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Example 3 In order to illustrate the application of (11), we find E at P(1, 1, 1) caused by four identical 3-nC (nanocoulomb) charges located at P₁(1,1,0), P₂(-1,1,0), P3(-1,-1,0), and P4 (1,-1,0), as shown in Figure 2.4.

Example 3 In order to illustrate the application of (11), we find E at P(1, 1, 1) caused by four identical 3-nC (nanocoulomb) charges located at P₁(1,1,0), P₂(-1,1,0), P3(-1,-1,0), and P4 (1,-1,0), as shown in Figure 2.4.

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
Example 3 In order to illustrate the application of (11), we find E at P(1, 1, 1) caused by four P₂(-1,1,0), identical 3-nC (nanocoulomb) charges located at P₁(1,1,0), P3(-1,-1,0), and P4 (1,-1,0), as shown in Figure 2.4. P4 (1,-1,0) Solution: Or X E = 26.96 az 1 26.96 / 1 + 1 = + 12 P3 (-1,-1,0) r-r4 r-r3 2ax + az √5 P(1, 1, 1) We find that r = ax + ay + a₂, r₁ = ax + ay, and thus r- r₁ = az. Thus magnitudes are: Ir-r₁ = 1, |r-r₂| = √√5, |r-r3|= 3, and [r - r4|= √5. Because Q/4€ 3 x 10-⁹/(4 x 8.854 x 10-¹2) = 26.96 V m, we may now use (11) to obtain 1 (√5)² r-r₁ P₁ (1, 1, 0) r-r2 . P₂ (-1, 1, 0) + 2a + 2ay + az 1 2ay + az 3 + 32 E = 6.82ax + 6.82a, + 32.8a, V/m 1 √5 (√5) =
Transcribed Image Text:Example 3 In order to illustrate the application of (11), we find E at P(1, 1, 1) caused by four P₂(-1,1,0), identical 3-nC (nanocoulomb) charges located at P₁(1,1,0), P3(-1,-1,0), and P4 (1,-1,0), as shown in Figure 2.4. P4 (1,-1,0) Solution: Or X E = 26.96 az 1 26.96 / 1 + 1 = + 12 P3 (-1,-1,0) r-r4 r-r3 2ax + az √5 P(1, 1, 1) We find that r = ax + ay + a₂, r₁ = ax + ay, and thus r- r₁ = az. Thus magnitudes are: Ir-r₁ = 1, |r-r₂| = √√5, |r-r3|= 3, and [r - r4|= √5. Because Q/4€ 3 x 10-⁹/(4 x 8.854 x 10-¹2) = 26.96 V m, we may now use (11) to obtain 1 (√5)² r-r₁ P₁ (1, 1, 0) r-r2 . P₂ (-1, 1, 0) + 2a + 2ay + az 1 2ay + az 3 + 32 E = 6.82ax + 6.82a, + 32.8a, V/m 1 √5 (√5) =
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