuuctance and equat tn 0 Examplkresistor a Fwd the reststance ct the reststcr shcwn T 5chth Lap ace s Equatmn assummg nu uepenuence an X Dry v z 752 J L 2 The eTectncneTu tsthe negatva grauTent ct the pctenttaT E vv 4 f2 Via 3 The current densm S mmedtate y knuvvn J CE 4 The current uensTty can be mtegrated tu nd the tutaT current V the resTstur aectmnacneuc reus JDds surface where ds is in the direction of the desired current here az r 2n VS rz I0 JoO39T a2 Dpdpd az O39TVS 5 At this point the total current I is known in terms of the potential difference applied across the resistor Vs The ration Vsl is the resistor s resistance R i L where A nr2 is the resistor s crosssectional area Steps 2 through 5 provide an outline of a general technique with which to find a resistor s resistance Here E is found from the potential but as we ve done it is also possible to determine the electric field from a charge distribution using Coulomb s law and superposition or with Gauss law El JIIUJDds RVi R J a I Joule s law g The differential work done on a charge q by an electric field E is dWF DdqE Ddl If the charge were in a vacuum the work done on charge q would be reflected in an increase in kinetic energy of q The charge would accelerate the speed would increase The mechanism of energy transfer would be between the field which doing work would lose energy to that of the charge which having work done on it would gain energy Vacuum If the charge is one of the free charges contributing to electrical conduction in a material it will not be free to continually accelerate It will be scattered that is it will bump into and bounce off the molecules of the material In these scattering events kinetic energy is transferred from the charge slowing it down The kinetic energy lost by the charge will be transferred to the material s molecules causing them to vibrate more energetically the material will grow warmer material At some point a balance is achieved between the energy the charges gains from the electric field to that they lose through scattering At this point of balance the charges as Electromagnetic Fields 55 a group will no longer have any net acceleration The energy they receive from the electric eld is then passed on to the material in the form of heat energy lattice vibrations Since the mobile charge no longer experiences any net acceleration its motion can be described by an average velocity v At this point the charge can viewed as acting simply as an intermediary transferring power from the electric field which loses energy to the material which gains energy Consider the case where a voltage source sets up the electric eld For this case the electric field continuously loses energy but does not grow smaller in magnitude since the voltage source provides energy to maintain the electric field if the electric eld did not receive energy from the voltage source its magnitude would decrease In this case the energy from the voltage source maintains the electric field which does work on the moving charge and this work is transferred to the material through scattering from the material s atoms lattice scattering Overall we can view this as power from the voltage source to the conducting material Looking at the individual charge q in an electric eld E moving with an average velocity v The time rate of work done by the area A electric eld on the charge is x V L dW d 7mi ED dt dt q V i PqEDv This is the power that is transferred from the electric eld E to thermal energy with the charge q acting as an intermediary Consider now the power associated with a differential charge dq dP dqE Dv pvdvE Dv dP 7 E n F W M V Here p is the power absorbed per unit volume Joule s law which is the point form of the familiar P Vl relationship giving the power absorbed by lumped circuit elements p EDpvvEDJ pEDJ Electromagnetic Fields 55 Example determining resistance using Joule s law it Take the resistor that has previously been considered 2 VS zL 0 20 T444 1 Using Ohm slaw 2 2 PVIL2R RL R P In terms of fields Jesdlf Ville R j cEzdv E if cE gdsjz surface In this case the electric field has been found to be VV 32 az Oz L L Finding the resistance with R V2P 0 VS 2 J39E gdl2 211 L az gdzaz V52 L L v nec EZdV II T IG2 pdpckpdz c jz L271 cnpz 5A z0 o p And now with R PI2 Electromagnetic Fields 57 v 2 2 2 6A pdpckpdz c j LG 39 L 2 R z0 o p0 I39 r OHM 6A Example cylindrical resistor with Laplace it Find the resistance between two PECs forming concentric cylinders inner radius a outer radius b of length L Let the voltage of the outer cylinder be 0 and the voltage of the inner cylinder be V0 Let the conductivity