University of AnbarCollege of EngineeringDepartment of Electrical EngineeringEx:-A point charge Ql=2Mc is located at P1(-3,7,-4) in free space whileQ2=-5Mc is at P2(2,4,-1) at the point P3(12,15,18) fund Ē, JE|, ãE theElectro-magnetics IDr. Ahmed A. AbbasThen R1 = 15āx + 8ãy + 22āzForce on 25nc at this point (P3).|Ř1|= V152 + 8² + 22² = 27.815āx8ãy22āzā, == 0.539āx + 0.287āy + 0.79āz27.827.827.82 x 10-6. Ē1 = 9 x 10(0.539āx + 0.287ãy + 0.79āz)773Ē2 = K-Q2z āz but R2 = 10āx + 11āy + 19āz|R| = V102 + 112 + 192 = 24.1-5 x 10-6 1011. E2 = 9 x 10°19az24.1-ах +-āy +58224.124.1Can you explain thisstep in detail?!.E = Ē1+ Ē2 = -19.5 āx – 28.5 ãy – 42.4āz= -0.356 āx - 0.521 ãy - 0.776 āzF = QË

It is asked to explain the step of addition of electric field at a point. To show the calculation…

Ex:-A point charge Ql=2Mc is located at P1(-3,7,-4) in free space while Q2=-5Mc is at P2(2,4,-I) at the point P3(12,15,18) find Ē, JE], ãE the University of Anbar College of Engineering Department of Electrical Engineering Electro-magnetics I Dr. Ahmed A. Abbas Then R1 = 15āx + 8ãy + 22āz Force on 25nc at this point (P3). |Ř1|= V152 + 8² + 222 = 27.8 15āx 8ãy a, = IRI 22āz = 0.539āx + 0.287āy + 0.79āz 27.8 it 27.8 27.8 2 x 10-6 . E1 = 9 x 10 (0.539āx + 0.287ãy + 0.79āz) 773 Q2 -āz but R = 10äx + 11āy + 19āz R Ē2 = K- |R| = V102 + 112 + 192 = 24.1 -5 x 106 10 11 19 . Ē2 = 9 x 10° āx + āy + 582 (24.1 24.1 24.1 Can you explain this step in detail?! :E = Ē1+ Ē2 = -19.5 āx – 28.5 ãy – 42.4āz = -0.356 āx-0.521 ãy - 0.776 āz F = QË

Ex:-A point charge Ql=2Mc is located at P1(-3,7,-4) in free space while Q2=-5Mc is at P2(2,4,-I) at the point P3(12,15,18) find Ē, JE], ãE the University of Anbar College of Engineering Department of Electrical Engineering Electro-magnetics I Dr. Ahmed A. Abbas Then R1 = 15āx + 8ãy + 22āz Force on 25nc at this point (P3). |Ř1|= V152 + 8² + 222 = 27.8 15āx 8ãy a, = IRI 22āz = 0.539āx + 0.287āy + 0.79āz 27.8 it 27.8 27.8 2 x 10-6 . E1 = 9 x 10 (0.539āx + 0.287ãy + 0.79āz) 773 Q2 -āz but R = 10äx + 11āy + 19āz R Ē2 = K- |R| = V102 + 112 + 192 = 24.1 -5 x 106 10 11 19 . Ē2 = 9 x 10° āx + āy + 582 (24.1 24.1 24.1 Can you explain this step in detail?! :E = Ē1+ Ē2 = -19.5 āx – 28.5 ãy – 42.4āz = -0.356 āx-0.521 ãy - 0.776 āz F = QË

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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