Three charged particles are placed at each of three corners of an equilateral tr

Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.6 . Two of the particles have a negative charge: = -7.2 and = -14.4 . The remaining particle has a positive charge, = 8.0 . What is the net electric force acting on particle 3 due to particle 1 and particle 2? Find the net force acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude and a direction measured from the positive x axis. Express the magnitude in newtons and the direction in degrees to three significant figures.

Explanation / Answer

Relevant equations Coulomb's Law: F = [k*q1*q2]/r2 3. The attempt at a solution F13 (on q3 due to q1) = [k *q3*q1]/r2 = [(9*109)(8*10-9)(8*10-9)]/(.035)2 = 4.702*10-4 N F23 (on q3 due to q2) = [k * q3*q2]/ r2 = [(9*109)(8*10-9)(16*10-9)]/(.035)2 = 9.404*10-4 N F13x = F13 * cos(60) = + 2.351*10^-4 N F13y = F13 * sin(60) = + 4.072*10^-4 N F23x = F23 * cos(0) = + 9.404 *10^-4 N F23y = 0 Fx = 2.351*10^-4 + 9.404*10^-4 = 1.176*10^-3 N Fy = 4.072*10^-4 N SF = 10^-3v(1.176)2 + (0.4072)2 = 1.2445*10^-3 N ? = arctan (0.4072/1.176) = 19.1 The velocity of the body will be directed with the force as the body was initially at rest F gives the direction of acceleration a a = (change in velocity ) /time a=(v-0)/t a is directed towards F but a, v, and t are unknowns

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