of the material between the PECs be 0 from symmetry V should not vary With and the model used takes the length to be large so that variations 2 can be neglected Vb o Va vo conductivity n Laplace s equation has been solved for cylindrical coordinates with these boundary conditionssee pages 4748 of these notes 39 i V ilnbln Vo lnbln In quot In In quot EVVa3ilnblnp V 1a mm p Electromagnetic Fields 58 b v 1 J39EDdl Ln 5 adePap R X a 27 zDL sull 5 Layla a spam V0 dP V0 0 b i In b In a n R ln 239 P ln ba 27 ZDL V V Boundary conditions it Consider two materials sharing a common boundary Material 1 has conductivity 01 and permittivity s1 and material 2 has conductivity 02 and permittivity 82 The laws of electromagnetics require certain relations for the J D and E vectors at the boundary Boundam conditions for J z Iout D J Dds O surface 6 FEglOn 2 J39J39J1Dds JZDds HJDds dzfJ 731 v top bottom sides quot J r r39 o E x Since the sides are infinitesimal in area the integral over the sides is zero 27 r 27 r j J39J updpd az l J2 Dpdpd az 0p0 0p0 o E o JZ m2 J2 m2 o JZZ or physically JN J2N ut J 12 The result of conservation of magnetic flux can be expressed compactly in vector notation using the unit normal an to the interface directed from region 2 m region 1 an a2 above anJ1 J2 O Electromagnetic Fields 59 Euundan Eundmuns fur D veglcn 1 Wm c D Dds amide 9s we regtan 2 om Hm nos H D2 nos H D ads to one so Smce the stoes are mttmtestmat m area the mtegrat quot uverthe stoes ts zeru um I 10 Emma 1 0 moomom urn HM o 10 4 70n1 Du 7D pt DrphysmaHy Dm VD px mu Hu Wot vectur normat to the tntertace otrecteo frum rEgmn 2 mm rEgmrM an a abuve in v Euundan conottonstor E Faraday s avv states thatthe vottage mduced about a terms of a hne mtegrat of the etectnc e d l fthe uup ts mttmtestmat a mute amuunt uf ux cannot be Endused and Faraday reads H at o u W to vectur normat to the tntertace otrecteo frum rEgmn 2 mm rEgmrM an a ao we a gtltE r 5 u mEEIdI jE DdyaV IEZEIdyaVZEI oath w w Ewt c Ezv u Ew Ezv stoomoonetc Fte ds This result could be expressed more physically by noting that the ay direction is tangential to the boundary E1T E2T T anXE139E2O Planes and normals A plane can be defined as all the vectors that are normal to a given vector NDrr NxaX Nyay Nzaz D xx aX yy ay zz az 0 NXxx Nyyy Nzzz 0 Nxx Nyy sz Nxxo Nyyo sz0 equation of plane an N N unit vector normal to plane Example 2x 5y 32 11 equation of plane an 2aX 5ay 3az 22 52 32 unit vector normal to plane Conductors and the perfect electrical conductor approximation Good conductors have a large conductivity 0 A perfect electrical conductor PEC has infinite conductivity This implies that the electrical field inside a PEC is zero That this must hold can be seen from the point form of Ohm s law J6E The magnitude of J must remain finite for a PEC with infinite 0 the magnitude of E must be zero Of course this condition is not exactly met within a good conductor but it is nearly so as one can see in the example below Example field required for DC currents in copper conductors 30 A is the maximum current permitted by the National Electric Code assuming 30 C ambient temperature for 10 AWG copper conductors What is the electric field inside a 10 AWG conductor carrying its maximum permissible current radius for 10 AWG conductor 129 mm J for 30 A in 10 AWG conductor 57 106 Am2 E Jo 01 Vm For a wire more on the order of that used for electronics consider the field required for a 10 mA current in a 24 AWG copper conductor Electromagnetic Fields 6 radius for 24 AWG conductor 0256 mm J for 10 mA in 24 AWG conductor 486104 Am2 E Jo 084 mVm Boundary conditions for perfect electric conductors PECs it 1 Since E 0 inside a PEC Etaquot 0 inside a PEC This requires that the tangential component of the electric field at the boundary of a PEC is zero The tangential component of the electric flux density at the boundary of a PEC is zero as well ET O and DT O 2 Since E 0 inside a PEC D 0 inside a PEC as well This combined with the fact that Dian O at the boundary of a PEC implies both the electric field and the electric flux density vector are always normal to the surface of a PEC DN ps and EN pss Incremental resistor it Additional insight can be gained from a discussion of the incremental resistor This can aid in numerical approaches which typically involves breaking regions into small pieces that can be approached more readily Looking at the relation between potential difference and electric field b j E Ddl vab a By incorporating geometry and material properties on can obtain a relationship between current and voltage for a small region of