Three point charges, q1 = +8?C, q2 = -4?C, and q3 = +2?C, are placed at the corn

Question

Three point charges, q1 = +8?C, q2 = -4?C, and q3 = +2?C, are placed at the corners of an equilateral triangle, 80mm on each side. Assume that the base of the triangle is formed by a line joining the 2?C and the -4?C charges. What are the magnitude and direction of the resultant force on the +8?C charge?
The textbook answer is 39N, 330 degrees.

Explanation / Answer

F = kQq/r^2 F1 = 9 x 10^9 x -4 x 8 x 10^-12/(80 x 10^-3)^2 = -45N F2 = 9 x 10^9 x 2 x 8 x 10^-12/(80 x 10^-3)^2 = 22.5N Add x and y components: F1> x: -45cos60 y: -45cos30 F2> x: 22.5sin30 y: 22.5cos30 Add these up and take resultant: 39N Direction tan^-1 of y component/x component = -30degrees = 330degrees

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