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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 1.5 The Electric Field 1.5-2 The Electric Field is a force field acting on charges by virtue of the property of charge. Q1Q2 Coulomb’s Law F1 a 21 F2 2 a12 4 R 0 Q 2 R F1 a21 Q1 Q2 Q1 F2 a12 2 4 0 R 0 permittivity of free space 10 –9 F/m 36 1.5-3 D1.13(b) Q 2a a Q Q2/40(2a2) Q2/40(4a2) 2/4 (2a2) Q 0 Q Q 4 0 Q From the construction, it is evident that the resultant force is directed away from the center of the square. The magnitude of this resultant force is given by 1.5-4 2 Q 2 4 0 2a 2 1 1 2 2 2a 4a 0.957 2 N a cos 45 Q 2 4 0 4a 2 1.5-5 Electric Field Intensity, E is defined as the force per unit charge experienced by a small test charge when placed in the region of the field. F E E Lim qE q0 q –q q –qE Thus Fe qE Units: N N–m V C C•m m Sources: Charges; Time-varying magnetic field 1.5-6 Electric Field of a Point Charge Qq F a 2 R 4 0 R (Coulomb’s Law) q Q q a 2 R 4 0 R q E due to Q E due to Q R Q Q 4 0 R 2 aR aR 1.5-7 Constant magnitude surfaces are spheres centered at Q. Direction lines are radial lines emanating from Q. E aR R Q E due to charge distributions (a) Collection of point charges Q1 E n Qj 2 4 R 0 j j 1 a Rj aRn R1 R2 R3 Q2 Q3 Qn Rn aR3 aR2 aR1 1.5-8 Ex. z Q (> 0) d d2 + x2 a e d Q (> 0) y d2 + x2 x Electron (charge e and mass m) is displaced from the origin by D (<< d) in the +x-direction and released from rest at t = 0. We wish to obtain differential equation for the motion of the electron and its solution. 1.5-9 For any displacement x, F 2 Qe 4 0 d x 2 2 Qex 2 0 d 2 x 2 32 is directed toward the origin, and x D d. Qe x F– ax 3 2 0 d cos a ax ax 1.5-10 The differential equation for the motion of the electron is d2 x Qe x m 2 – dt 2 0 d 3 d2 x Qe x 0 dt 2 2m 0 d 3 Solution is given by Qe x A cos tB 3 2m 0 d 1.5-11 dx Using initial conditions x D and 0 at t = 0, dt we obtain x D cos Qe t 2m 0 d 3 which represents simple harmonic motion about the origin with period 2 Qe . 2m 0 d 3 1.5-12 (b) Line Charges Line charge density, rL (C/m) dl P dS (c) Surface Charges Surface charge density, rS (C/m2) (d) Volume Charges Volume charge density, r (C/m3) dv 1.5-13 Ex. Finitely-Long Line Charge z a rL rL0 4 0 C m dz z f x –a r2 z2 y r a E ar dE 2 E 1.5-14 rL dz 4 0 r z 2 2 4 0 a z 0 2 cos a ar rL 0r dz r 2 z 2 32 ar rL 0r 2 0 z 2 2 2 ar r r z z 0 rL 0a a 2 0 r r 2 a2 ar 2a r r 2 a2 r L0 For a , E ar 2 0 r ar 1.5-15 Infinite Plane Sheet of Charge of Uniform Surface Charge Density z a y2 z 2 z y rS0 x dy y 1.5-16 dEz 2 rS0 dy 2 0 y 2 z 2 cos a rS0 z dy 0 y 2 z 2 rS0 z Ez 0 rS0 z 0 rS0 2 0 dy y0 y 2 z 2 1 2 da a 0 z 1.5-17 rS0 E a z for z 0 2 0 rS0 an 2 0 z<0 rS0 – az 2 0 rS0 + z>0 + rS0 az 2 0 + + + z=0 z 1.5-18 D1.16 Given E(3,5,1) 0 V m E(1, – 2,3) 6a z V m E(3, 4,5) 4a z V m rS1 rS2 rS3 z=0 z=2 z=4 1.5-19 1 rS 1 rS 2 rS 3 0 2 0 1 rS 1 rS 2 rS 3 6 2 0 1 rS 1 rS 2 rS 3 4 2 0 Solving, we obtain (a) rS 1 4 0 C m 2 r 6 C m (b) S 2 0 2 (c) rS 3 2 0 C m 2 (d) E 2,1, 6 4az V m