resistive material the incremental resistor The sides of the incremental resistor are parallel to the current flux so that any charge entering the element also leaves the element Also the ends of the incremental reluctor are equipotential surfaces Electromagnetic Fields 62 A UEEdv JEdv JEJDE J andEarE parattet Persquare resistance Suppuse the trtererherttat teststartee ts tWe utrhertstehat th eharaeterwtth a thtcknesst Al Al t th AR i h ts ease MA MAW AV SupposetunhertthatAzwvvt th whteh ease Rn AR 7 At Ut Where Rn ts referred D as reststahee per square Thts humehetature ts used th PCB and to rhahutaeturthg Where must s tmmures are 2D th nature Example resistance of PCB trace Eteumaenette Ftetds 2oz copper 2 oz copper spread over 1 ft2 has a typical thickness of 681 105 m b i m pcu 8960 kgm3 7 M v 212 In 00254 m 1ft 7 t 1ft 1 In 1b 1kg 1mm t 12in 2 00254in 2 6398110 m 1 ftii 8960 kgm3 1ft 1 In For so 58107 Sm the persquare resistance for copper is approximately 0253 mg per square Given this persquare resistance an estimate of the DC trace resistance can quickly be obtained for this 20square trace Rquotace 20 RD 506 mg Resistivity estimates of irregular 2D shapes curvilinear squares The approach taken above in persquare resistance can be generalized for application regarding resistors with more general crosssections Here shapes are chosen so that the average width is equal to the average height The boundaries of these curvilinear squaresquot are chosen so that their sides are parallel to the current flow so that the current quotinquot is equal to the current quotoutquot and their ends are equipotential surfaces to allow a potential difference to be defined Example curvilinear squares Assuming a thickness t use curvilinear squares to graphically estimate the resistance of the 2D resistor shown 64 Electromagnetic Fields flux lines r equipotential surfaces slat 45Iot Symmetry can be used to reduce the work to two identical resistance in parallel it can be seen that the resistance ofthe 2D resistance is 1 1 0744 7 751 451 41391 2 at at This result seems reasonable since the resistor is a bit wider than it is long Resistance of inhomogeneous materials What are the options when the material s conductivity varies with position Case one Sides parallel to current flux lines and ends on equipotential surfaces If two materials are separated by surfaces parallel to current flux lines and if they share common potential differences then the two resistances have common potentials respective currents add and the resistances can be treated as in parallel In this case find the resistance of each resistor separately and then treat them as in parallel to find the resistance of the combination Electromagnetic Fields 65 Case two ends on equipotential surfaces and common currents passing through the two materials In this case the resistances have common currents potential differences add and the resistances can be treated as in series Consider a case with two different materials 01 In this case find the resistance of each resistor separately and then treat them as in series to find the resistance of the combination General case of nonhomogeneous conductivity Case three Boundaries neither parallel nor perpendicular to current density flux lines Consider twodimensional resistors in xyz The general cell is shown below Electromagnetic Fields 66 The basic relation used is conservation of charge under the condition that one would not expect the interior ofa resistive body to be capable of storing charge That is for a resistor each region must satisfy IBM 0 Use the formula I J garea 17E area Looking at the right surface I vO vR 7 A t 7 A t out right 7A 172 4 2 V V Ioutmgm 70 2 R 71 74 Performing similar calculations at the upper the left and the bottom surfaces we would obtain I Mtm wtmvz i39v t 7 2 1760 3 Rearranging one obtains 2Vo 71 72 73 74 39 VR 71 74 39 Vu 71 72 39VL 72 73 39 VE 73 74 0 Solving for VD V Vu 71 72 VL 72 73 VE 73 74 VR 71 74 o 2171 172 73 174 Electromagnetic Fields 57 This is the equation that can be used to find the potentials in a twodimensional inhomogeneous resistor The 2D resistor is of extreme importance Most of the resistors on the planet are 2D resistors in integrated circuits Beyond this this technique can readily be generalized to the general inhomogeneous 3D resistor Consider the two dimensional resistor below 1 5 e9 j4 w 10 520 1V 1 1 1 1 n m lW l m l i i l 4 8 lu inzin IL1 77717877I23 ov The solution will involve writing 23 node equations Begin with N1 Youtside 4w1vw2w The other 22 node voltage equations Electromagnetic Fields 68 n2 8V2 21V 2v1 2v6 2V3 