Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
4- Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let € = 0. a) Find F2, the force on Q2: This force will be Q122 R12 F2 = 470 R1213 (8 x 10-6)(-5 x 10-6) (4ax +10ay) 47 €0 (116)1.5 (-1.15ax -2.88ay) mN b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: Q3 Q1R13 Q₂R23 + 470 R1313 R233 where R₁3 = (x - 2)a, + (y – 5)ay and R₂3 = (x − 6)ax + (y - 15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5.x, 8). With this restriction, the force becomes: F3 = or Q38[(x-2)ar +2.5(x - 2)ay] 5[(x-6)a, +2.5(x - 6)ay] F3 = 47 €0 [(x-2)²+(2.5)²(x − 2)²]¹.5¯¯ [(x − 6)² + (2.5)²(x − 6)²]¹.5 where we require the term in large brackets to be zero. This leads to which reduces to - 8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 — 5(x − 6)[((2.5)² + 1)(x − 2)²]¹5 = 0 - = 8(x − 6)²-5(x - 2)² = 0 - 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) X = = 21.1

4- Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let € = 0. a) Find F2, the force on Q2: This force will be Q122 R12 F2 = 470 R1213 (8 x 10-6)(-5 x 10-6) (4ax +10ay) 47 €0 (116)1.5 (-1.15ax -2.88ay) mN b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: Q3 Q1R13 Q₂R23 + 470 R1313 R233 where R₁3 = (x - 2)a, + (y – 5)ay and R₂3 = (x − 6)ax + (y - 15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5.x, 8). With this restriction, the force becomes: F3 = or Q38[(x-2)ar +2.5(x - 2)ay] 5[(x-6)a, +2.5(x - 6)ay] F3 = 47 €0 [(x-2)²+(2.5)²(x − 2)²]¹.5¯¯ [(x − 6)² + (2.5)²(x − 6)²]¹.5 where we require the term in large brackets to be zero. This leads to which reduces to - 8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 — 5(x − 6)[((2.5)² + 1)(x − 2)²]¹5 = 0 - = 8(x − 6)²-5(x - 2)² = 0 - 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) X = = 21.1

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...
See similar textbooks
icon
Related questions
Question
(Electromagnetic field) I want a detailed solution for this solution
4- Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let e = €0. a) Find F2, the force on Q2: This force will be Q1 Q2 R12 F2 = 470 R12³ b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: Q1R13 Q2R23 + R133 R23³ where R₁3 = (x - 2)a¸ + (y – 5)ay and R23 = (x − 6)ax + (y - 15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5.x, 8). With this restriction, the force becomes: F3 or (8 × 10-6)(−5 × 10−6) (4ax + 10ay) (116)1.5 47 €0 = F3 = which reduces to Q3 470 where we require the term in large brackets to be zero. This leads to Q3 8[(x-2)a, - 2)ay] 5[(x-6)a, +2.5(x - 6)ay] 47€0 [[(x - 2)² + (2.3)²(x - 2)²3³-³3- ((x - 5²4 (23) ²(x - 621-3 8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 – 5(x − 6)[((2.5)² + 1)(x − 2)²]¹-5 = 0 8(x − 6)² - 5(x - 2)² = 0 - (-1.15a, -2.88ay) mN X = 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) = 21.1
Transcribed Image Text:4- Let Q1 = 8 μC be located at P₁ (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let e = €0. a) Find F2, the force on Q2: This force will be Q1 Q2 R12 F2 = 470 R12³ b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: Q1R13 Q2R23 + R133 R23³ where R₁3 = (x - 2)a¸ + (y – 5)ay and R23 = (x − 6)ax + (y - 15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15 — 5)/(6-2) = 2.5. Therefore, we look for P3 at coordinates (x, 2.5.x, 8). With this restriction, the force becomes: F3 or (8 × 10-6)(−5 × 10−6) (4ax + 10ay) (116)1.5 47 €0 = F3 = which reduces to Q3 470 where we require the term in large brackets to be zero. This leads to Q3 8[(x-2)a, - 2)ay] 5[(x-6)a, +2.5(x - 6)ay] 47€0 [[(x - 2)² + (2.3)²(x - 2)²3³-³3- ((x - 5²4 (23) ²(x - 621-3 8(x − 2)[((2.5)² + 1)(x − 6)²]¹.5 – 5(x − 6)[((2.5)² + 1)(x − 2)²]¹-5 = 0 8(x − 6)² - 5(x - 2)² = 0 - (-1.15a, -2.88ay) mN X = 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) = 21.1
Expert Solution
steps

Step by step

Solved in 4 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Capacitor
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS