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Suppose that the charges in Example 1.3 are situated as shown in the figure. Find the resultant electric force on the charge at A.

Suppose that the charges in Example 1.3 are situated as shown in the figure. Find the resultant electric force on the charge at A.

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)...
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2. Suppose that the charges in Example 1.3 are situated as shown

in the figure. Find the resultant electric force on the charge at A.

R - what is the electric
Transcribed Image Text:R - what is the electric
EXAMPLE 1.3 Force of Adjacent Charges - Four Charges 1. > Four point charges (two with q = 2.50 × 10-6 C and two with q = -2.50 x 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. Given: Find: qa = 9B = +2.50 × 10-6 C 9c = 9p = -2.50 × 10-6 C r = 1.0 m FB on A = ? Fc on A = ? FD on A = ? Fresultant = ? Solution: For FBon A a. |(+2.50 × 10-6 C)(+2.50 × 10-6 C)| FB on A = k (r)2 (9 × 10º N · m²/C²) (1.0 m)² |+6.25 × 10-12 C| 1 m² FB on A is repulsive. Since the charges are both positive, the charge at B will nch the charge away from it. Therefore, FB on A is directed to the left. FB on A = (9 × 10° N · m²/C²) = 0.06 N 4/11 b. For Fc on A |(+2.50 × 10-6 C)(-2.50 × 10-6 C)| (1.0 m)² Fc on A = k (r)2 = (9 × 10° N · m²/C²) |-6.25 × 10-12 C| 1.0 m² Fc on A = (9 × 10° N · m² /C²) = 0.06 N 2 GENERAL PHYSICS 2 LEARNING MATERIAL 2 Fc on 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. c. For Fp on Ar |(+2.50 × 10-6 C)(-2.50 × 10-6 C)| FD on A = k = (9 x 10° N · m²/C²)· (r)² (1.414 m)² Note: We used 1.414 m for distance between D and A by applying the Pythagorean Theorem. 1.00 m is only for the sides. |-6.25 × 10-12 C| Fp on A = (9 × 10° N · m² /C²) = 0.03 N 2.00 m² FD on A is attractive since the charges are opposite. The charge at D will attract the charge at A toward it. Therefore, FD on A is directed down at 45° with +x-axis. The diagram for the forces is shown in the figure. Now, to find for the resultant force Fresultant, let's use the component method again. Force Horizontal Component Vertical Component FB on A = 0.06 N 0.06 N(cos 180°) = –0.06 N 0.06 N(sin 180°) = 0 N Fc on A = 0.06 N 0.06 N(cos 270°) = 0 N 0.06 N(sin 270°) = –0.06 N Fp on A = 0.03 N 0.03 N(cos 315°) = +0.02 N 0.03 N(sin 315°) = –0.02 N Fresultant F = -0.04 > F, = -0.08 N Solving for the magnitude of the resultant force, Fp on A Fresultant = + Fresultant = - (-0.04 N)² + (-0.08 N)² = 0.08 N FD on A Solving for the direction, -0.08| |-0.04| Fresultaar Fc on A 0 = tan-1 = tan-1 = 64.5° r Thus, the resultant force is 0.08 N, 64.5° from the negative x-axis.
Transcribed Image Text:EXAMPLE 1.3 Force of Adjacent Charges - Four Charges 1. > Four point charges (two with q = 2.50 × 10-6 C and two with q = -2.50 x 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. Given: Find: qa = 9B = +2.50 × 10-6 C 9c = 9p = -2.50 × 10-6 C r = 1.0 m FB on A = ? Fc on A = ? FD on A = ? Fresultant = ? Solution: For FBon A a. |(+2.50 × 10-6 C)(+2.50 × 10-6 C)| FB on A = k (r)2 (9 × 10º N · m²/C²) (1.0 m)² |+6.25 × 10-12 C| 1 m² FB on A is repulsive. Since the charges are both positive, the charge at B will nch the charge away from it. Therefore, FB on A is directed to the left. FB on A = (9 × 10° N · m²/C²) = 0.06 N 4/11 b. For Fc on A |(+2.50 × 10-6 C)(-2.50 × 10-6 C)| (1.0 m)² Fc on A = k (r)2 = (9 × 10° N · m²/C²) |-6.25 × 10-12 C| 1.0 m² Fc on A = (9 × 10° N · m² /C²) = 0.06 N 2 GENERAL PHYSICS 2 LEARNING MATERIAL 2 Fc on 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. c. For Fp on Ar |(+2.50 × 10-6 C)(-2.50 × 10-6 C)| FD on A = k = (9 x 10° N · m²/C²)· (r)² (1.414 m)² Note: We used 1.414 m for distance between D and A by applying the Pythagorean Theorem. 1.00 m is only for the sides. |-6.25 × 10-12 C| Fp on A = (9 × 10° N · m² /C²) = 0.03 N 2.00 m² FD on A is attractive since the charges are opposite. The charge at D will attract the charge at A toward it. Therefore, FD on A is directed down at 45° with +x-axis. The diagram for the forces is shown in the figure. Now, to find for the resultant force Fresultant, let's use the component method again. Force Horizontal Component Vertical Component FB on A = 0.06 N 0.06 N(cos 180°) = –0.06 N 0.06 N(sin 180°) = 0 N Fc on A = 0.06 N 0.06 N(cos 270°) = 0 N 0.06 N(sin 270°) = –0.06 N Fp on A = 0.03 N 0.03 N(cos 315°) = +0.02 N 0.03 N(sin 315°) = –0.02 N Fresultant F = -0.04 > F, = -0.08 N Solving for the magnitude of the resultant force, Fp on A Fresultant = + Fresultant = - (-0.04 N)² + (-0.08 N)² = 0.08 N FD on A Solving for the direction, -0.08| |-0.04| Fresultaar Fc on A 0 = tan-1 = tan-1 = 64.5° r Thus, the resultant force is 0.08 N, 64.5° from the negative x-axis.
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