1 5 9 14 19 n3 1 1 1 1 m3 1015m 20 n4 4v2 1v2v3v8 1V i i and so on 1 l l l l 15 21 L4 8 1172774177 22 J 1 23 0V To find resistance divide applied voltage by total current Electromagnetic Fields 69 In order to do this one must find the current density from the conductivity and the electric field In this method a linear approximation for the electric field is found from the node potentials EV13 O V The total current can be calculated as a sum of cell currents over an area that encloses all the current For example to find the current that flows between N13 N18 and N23 and the O V electrode I1 2I3J1A1 J2Az JsAs 0391E1A1 02E2A2 UsEsAs l 0 a V18 OtA a A 2 A A 2 Note Cells 13 and 23 have half their widths in material with zero conductivity For V13 0218 V V18 0198 V and V23 0189 V 04015 0t so that R 1 V 2490t The total current could also be found at the 1V electrode Electromagnetic Fields 70 1 I2I3I4 at1V2 at1V3 at 1 v1 at 1V 2 Electromagnetic Fields 2496t g 0871 V 0862 V 0855 V 2 04020t so that R V1 0875V V2 V3 V4 Permittivity The ultimate source of the electric field E is charge In free space the ratio of D to E is so 8854103912 Fm In matter their ratio is affected by the ease by which the material s bound charges are polarized ln matter bound charge can polarize which produces a component of the D due to polarization DsEsrsoEsoEsr 1s E D8 EP E In a material with bound charge permittivity is a measure of how easily bound charge is separated whether by the orientation of dipoles by distortion of permanent dipoles or by displacement of electrons by an external electric field The permittivity of a material is a measure of the ease in which the bound charge can be polarized of how easily bound charge is separated under the influence of an external field The easier the charge is separated the larger 8 charge stored on upper plate area of one 7 ofthe plates q A d 39 AE 39 s 39 lE 39 v qp 39 voltage across plates plate separation T polarized bound charge stomd charge producing Ep on lower plate Polarization refers to bound charge separation which can be split into three different mechanisms electronic ionic and orientational Electronic polarization is shared by all atoms and molecules electron clouds can be displaced from the nucleus Ionic polarization is the displacement of permanent dipoles such as salts this mechanism involves the movement of ions Orientational polarization involves the rotation of entire polar molecules such as water Electronic Polarization All materials composed of positive nuclei and negative electrons participate in the electronic polarization mechanism Since electrons are much lighter than nuclei For instance the mass of an electron me 911103931 kg and the mass of a proton is 1671O3927 kg a ratio of about 1800 Since most nuclei comprise more thatjust one proton one can appreciate that nuclei are much more massive than are electrons When nuclei and electrons are subjected to an electric field that is rapidly timevarying the difference in inertia between the electrons and the nuclei results in nearly the entire relative motion to be due to the electrons motion Therefore the mass associated with this process is very nearly just the electronic mass Ionic Polarization Electromagnetic Fields 72 The electronic bonding in many materials has an ionic component due to differences in electron affinities of the material s constituents For example when sodium Na and chlorine Cl form sodium chloride NaCl or common table salt the chloride ion attracts the bond electrons more strongly than do the sodium ions the result being that NaCl is an ionic solid with the Na ion being positively charged and the chloride ion being negatively charged NaCl39 When ionic solids are subjected to a rapidly timevarying electric field these two ions are displaced in opposite directions due to the Coulomb force The mass associated with this process includes some relatively small portion of the ions mass which is much higher than that associated with electronic polarization Orientational Polarization This mechanism is associated with the movement and rotation of permanent dipoles the outstanding example of which is the water molecule Orientation polarization in this case refers to the movement and rotation of the Jr dipole not just distortion as is the case in ionic polarization The result is that a much greater mass is typically associates with orientational polarization than is with either electronic or with ionic polarization Variation of polarization with frequency The three polarization mechanisms involve different characteristic inertias and so have different time constants