No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]
Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, Urbana, Illinois, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India 3.2 Gauss’ Laws and the Continuity Equation 3.2-3 GAUSS’ LAW FOR THE ELECTRIC FIELD S D • dS V dv z (x, y, z) x Dx x x y z Dx x y z Dy z x Dy z x y y y Dz z z x y Dz z x y x y z z y y x 3.2-4 D x x x Dx x y z Dy Dy Δ z Δ x y +Δy y Lim x y z Dz z z Dz z x y x y z x y z Lim x x y z y z 3.2-5 Dx Dy Dz x y z Longitudinal derivatives of the components of D • D Divergence of D = Ex Given that 0 for – a x a 0 otherwise Find D everywhere 3.2-6 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • x=–a x=0 x=a Noting that = (x) and hence D = D(x), we set and 0, so that y z Dx Dy Dz Dx • D x y z x 3.2-7 Thus, • D = gives Dx (x) x which also means that D has only an xcomponent Proceeding further, we have x Dx – x dx C where C is the constant of integration Evaluating the integral graphically, we have the following: 3.2-8 –a a x x – ( x ) dx 0 a –a 0 a x From symmetry considerations, the fields on the two sides of the charge distribution must be equal in magnitude and opposite in direction Hence, C = – 0a 3.2-9 Dx 0 a –a a x – 0a – 0 a a x D 0 x a x 0 a a x for x –a for – a x a for x a 3.2-10 GAUSS’ LAW FOR THE MAGNETIC FIELD D • dS = dv S From analogy V • D S B • dS = = V dv • B 0 • B 0 Solenoidal property of magnetic field lines Provides test for physical realizability of a given vector field as a magnetic field 3.2-11 LAW OF CONSERVATION OF CHARGE d dv 0 J • dS S dt V • J t ( ) 0 • J t 0 Continuity Equation 3.2-12 SUMMARY B E – t D H J t • D (1) (2) (3) • B 0 (4) • J 0 t (5) (4) is, however, not independent of (1), and (3) can be derived from (2) with the aid of (5)