Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1513027 • Letter: C
Question
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is |F | = K |QQ'/d^2 where K = 1/4 pi episilon_0 Episilon_0 = 8.854 Times 10^12 C^2/(N m^2) is the permittivity of free space. Consider two point charges located on the x axis: one charge. q_1 = -19.0 nC, is located at x_1 = -1.680 m: the second charge. q_2 = 38.5 nC. is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 53.5 nC placed between q_1 and q_2 at x_3 = -1 085 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.Explanation / Answer
F = k[-19*53.5/(1.68-1.085)^2 - 83.5*53.5/(1.085)^2 ]*10^-18 = -5.926*10^-5 N