The result is that all materials have a frequency dependent permittivity It is important to know at what frequencies these polarization mechanisms begin to vary Think of bound charge in terms of a massspringdamper system The spring constant represents the coulomb force the mass represents the mass associated with the bound charge and the damper represents loss mechanisms time domain eom m t b18lt kx ft sdomain eom X ms2 bs k F X 1 H s F ms2 bs k 1m 1m ft s2 s S22gwnswn2 m m These results agree with intuition The system acts as a lowpass filter with the break frequency growing smaller as the mass increases Electromagnetic Fields 73 Eteetmhs m 31 kg u artzatmn ah eteetreh ts 9 kg Eteetmhte are hghtimuch hghterthan hueteehs The mass assumated thh M The mass Ufa prutuns and neutrons are appruxtmatetyt Emu p vtstme spectrum Whmh starts at 470 Hz mm m nequ Waveka emu Hana12 mm m elecum mum Jam 1 1022 m It 105 mm 19 102 m mm m2 Mme 1111 102 IEF m UmtAhhemm THz exam A Augsle MeV MegaltaxM mnelecxmnVam an mm m mmmzmx w meeumme MHZ magnum mm kHz mm m unmet km met EM Spectrum from NASA thh respem tn tts huetet Thts ts the mechamsm behmd rameWS a d ts what aHuvvs pHSmS te separate whtte hght mm a spectrum uf enter The ts Why tn ether eptteat prunessesbecume mf cmtfurxrrays Atxrrayfrequenmes aH petahzatmh mechamsms are tehg gDnE Whmh creates ehauehges fur fucusmg maghttythg ete arHEIrHE suhd t a mquot are tewm tmesthe mass uf a sthgte eteetmh Thts tmphes a break frequency Wu ur three decades tewer thah thatter eteetmhte petahzatteh E teetmhaehete Fte ds 74 This is just what is typically seen ionic polarization begins leaving in the vicinity of infrared IR frequencies beginning around 1012 to 1013 Hz orientational polarization associated with the movement of the permanent electric dipole of a polar molecule This mechanism begins leaving in the microwave regions a few GHz Water is a good example of the frequency dependence of orientational polarization and its subsequent affect on permittivity as frequency varies At DC water has a relative permittivity of 80 and at 100 GHz its relative permittivity is around 10 Dielectric strength and dielectric breakdown g A related property is the material39s dielectric strength A material s dielectric strength is the maximum electric field that the material can withstand without damage At sufficiently strong electric fields electrons break their bonds become free and then accelerate due to the applied field At this point the material no longer is an effective insulator Moreover these electrons can in turn collide with other bound electrons breaking their bonds Once their bonds are broken these newly free electrons accelerate under the strong electric field and when they collide with a neighboring molecule other electrons become free This chain reaction can very quickly create large numbers of free electrons The process is referred to as dielectric breakdown The heat generated by these energetic scattering events usually results in catastrophic damage to the dielectric material air glass quanz polystyrene mica Electromagnetic Fields dielectric strength K Vlm 3106 30106 40106 50106 200106 75 Capacitance A capacitance exists between any two conductors separated by an insulator The term capacitor comes from its capacity to store electric charge The capacitance of a capacitor is depends on the geometry of the conductors and the physical properties of the insulator in particular its permittivity The capacitance of a capacitor is equal to the ratio of the charge on the plates to the voltage across its plates Below is a parallel plate capacitor with air or vacuum as the insulator charge stored on upper plate area of one lew of the plates q A C 1 d air or vacuum so E V V 39 voltage plate separation T q across plates charge stored on lower plate In circuits the capacitor is thought of in terms of its currentvoltage relation dj cvc iC dt dt dt dt where C is assumed to be independent of time What happens when some material other than air or vacuum is placed between capacitance plates The material will polarize in response to the electric field which will act to reduce the total electric field charge stored on upper plate area of one l of the plates q A d Ep s 1 v q 39 voltage L 1 plate separation Polarized bound across p a as charge ared charge producing Ep on lower plate Since the voltage between the plates is the line integral of the total electric field V will be reduced for a given charge q stored on the plates The presence of a dielectric increases the capacitor s capacitance compared to that when air or vacuum is the insulator Electromagnetic