Energy and Potential • Exercise: Calculate the work done in moving a 4 C charge from B(1, 0, 0) to A(0, 2, 0) along the path y=2-2x, z=0 in the E-field: a. Ē = 5a, V/m b. Ē= 5xx V/m c. Ē = 5xax + 5ya, V/m • Solution: • W = -Q SĒ. dL dī= =dxax + dyay + dzāz (a): 5ax. (dxax + dyay + dzāz • Exercise E. dL • E. dL = 5dx • W = -4 ₁ 5dx • W = 20 [x]! • W = 20 J = • Exercise (b): . E.dL= • Ē.dī <= 5xdx • W = -4 ₂₁5xdx O ET. 5xax. (dxax + dyay + dzaz) • W = 20 • W = 10. J Exercise (c): Ē.dī = (5xax + 5yāy). (dxāx + dyāy + dzāz) • Ē.dL = 5xdx + 5ydy (₁5xdx + W = -4 • W = 20 20 (+1) 5ydy) • W = 20(0.5 - 2) .W-30 J

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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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
• Exercise: Calculate the work done in moving a 4 C charge from B(1,
0, 0) to A(0, 2, 0) along the path y=2-2x, z=0 in the E-field:
a. Ē = 5ax V/m
b. Ē = 5xx V/m
c. Ē = 5x + 5ya, V/m
• Solution:
• W = -Q SĒ. dL
dī=
=dxax+dyay+dzāz
• Exercise (a):
E. dL
• E. dL = 5dx
=
• W = -4 ₁ 5dx
• W = 20 [x]!
• W = 20 J
.
• Exercise (b):
E.dL=
• Ē.dī <= 5xdx
5ax. (dxax + dyay + dzāz
• W = -4 ₂₁5xdx
O ET.
• W = 20
5xax. (dxax + dyay + dzaz)
• W = 10. J
• W = -4
Exercise (c):
Ē.dī = (5xāx + 5yāy). (dxāx + dyāy + dzāz)
• Ē.dL = 5xdx + 5ydy
(₁5xdx +
• W = 20
20 (+1)
5ydy)
• W = 20(0.5 - 2)
.W-30 J
Transcribed Image Text:Energy and Potential • Exercise: Calculate the work done in moving a 4 C charge from B(1, 0, 0) to A(0, 2, 0) along the path y=2-2x, z=0 in the E-field: a. Ē = 5ax V/m b. Ē = 5xx V/m c. Ē = 5x + 5ya, V/m • Solution: • W = -Q SĒ. dL dī= =dxax+dyay+dzāz • Exercise (a): E. dL • E. dL = 5dx = • W = -4 ₁ 5dx • W = 20 [x]! • W = 20 J . • Exercise (b): E.dL= • Ē.dī <= 5xdx 5ax. (dxax + dyay + dzāz • W = -4 ₂₁5xdx O ET. • W = 20 5xax. (dxax + dyay + dzaz) • W = 10. J • W = -4 Exercise (c): Ē.dī = (5xāx + 5yāy). (dxāx + dyāy + dzāz) • Ē.dL = 5xdx + 5ydy (₁5xdx + • W = 20 20 (+1) 5ydy) • W = 20(0.5 - 2) .W-30 J
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