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8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: (12 x 10-) (-2a, -6a₂) [(3 x 10-9) (2a, - 3ay)] (4+36)1.5 + 4740 2740 (4+9) -3.9a, 12.4ay-2.5a, V/m 9- E= = dW=-qE-dL = -4(400a, - 300ay +500a₂). (4 x 10-³) √√3 (0.2 x 10-)ar 2€0 2]-[10.2 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+az: We write (400-300+500) = -1.39J (a +ay+a) √√3 (10-³)

8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: (12 x 10-) (-2a, -6a₂) [(3 x 10-9) (2a, - 3ay)] (4+36)1.5 + 4740 2740 (4+9) -3.9a, 12.4ay-2.5a, V/m 9- E= = dW=-qE-dL = -4(400a, - 300ay +500a₂). (4 x 10-³) √√3 (0.2 x 10-)ar 2€0 2]-[10.2 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+az: We write (400-300+500) = -1.39J (a +ay+a) √√3 (10-³)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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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
8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- E= = (12 x 10-) (-2a, -6a₂) 4740 (4+36)1.5 -3.9a, 12.4ay-2.5a, V/m + [(3 x 10-9) (2a, - 3ay)] 2740 (4+9) dW=-qE-dL = -4(400a, - 300ay +500a₂). (4 x 10-³) √3 = 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+az: We write (0.2 x 10-)ar 2€0 (400-300+500) = -1.39J 2]-[¹0.2 (a +ay+a) √√3 (10-³)
Transcribed Image Text:8- Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2.0.6); uniform line charge density, 3nC/m at x = -2, y = 3; uniform surface charge density, 0.2nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- E= = (12 x 10-) (-2a, -6a₂) 4740 (4+36)1.5 -3.9a, 12.4ay-2.5a, V/m + [(3 x 10-9) (2a, - 3ay)] 2740 (4+9) dW=-qE-dL = -4(400a, - 300ay +500a₂). (4 x 10-³) √3 = 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+az: We write (0.2 x 10-)ar 2€0 (400-300+500) = -1.39J 2]-[¹0.2 (a +ay+a) √√3 (10-³)
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