Fields 76 Calculation of capacitance For the calculation of capacitance Gauss law w q is used to express the capacitance in terms of electric flux rather than charge l D Dds cistored 39om conductor C on conductor conductor surface between between I E D conductors conductors between conductors Exampleparallel plate capacitor g The assumption in the parallel plate model is 2 that the plate separation is considered 2 d 1 I A sufficiently small to be negligible compared to a v plate area 2 q 39 The plate area is very large so that the fields in the interior can be assumed to be those that would be obtained from infinite planes of charge Then if the plate area is large compared to the plate separation the region in which these approximations do not hold which would be along the edges where the plates do not appear as approximately infinite is small compared to the total plate area These small effects the fringing fieldsquot or edge effects are neglected in the parallel plate model This assumption allows one to consider the charge to be evenly distributed on the plates and to consider the field lines to run straight between the plates Solution outline The surface charge density on the upper plate can be calculated q ps K Assuming the plates to be PECs the electric flux density vector and the electric field vector can be calculated between the plates using Gauss law or using results obtained when discussing boundary conditions D p re ltazgt E ltazgt Knowing the electric field the potential difference V can be found from a line integral 0 0 q q 0 qd V EDdl a Ddza z 7 2L LA 2 2 Ash Ag C ig V qdAg of Electromagnetic Fields 77 Example cylindrical capacitor g Find the capacitance perunitlength of the capacitor shown Give all calculations in detail all integrals with limits vector notation used properly and all steps justified permittivity 5 Hints 1 assume linear charge density pL on inner conductor 2 use Gauss law to find D and E between plates 3 take line integral of E between plates to find V 4 take ratio to obtain C Electromagnetic Fields 78 Incremental capacitor g The element relationship for the incremental capacitor is the ratio of the charge using Gauss equivalent to electric flux to potential difference AC 7 7 AV EM EM 8AA l AC l M 51 The incremental capacitor provides an excellent means by which to explore the reasons behind the differences in how resistances and capacitances combine in series and in parallel Let s start with the incremental capacitor Suppose Capacitors in parallel two are in parallel with their ends on the same mm A1p2 equipotential surfaces How do they combine Individually we have 3 AV AC and AC2 SZMZ M M what is their combined capacitance 7 lt 361111 Ac amp D1AA1 DZAA2 81EAA182EAA2 AV EM EM Capacitors in series AC 81AA12 M2 AC AC2 Capacitance in parallel add just as we knew they did Now how about capacitors in series Aw DAA DAA AV E1A E2N 2M12A 81 81 AC4 7qu MI AC AC3921 AA 81M SZAA Electromagnetic Fields 79 Compare these rules for combining capacitances to those for combining resistances Capacitance Resistance Equivalent capacitance for parallel Equivalent resistance for parallel capacitances is the sum of the individual resistances is the reciprocal of the sum of capacitances the reciprocals Equivalent capacitance for series Equivalent resistance for series capacitances is the reciprocal of the sum resistances is the sum of the individual of the reciprocals resistances What are the reasons behind these differences The reasons are in the definitions of resistance and capacitance Resistance is potential difference over flux current while capacitance is flux electric flux over potential difference R potential difference flux electric flux flux current potential difference The flux for elements in parallel add and the potential difference is common For capacitance the flux is in the numerator and the potential difference is in the numerator Therefore capacitances in parallel add For resistance the flux current is in the denominator and the potential difference is in the numerator Therefore the reciprocal of the individual resistances add their sum being the reciprocal of the equivalent resistance The potential for elements in series add and their flux is common For resistance the potential is in the numerator and the flux is in the numerator Therefore resistances in series add For capacitance the potential is in the denominator and the flux is in the numerator Therefore the reciprocal of the individual capacitances can be added their sum being the reciprocal of the equivalent capacitance Electromagnetic Fields 80 Persquare capacikance