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Energy and Potential An electric field is expressed in rectangular coordinates by E = 6x²ax+6yay + 4az V/m Find: a. VN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2) b. VM if V=0 at Q(4,-2,-35) c. Vif V=2 at P(1, 2,-4).

Energy and Potential An electric field is expressed in rectangular coordinates by E = 6x²ax+6yay + 4az V/m Find: a. VN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2) b. VM if V=0 at Q(4,-2,-35) c. Vif V=2 at P(1, 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
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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
Energy and Potential • An electric field is expressed in rectangular coordinates by E = 6x²ax+6yay + 4az V/m • Find: a. VMN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2) b. VM if V=0 at Q(4,-2,-35) c. Vif V 2 at P(1, 2,-4). • Solution: • Exercise (a): • Ē.dī= (6x²ax + 6yay + 4az). (dxax + dyāy + dzāz) • Ē. dL = 6x² dx + 6ydy + 4dz • V = -√ E.dL . . Exercise (b): • E.dī= (6x¹ax + 6ya, +4āz). (dxax + dya, + dzā₂) • E.dL = 6x²dx + 6ydy + 4dz • VMO = -26x² dx + 2₂6ydy + 4dz -[2x³12-[3y²1₂-[42]-35 • VMO • VMN = -√²₂ 6x² dx + 6ydy + f₂¹4dz VMN-[2x³]23 - [3y2]63 - [4z]z¹ = VMN-70-81 + 12 • VMN-139 V ● ● VMO . . VMO -120 V . = - Exercise (c): 112-96-136 • E. dL = (6x²ax + 6yāy +4āz). (dxāx+ dya, + dzāz) • E. dL = 6x²dx + 6ydy + 4dz VNP −₁³6x²dx + √₂³ 6ydy + f²354dz VNP-[2x³]3- [3y²123- [42]24 VNP = 56 – 15 – 24 VNP = 17 V = • VMO = VM-Vo • VM-[16+ 108-4]-0 • VM = -120 V • VNP =Vn−Vp Vn =VNP+Vp . • VN=17+2 VN = 19 V
Transcribed Image Text:Energy and Potential • An electric field is expressed in rectangular coordinates by E = 6x²ax+6yay + 4az V/m • Find: a. VMN if points M and N are specified by M(2, 6,-1) and N(-3,-3, 2) b. VM if V=0 at Q(4,-2,-35) c. Vif V 2 at P(1, 2,-4). • Solution: • Exercise (a): • Ē.dī= (6x²ax + 6yay + 4az). (dxax + dyāy + dzāz) • Ē. dL = 6x² dx + 6ydy + 4dz • V = -√ E.dL . . Exercise (b): • E.dī= (6x¹ax + 6ya, +4āz). (dxax + dya, + dzā₂) • E.dL = 6x²dx + 6ydy + 4dz • VMO = -26x² dx + 2₂6ydy + 4dz -[2x³12-[3y²1₂-[42]-35 • VMO • VMN = -√²₂ 6x² dx + 6ydy + f₂¹4dz VMN-[2x³]23 - [3y2]63 - [4z]z¹ = VMN-70-81 + 12 • VMN-139 V ● ● VMO . . VMO -120 V . = - Exercise (c): 112-96-136 • E. dL = (6x²ax + 6yāy +4āz). (dxāx+ dya, + dzāz) • E. dL = 6x²dx + 6ydy + 4dz VNP −₁³6x²dx + √₂³ 6ydy + f²354dz VNP-[2x³]3- [3y²123- [42]24 VNP = 56 – 15 – 24 VNP = 17 V = • VMO = VM-Vo • VM-[16+ 108-4]-0 • VM = -120 V • VNP =Vn−Vp Vn =VNP+Vp . • VN=17+2 VN = 19 V
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