9- Let E = 400ax - 300ay +500a; in the neighborhood of point P(6. 2, -3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by: a) ax + ay+a: We write dW=-qE-dL=-4(400a, -300ay +500a.).. (4 x 10-³) √√3 (400-300+500) = -1.39 J (a+ay+a) (10-3) √√3 b) -2ax +3ay-a:: The computation is similar to that of part a, but we change the direction: (-2ax + 3ay - a-) dW = -qE. dL = -4(400a, - 300ay +500a-).. (4 x 10-³) √14 (-800-900-500) = 2.35 J (10-³)
9- Let E = 400ax - 300ay +500a; in the neighborhood of point P(6. 2, -3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by: a) ax + ay+a: We write dW=-qE-dL=-4(400a, -300ay +500a.).. (4 x 10-³) √√3 (400-300+500) = -1.39 J (a+ay+a) (10-3) √√3 b) -2ax +3ay-a:: The computation is similar to that of part a, but we change the direction: (-2ax + 3ay - a-) dW = -qE. dL = -4(400a, - 300ay +500a-).. (4 x 10-³) √14 (-800-900-500) = 2.35 J (10-³)
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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(electromagnetic field)I want a detailed solution because my teacher changes the numbers. I want a detailed solution. Understand the solution
Transcribed Image Text:9-
Let E = 400ax - 300ay +500a; in the neighborhood of point P(6. 2, -3). Find the incremental work
done in moving a 4-C charge a distance of 1 mm in the direction specified by:
a) ax +ay+a:: We write
dW=-qE-dL=-4(400a, -300ay +500a.)..
(4 x 10-³)
√√3
(400-300+500) = -1.39 J
(a+ay+a) (10-3)
√√3
b) -2ax +3ay-a:: The computation is similar to that of part a, but we change the direction:
(-2ax + 3ay - a₂)
dW = -qE. dL = -4(400a, - 300ay +500a-)..
(4 x 10-³)
√14
-(-800-900-500) = 2.35 J
(10-³)
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