uppuse the maementa capamtance stvvur Aw dwmensmna m eparamerwuh a thmknesst 4 AW mms case AC w l w Suppuse funher thatAl M m mm ease cn AC El 39 Aw Examp urvilinear squares forirregularZD capacitors Asfur resws tances ms appruach can be generahzed fur capamtances thh genera erussseeuuns Fur rregu ar shapes perfem squares are nut pusswb e anu curvmnear squares havmg an average vwdth Equa m an average hewght are used The ppunuanes ufthese uMhnEar squares are husen su matmeu sues are paraHE m the ux and thaw ends are Eqmputentwa surfaces ufthe 2D capamtance Shawn llux lines equipmmill audios can be seen that the reswsiance ufthe 2D reswstance s CZE5quot45quot4quot 13421 aemmaenmrems 3 General case of nonhomogeneous permittivity Case three Boundaries neither parallel nor perpendicular to electric flux lines Consider twodimensional capacitors in xyz The general cell is shown below The basic relation used is Gauss law assuming an uncharged dielectric Consequently each region must satisfy wequot 0 Use the formula W Dgarea sEarea Looking at the right surface Vo VR 8g 8 a Woutright A 1 2 4 2 V V Wout right o 2 R t81 84 Performing similar calculations at the upper the left and the bottom surfaces we would obtain wont tamp 84 t8182 H82 83 t83 84 0 Rearranging one obtains 2VO8182 83 84 VR8184 VU8182 VL82 83 VB83 84 0 Solving for V0 V VUS182 VL8283 VB83S4 VRS1S4 O 251525354 Electromagnetic Fields 82 This is the equation that can be used to find the potentials in a twodimensional inhomogeneous capacitor The 2D case is importance since the majority of capacitors are in integrated circuits the vast majority of which are 2D capacitors fabricated with planar deposition processes As with the resistive case this technique can readily be generalized to the general inhomogeneous 3D capacitor Consider the two dimensional capacitor below The permittivity of the shaded regions is 48 and that of the unshaded regions is s O39I l l l 4444 1f A 5 03 The solution will involve writing 8 node equations Begin with N1 4v1 1 v v5 2v2 outside ED ltlt 5 an ltlt 8 The other 7 node voltage equations 391 j 5 n2 14V2 21V 2V1 5V6 5V3 n3 14V3 21V 5V2 5V7 3V4 n4 4V2 1V 2V3 VB n5 4V5 1V 2V6 To find capacitance divide total electric flux by the applied voltage 139V1 139V4 i stT1V21V3T b1 1v OR W V5 V3 C i t 725V25V 7 b2 1v 82 6 7 2 Electromagnetic Fields 83 Energy density Coulomb s law has been used to show the electrostatic potential energy in the interaction between two charges is We qq 47rsR Where R is the distance between the charges For multiple charges the energy can be expressed as q q 1 q 1 W I J 7 I J e iltj47rs R 4778 R In terms of distributions 1 prpr39 we Em39m39mdvdv we gwmrwamv gmw DD Vrdv using the vector identity V D VD VV DD D D we gmw DD Vrdv awn dv gj omvw dv using the divergence theorem for the first integral 1 1 we Es ceVD Dds 5WD DVV dv for p 0 only within afinite region thenV O at least as 1 and D Oat least as 12 r r which results in the first integral going to zero as r a 00 e W JJDDVVdvJUDDEdv where we is the electrostatic energy density We Other forms given that D 8E 18E2 iD2 Wei 2 28 Electromagnetic Fields 84 Example calculation of capacitance using energy g Energy storage provides another path by which capacitance can be calculated From electrical circuits We 1CV2 C 2 vye 2 V In terms of fields 1 2 i D D E dv D D E dv villi gt will gt C 2 2 j E Ddl j E Ddl between between plates plates Taking the parallelplate capacitor as an 2 example where the fields have already 2 d 1 I A been calculated using the parallel plate a v model assumptions and Gauss law 2 0 q I 39 q q D a and E a A 2 As 2 D DEdv Roma q 412 Ad q7ch volume A 8 j Emdl2 hazwdzazf dj d between p ates i d C Electromagnetic Fields 85 Energy and force g lfW is the total system potential energy and is a function of some dimension l dW F Ddl The force can be seen as a change in potential energy per unit length The direction of the force is determined from whether energy increases or decreases with the direction of movement The force exerted by the field is in the direction of energy decrease As discussed in lecture F VW Consider the parallel plate capacitance Since the plates are oppositely charged it is evident that there will be an electrostatic attraction between the plates after all opposite charges attract Two cases will be considered The first is the case of constant stored charge during movement The second is the case of the charge stored on the plates re We can find it by using this idea of energy 121lt12q2 We CV C 2 2 C 20 Case one is the case where the charge stored does not change The electrical energy stored