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Drill 2.5 [1] Two infinite uniform line charges of 5 nC/m charge density lie along x and y axis in free space. a) PA(0, 0, 4)? b) Ра(0, 3, 4)? Solution: a) At P,(0, 0, 4): Total E at PA(0, 0, 4) is E, + E RA = (0 - 0)a, +(0 - 0)a,+(4 - 0)a, = 4a, |RA = 16 = p for both line charges RA %3D a,= RA 4а, 16 for both line charges pl. = SnC/M pL E a, 2πεορ 5x10-9 4а, pl. = SnC/M E, 10 16 = 22.5a, V/m = E, (16) 36 E= 22.5a, + 22.5a, = 45a, V/m b) At Pg(0, 3, 4) : H.W. Ans. :10.8a, + 36.9a, V/m

Drill 2.5 [1] Two infinite uniform line charges of 5 nC/m charge density lie along x and y axis in free space. a) PA(0, 0, 4)? b) Ра(0, 3, 4)? Solution: a) At P,(0, 0, 4): Total E at PA(0, 0, 4) is E, + E RA = (0 - 0)a, +(0 - 0)a,+(4 - 0)a, = 4a, |RA = 16 = p for both line charges RA %3D a,= RA 4а, 16 for both line charges pl. = SnC/M pL E a, 2πεορ 5x10-9 4а, pl. = SnC/M E, 10 16 = 22.5a, V/m = E, (16) 36 E= 22.5a, + 22.5a, = 45a, V/m b) At Pg(0, 3, 4) : H.W. Ans. :10.8a, + 36.9a, V/m

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Drill 2.5 [1] Two infinite uniform line charges of 5 nC/m charge density lie along x and y axis in free space. a) PA(0, 0, 4)? b) Pg(0, 3, 4)? Solution: a) At P,(0, 0, 4): Total E at PA(0, 0, 4) is E, + E2 RA 3 (0 - 0jа, +(0- 0ја, +(4 - 0ја, 4а, |RA = 16 = p for both line charges R a, = R 4a 16 for both line charges pl. = SnC/M pL E a, 2πεορ 5x10-9 4а, pl. = SnC/M E, 10 16 = 22.5a, V/m = E, (16) 36R E = 22.5a, + 22.5a, = 45a, V/m b) At Pg(0, 3, 4) : H.W. Ans. :10.8a, + 36.9a, V/m
Transcribed Image Text:Drill 2.5 [1] Two infinite uniform line charges of 5 nC/m charge density lie along x and y axis in free space. a) PA(0, 0, 4)? b) Pg(0, 3, 4)? Solution: a) At P,(0, 0, 4): Total E at PA(0, 0, 4) is E, + E2 RA 3 (0 - 0jа, +(0- 0ја, +(4 - 0ја, 4а, |RA = 16 = p for both line charges R a, = R 4a 16 for both line charges pl. = SnC/M pL E a, 2πεορ 5x10-9 4а, pl. = SnC/M E, 10 16 = 22.5a, V/m = E, (16) 36R E = 22.5a, + 22.5a, = 45a, V/m b) At Pg(0, 3, 4) : H.W. Ans. :10.8a, + 36.9a, V/m
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