Homework Help is Here – Start Your Trial Now!
Science
Physics
3. Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. +q b. 1.00 m A В 1.00 m D b. +9

3. Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. +q b. 1.00 m A В 1.00 m D b. +9

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
Question

Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown, find the resultant electric force on the charge at A. The diagrams are attached herewith.

Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. 3. +q 1.00 m A В 1.00 m D b. +q
Transcribed Image Text:Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. 3. +q 1.00 m A В 1.00 m D b. +q
Four point charges (two with q=2.50×10-6C and two with q=-2.50×10-6 C) are situated at the corners of a square of side 1.00 m as shown. Find the resultant force that the charge at A will experience due to the charges at the other corners of the square. +9 3. +q Given: 9,=9B=+2.50×10¬6 C r=1.0 m Ic=9p=-2.50×10-6C Solution: Eq. (1.1) is used to solve for the magnitude of the electric force exerted by each of the charges at the other corners of the square on the charge at A. Fs on A = k- |(+2.50×10¬°C)(+2.50×10¯“C)| = 19x10° N-m²/C"| (1.00 m)² = 0.05625 N20.06 N Feom A = k |(+2.50×10“CX-2.50×10°C =19x10° N m²/C²| (1.00 m)? = 0.05625 N20.06 N Solve first for the distance between D and A. Using the Pythagorean theorem, 'D on A = ("B on A V(1.00 m)’ +(1.00 m)² = 1.414 m~1.41 m. Therefore, Fo on A = k- =[9x10° N m²/C²| (+2.50×10°C)X-2.50×10°C|| (1.414 m)² = 0.02813 N20.03 N. FB on A is repulsive. Since the charges are both positive, the charge at B will push the charge at A away from it. Therefore, FB on A is directed to the left. Fcon A is attractive since the charges are opposite. The charge at C will attract the charge at A toward it. Therefore, Fc on A is directed down. Fpon A is attractive since the charges are opposite. The charge at D will attract the charge at A toward it. Therefore, F on A is directed down at 45° with the +x-axis, with q, placed at the origin of the x-y plane as shown in the diagram on the right. All other forces are also shown in this diagram. Fooma Foona Fcon A The table below presents the respective horizontal and vertical components of these forces and the resultant electric force. Force Horizontal Component Vertical Component FoonA = 0.05625 N –0.05625 N FconA=0.05625 N –0.05625 N Foona = 0.02813 N +0.02813 N (cos 45°) =+0.0199 -0.02813 N (sin 45°)=-0.0199 N Resultant electric force F EF,=-0.03635 N EF,=-0.07615 N Solving for the magnitude of the resultant electric force, F = V(-0.03635 N)' +(-0.07615 N)² = 0.0844 N. Solving for the direction, |EF, 0 = tan EF TA --0.07615 N 64,5° = tan |-0.03635 N = 64.5°. Thus, the resultant force is 0.0844 N directed down at 64.5° with the negative x-axis as shown.
Transcribed Image Text:Four point charges (two with q=2.50×10-6C and two with q=-2.50×10-6 C) are situated at the corners of a square of side 1.00 m as shown. Find the resultant force that the charge at A will experience due to the charges at the other corners of the square. +9 3. +q Given: 9,=9B=+2.50×10¬6 C r=1.0 m Ic=9p=-2.50×10-6C Solution: Eq. (1.1) is used to solve for the magnitude of the electric force exerted by each of the charges at the other corners of the square on the charge at A. Fs on A = k- |(+2.50×10¬°C)(+2.50×10¯“C)| = 19x10° N-m²/C"| (1.00 m)² = 0.05625 N20.06 N Feom A = k |(+2.50×10“CX-2.50×10°C =19x10° N m²/C²| (1.00 m)? = 0.05625 N20.06 N Solve first for the distance between D and A. Using the Pythagorean theorem, 'D on A = ("B on A V(1.00 m)’ +(1.00 m)² = 1.414 m~1.41 m. Therefore, Fo on A = k- =[9x10° N m²/C²| (+2.50×10°C)X-2.50×10°C|| (1.414 m)² = 0.02813 N20.03 N. FB on A is repulsive. Since the charges are both positive, the charge at B will push the charge at A away from it. Therefore, FB on A is directed to the left. Fcon A is attractive since the charges are opposite. The charge at C will attract the charge at A toward it. Therefore, Fc on A is directed down. Fpon A is attractive since the charges are opposite. The charge at D will attract the charge at A toward it. Therefore, F on A is directed down at 45° with the +x-axis, with q, placed at the origin of the x-y plane as shown in the diagram on the right. All other forces are also shown in this diagram. Fooma Foona Fcon A The table below presents the respective horizontal and vertical components of these forces and the resultant electric force. Force Horizontal Component Vertical Component FoonA = 0.05625 N –0.05625 N FconA=0.05625 N –0.05625 N Foona = 0.02813 N +0.02813 N (cos 45°) =+0.0199 -0.02813 N (sin 45°)=-0.0199 N Resultant electric force F EF,=-0.03635 N EF,=-0.07615 N Solving for the magnitude of the resultant electric force, F = V(-0.03635 N)' +(-0.07615 N)² = 0.0844 N. Solving for the direction, |EF, 0 = tan EF TA --0.07615 N 64,5° = tan |-0.03635 N = 64.5°. Thus, the resultant force is 0.0844 N directed down at 64.5° with the negative x-axis as shown.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Chemical Thermodynamics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS