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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 В 1.00 m C 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 В 1.00 m C D b- +9

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter1: Units And Measurement
Section: Chapter Questions
Problem 15P: Use the orders of magnitude you found in the previous problem to answer the following questions to...
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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.

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.
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