What is the magnitude of the resultant force on the positive charge (a = 1 m, b

Question

What is the magnitude of the resultant force on the positive charge (a = 1 m, b = 0.6 m, q = 2.3 × 10-9C)?

9:59 AM a flipitphysics.com AT&T; 4G 94% Fliplt macmilian leaming Physics 1161 SP16 Eastern illinols University Unit 2: PrelectureCheckpoint Hamework Homework: Hwk2: Coulomb's Law And Electric Fields Four charges are placed on the corners of a rectangle What is the magnitude of the resultant force on the positive charge (a-1 m, b 0.6 m, q = 2.3x 10°C)? 1.07E-7 Your submissions: 1.07E-7 Computed vale:10-7 Submitted: Monday, January 25 at 9:48 AM Copyright © 2016 Freeman mrth Publither-a divison of Macmillan Learning Abou·Tech Support | Find Your Local Sales Rro Terms Of Use Privacy Policy

Explanation / Answer

a) F = k(+q)(-q) / (r^2)
= (8.988 x 10^9)(2.3 x 10^(-9))(2.3 x 10^(-9)) / (1^2)
= 47.55 x 10^(-9)   F is on the x-axis

b) F = k(+q)(-q) / (r^2)
= (8.988 x 10^9)(2.3 x 10^(-9))(2.3 x 10^(-9)) / (0.6^2)
= 132.07 x 10^(-9)   F is on the y-axis

r^2 = a^2 + b^2
c) F = k(+q)(-q) / (r^2)
= (8.988 x 10^9)(2.3 x 10^(-9))(2.3 x 10^(-9)) / (1^2 + 0.6^2)
= 34.96 x10^(-9)
= tan^-1(b / a) = tan^-1(0.6 / 1.0) = 30.96
Fx = 34.96 x10^(-9) x cos(30.96) = 29.98 x 10^(-9)
Fy = 34.96 x10^(-9) x sin (30.96) = 17.98 x 10^(-9)

F = ( Fx^2 + Fy^2)
= [(47.55 x 10^(-9) + 29.98 x 10^(-9))^2 + (132.07 x 10^(-9) + 17.98 x 10^(-9))^2]
= 168.9 x 10^(-9)

the resultant force is 1.689 x 10^(-7) N

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