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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: E= (12 x 10-9) 4.740 (-2a,-6a₂) - ¹)]-[(0.² (4+36)13+(3× 10-9) (2ax − 3a,)] (4+9) = -3.9a, - 12.4a, -2.5a, V/m [(0.2 x 10-9)ax 240

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: E= (12 x 10-9) 4.740 (-2a,-6a₂) - ¹)]-[(0.² (4+36)13+(3× 10-9) (2ax − 3a,)] (4+9) = -3.9a, - 12.4a, -2.5a, V/m [(0.2 x 10-9)ax 240

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
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
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: E= x 2]-[(0.2 (4+36)1.3+(3× 10-9) (2ax - 3a,)] (4+9) [(12 % (12 x 10-9) (-2a,- 6a:)] 4740 =-3.9a, -12.4a, -2.5a, V/m (0.2 x 10-)ax 240
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: E= x 2]-[(0.2 (4+36)1.3+(3× 10-9) (2ax - 3a,)] (4+9) [(12 % (12 x 10-9) (-2a,- 6a:)] 4740 =-3.9a, -12.4a, -2.5a, V/m (0.2 x 10-)ax 240
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