changes as the plates move The electrical force exerted by the electrodes is a a a q2 q2 F VW aiaia77 7a e e Xax Yay ZOZJZSAY 25A y Another look Consider this same problem this time from the viewpoint of energy conservation Considerthe capacitor to be a system With constant charge on the plates energy energy can enter the system via the mechanical force where the mechanical force Fm mustjust balance the electrical force Fe Fm between the plates dWe 1q dV 1q2 Fmdy F 2 2 C q A e 2 q dy Fm dy E Fm iv 28A l q q2 F F a 7 m y28A e Electromagnetic Fields 86 1n 111s ease s1nee tne yunage rerna1ns unsiant energy 1s bES L expressed as m terms ur yunage Here the mm e1eetne energy 1 memes that supphed bythe suume gvve 7y am y dq aw ygm 7 Same problem another look Cuns1der energy eunseryanun W1ththe capamtur as the sysiem 11 1g aH 1111 1 an 1 the e1emn531furce Fe rFm dWep WVW dWmn qu yewde 1 77y g F g 2 u m y 1 1 Wyn wacv 2 u 2 y j Fm dy v my2 1 1eAy2 E y dyFmdy 2 FerFmraV 2 y2 Tne same reasumng can be used to ana1 ze the gressures at 1nrerraees 1nterraee nurrna1 tuf 1nterface1s unena 1e1g11nes Assummg thatthe rged1 Tnererure as tne mterface 1s rnuyeg tne ux gensny 1 L in enarge g1s euns1ant asamnaenene F1216 ln ferms cif energy dEnSltlES 2 2 q q U D2 e A r A777 3V 252A zeA2 3V 2g 2g Interfacetangemialto eld lines lnfnis e eleciric field duesn t cnange angi since fne plate Separatlun is alsci uns iant fne ycilfage is uns iant gm evdq dq1qu qu1 in terms cif energy dEnSltlES v1 i v 7 W crawl Eierei j nu F a Physics offorces involving dielectrics i Elecinc pressure is girecieg frcirn nigner perrniniyifyfci lciyyer perrniniyify Tne rnaterial vnfn lne nigner permittivity tendstu Expand aune Expense cifl e rnaterial vnfn lciwer permittivity Tne buundary pressure at fne interface is egual lei fne difference in energy densities Other applications and to look further Llrlktu rnany lrltErES Llrlg applicaticins nnp vwwv electrcislatic ccirn Linka sirnulaficins cif elecircisfafic applicaficins using CST Ela mmagnelll rieius Electrostatic actuation and sensing The above discussion provides some background for electric actuation How is this put to use Comb drives use electrostatic actuation coupled with spring elements to produce linear motion by varying a voltage 7 restoring spring element COMB DRWE Comb drives with gears can produce rotary motion a microengine 3 0 Why are two comb drives necessary in the microengine Sandia s website shows devices giving outofplane motion httn39lwww andia 39 39 39 39 html Electromagnetic FleldS 89 Exercise Pullin A wellknown phenomenon of electrostatic actuators is pullin in which the two plates suddenly come together Consider the static case and explore pullin Work with your neighbors to develop and discuss a physical explanation yo is equilibrium point for V 0 electrical conduciors spring spring elemeni g V damp39ng element y 1 5A Fe ay y 2V2 F ay ky y Fe sA22y2 Fe AW2y2 Electromagnetic Fields 90 Tl mirror array w J u H Over Ewan Hm w W New apphcatmns m dwgwta hght prunessmg DLF39 quotmmaza TEXASNSYRUMENYSM CROVMWROR m1 MNWW m um RF MEMS Switches u 51am that can suhd state sthches M warm RFMEMS THEORV Dsswemmo TECHNOLOGV av mama M REBsz PuBL vmv Corona Motor Sumr mamdc W Munrrm gurmnn M 5mm hams uh Mgr CORONA MoYoR mom SENSORSIND ACTUATORSIM 15 2267332 2mm aemamagnmnem 91 Energy Scavenging 7 spring anther L a 16 prnnfmlss Ly m w J LnJ W H electret ELECTROLET ENERGY SCAVENGING PEANOANDTAMBOSSO IEEE J OF MEMS 14 429435 2005 Sensors Capacitive sensors change their capacitance due to either changes in geometry plate separation plate area orand due to changes in permittivity TABLE OF CAPACITIVE SENSORS Capaci ve Principle of Operation Measurement Uses Sensors Pressure quot39 39 iuimeu uy iuiming pressure closed cavities which contain capacitor conductors typically one located on a membrane exposed to pressure to be measured Varying the pressure varies the plate separation and so varies the capacitance Displacement Geometry varies with One plate is usually xed with the other movable displacement Capacitance displacement sensors are made which have resolutions of less than 0 Accelerometer Geometry varies with Vibration control in harddisk drives acceleration Vibration detection In various consumer products MEMSbased capacitive accelerometers are widely used to deploy automobile air bags Electromagnetic Fields 92 Fruit Fly Sensor PCB t fs win bonded force sensor a g9 6 frmu springs sensor probe UV cum glue FRUW FLV ELECTROSTAT C COME SENSOR SUN ET AL IEEE J OF MEMS14411 2005 E Ectrumagnetm Hams 93
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