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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, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: (12 x 10-9) (-2ax - 6a-) 4740 (4+36)1.5 = -3.9a, 12.4a, - 2.5a, V/m E= + (3 x 10-9) (2ax - 3ay)" (4+9) 2.77 €0 (0.2 x 10-9)ax 2€0

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, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: (12 x 10-9) (-2ax - 6a-) 4740 (4+36)1.5 = -3.9a, 12.4a, - 2.5a, V/m E= + (3 x 10-9) (2ax - 3ay)" (4+9) 2.77 €0 (0.2 x 10-9)ax 2€0

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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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, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- [(12 x 10-⁹) (-2ax – 6az) 4740 (4+36)1.5 = -3.9a, 12.4a, - 2.5a, V/m E= == 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 + dW=-qE-dL=-4(400a, - 300ay +500a₂).. (4 x 10-³) √3 == dW=-qE dL = -4(400ax - 300ay +500a-).. (4 x 10-³) √14 W = -9 (3 x 10-) (2a,- 3ay) 27€0 (4+9) SE (400-300+500) = -1.39 J b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction: (-2ax +3ay-a-) √14 (10-³) -3 -6/²221 2]-[ (-800-900-500) = 2.35 J (0.2 x 10-9)ax 240 E.dL-6 10- Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, −1) in the field E = 2xax - - 3y²ay + 4a V/m along the straight-line path x = -3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path. (ax + ay + a) √√3 =-6 2xdx + 6 (10-³) −6 [ (2xax − 3y¹²ay + 4ª-) · (dxax +dyay+dza:) of ³y³²dy - 65 -6 4dz = -24J
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, 0.2 nC/m² at x = 2. The sum of the fields at the origin from each charge in order is: 9- [(12 x 10-⁹) (-2ax – 6az) 4740 (4+36)1.5 = -3.9a, 12.4a, - 2.5a, V/m E= == 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 + dW=-qE-dL=-4(400a, - 300ay +500a₂).. (4 x 10-³) √3 == dW=-qE dL = -4(400ax - 300ay +500a-).. (4 x 10-³) √14 W = -9 (3 x 10-) (2a,- 3ay) 27€0 (4+9) SE (400-300+500) = -1.39 J b) -2ax +3ay-a: The computation is similar to that of part a, but we change the direction: (-2ax +3ay-a-) √14 (10-³) -3 -6/²221 2]-[ (-800-900-500) = 2.35 J (0.2 x 10-9)ax 240 E.dL-6 10- Find the amount of energy required to move a 6-C charge from the origin to P(3, 1, −1) in the field E = 2xax - - 3y²ay + 4a V/m along the straight-line path x = -3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path. (ax + ay + a) √√3 =-6 2xdx + 6 (10-³) −6 [ (2xax − 3y¹²ay + 4ª-) · (dxax +dyay+dza:) of ³y³²dy - 65 -6 4dz = -24J
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