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6- Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, -1) in free space. a) Find E at P(0.5, 0, 0): This will be Ep 120 x 10-⁹ Απερ Ep RAP RBP IRAP3 RBP|3 + where RAP = 0.5ax - a; and RBP = 0.5ax +az. Also, [RAP] = [RBP| = √1.25. Thus: 120 x 10-ax 4T (1.25)1.50 = 772 V/m

6- Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, -1) in free space. a) Find E at P(0.5, 0, 0): This will be Ep 120 x 10-⁹ Απερ Ep RAP RBP IRAP3 RBP|3 + where RAP = 0.5ax - a; and RBP = 0.5ax +az. Also, [RAP] = [RBP| = √1.25. Thus: 120 x 10-ax 4T (1.25)1.50 = 772 V/m

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
6- Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, -1) in free space. a) Find E at P(0.5, 0, 0): This will be 7- Ep = RAP|3+ RBP|3 where RAP = 0.5ax - a; and Rgp = 0.5ax +az. Also, [RAP| = |RBP] =√1.25. Thus: 120 x 10-ax 4x (1.25)1.50 Ep Q=10 120 x 10-⁹ 4лео 1.1x 0 1² 1² 1² (²-4) (². 9x The region in which 4 <r < 5,0 <0 < 25°, and 0.97 << 1.17 contains the volume charge density of p, 10(r-4) (r-5) sin 0 sin(p/2). Outside the region, P, = 0. Find the charge within the region: The integral that gives the charge will be = = 10(-3.39)(.0266)(.626)=0.57 C = 772 V/m = (r-4) (r-5) sin 0 sin(/2) r² sine dr de do 2-10-9+2020)-200 (9) 3 sin
Transcribed Image Text:6- Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, -1) in free space. a) Find E at P(0.5, 0, 0): This will be 7- Ep = RAP|3+ RBP|3 where RAP = 0.5ax - a; and Rgp = 0.5ax +az. Also, [RAP| = |RBP] =√1.25. Thus: 120 x 10-ax 4x (1.25)1.50 Ep Q=10 120 x 10-⁹ 4лео 1.1x 0 1² 1² 1² (²-4) (². 9x The region in which 4 <r < 5,0 <0 < 25°, and 0.97 << 1.17 contains the volume charge density of p, 10(r-4) (r-5) sin 0 sin(p/2). Outside the region, P, = 0. Find the charge within the region: The integral that gives the charge will be = = 10(-3.39)(.0266)(.626)=0.57 C = 772 V/m = (r-4) (r-5) sin 0 sin(/2) r² sine dr de do 2-10-9+2020)-200 (9) 3 sin
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