MIT OpenCourseWare http://ocw.mit.edu Electromechanical Dynamics For any use or distribution of this textbook, please cite as follows: Woodson, Herbert H., and James R. Melcher. Electromechanical Dynamics. 3 vols. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-NonCommercial-Share Alike For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsAppendix BREVIEW OFELECTROMAGNETIC THEORYB.1 BASIC LAWS AND DEFINITIONSThe laws of electricity and magnetism are empirical. Fortunately they canbe traced to a few fundamental experiments and definitions, which are re-viewed in the following sections. The rationalized MKS system of units isused.B.1.1 Coulomb's Law, Electric Fields and ForcesCoulomb found that when a charge q (coulombs) is brought into the vicinityofa distribution of chargedensity p,(r') (coulombs per cubic meter), as shownin Fig. B.1.1, a force of repulsion f (newtons) is given byf = qE,(B. 1.1)where the electricfield intensity E (volts per meter) is evaluated at the position= qEFig. B.1.1 The force f on the point charge q in the vicinity of charges with density Pe(r')is represented by the electric field intensity E times q, where E is found from (B.1.2).Review of Electromagnetic Theoryr of the charge q and determined from the distribution of charge density byE(r) = e(r') -r) dV'. (B.1.2)4E(r) e =r -r'ljIn the rationalized MKS system of units the permittivity eo of free space isqo = 8.854 x 10- 12 _ --X 10- 9 F/m. (B.1.3)367rNote that the integration of (B.1.2) is carried out over all the charge dis-tribution (excluding q), hence represents a superposition (at the location rof q) of the electric field intensities due to elements of charge density at thepositions r'.A I-U h Ud s an exampp , suppose tlatiLe cargeLdistribution p,(r') is simply a point chargeQ (coulombs) at the origin (Fig. B.1.2);that is,p,= Q 6(r'), (B.1.4)where 6(r') is the deltafunction defined byqQI,.1__W4Xeo-]-r0(r')= 0, r' # 0, Fig. B.1.2 Coulomb's law for pointcharges Q (at the origin) and q (atS6(r') dV' = 1. (B.1.5) the position r).For the charge distribution of (B.1.4) integration of (B.1.2) givesE(r) = Qr (B.1.6)4rreo Ir"Hence the force on the point charge q, due to the point charge Q, is from(B. 1.1)f = qQr (B.1.7)4 ore0 Irl"This expression takes the familiar form of Coulomb's law for the force ofrepulsion between point charges of like sign.We know that electric charge occurs in integral multiples of the electroniccharge (1.60 x 10- 19 C). The charge density p., introduced with (B.1.2), isdefined asPe(r) = lim - I q,, (B.1.8)av-o 61 i__ _Appendix Bwhere 6V is a small volume enclosing the point r and Z, q, is the algebraicsum of charges within 6V. The charge density is an example of a continuummodel. To be valid the limit 6V --0 must represent a volume large enough tocontain a large number of charges q1,yet small enough to appear infinitesimalwhen compared with the significant dimensions of the system being analyzed.This condition is met in most electromechanical systems.For example, in copper at a temperature of 200C the number density offree electrons available for carrying current is approximately 1023 electrons/cm3.If we consider a typical device dimension to be on the order of 1 cm,a reasonable size for 6V would be a cube with 1-mm sides. The number ofelectrons. in 6 Vwould be 10", which certainly justifies the continuum model.The force, as expressed by (B.I.1), gives the total force on a single testcharge in vacuum and, as such, is not appropriate for use in a continuummodel of electromechanical systems. It is necessary to use an electricforcedensity F (newtons per cubic meter) that can be found by averaging (B.1.1)over a small volume.F = lim = lim I qjEj (B.1.9)av-o 6V 6v-o 6VHere q, represents all of the charges in 6V, E, is the electric field intensityacting on the ith charge, and f, is the force on the ith charge. As in the chargedensity defined by (B.1.8), the limit of (B.1.9) leads to a continuum model ifthe volume 6V can be defined so that it is small compared with macroscopicdimensions of significance, yet large enough to contain many electroniccharges. Further, there must be a sufficient amount of charge external to thevolume 6V that the electric field experienced by each of the test charges isessentially determined by the sources of field outside the volume. Fortunatelythese requirements are met in almost all physical situations that lead to usefulelectromechanical interactions. Because all charges in the volume 6 V ex-perience essentially the same electric field E, we use the definition of freecharge density given by (B.1.8) to write (B.1.9) asF = p,E. (B.1.10)Although the static electric field intensity E can be computed from (B.1.2),it is often more convenient to state the relation between charge density andfield intensity in the form of Gauss'slaw:soEE.n da = Pe dV. (B.1.11)In this integral law nis the outward-directed unit vector normal to the surfaceS, which encloses the volume V. It is not our purpose in this brief review toshow that (B.1.11) is implied by (B.1.2). It is helpful, however, to note thatReview of Electromagnetic TheoryFig. B.1.3 A hypothetical sphere of radius r encloses a charge Q at the origin. The integralof eoE, over the surface of the sphere is equal to the charge Q enclosed.in the case of a point charge Q at the origin it predicts the same electricfield intensity (B.1.6) as found by using (B.1.2). For this purpose the surfaceS is taken as the sphere of radius r centered at the origin, as shown in Fig.B.1.3. By symmetry the only component of E is radial (E7), and this is con-stant at a given radius r. Hence (B.1.11) becomes47rrEEo= Q. (B.1.12)Here the integration of the charge density over the volume V enclosed by Sis the total charge enclosed Q but can be formally taken by using (B. 1.4) withthe definition provided by (B.1.5). It follows from (B.1.12) thatE, = 4rEr, (B.1.13)a result that is in agreement with (B.1.6).Because the volume and surface of integration in (B.1.11) are arbitrary,the integral equation implies a differential law. This is found by making useof the divergence theorem*A .nda = V.AdV (B.1.14)to write (B.1.11) asfv(VU .oE -P) dV = 0. (B.1.15)* For a discussion of the divergence theorem see F. B. Hildebrand, Advanced CalculusforEngineers, Prentice-Hall, New York, 1949, p.
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