Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1537546 • 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 = and epsilon_0 = 8.854 times 10^-12 C^2/(N middot m^2) is the permittivity of free space. Consider two point charges located on the x axis one charge, q_1 = -20.0 nC, is located at x_1 = -1.696 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_1 = 47.5 nC placed between q_1 and x_3 = -1.205 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
Here F = summation kqq'/r^2
= [1/(4pi *8.854e-12)]47.5e-9*[-20e-9/(1.695-1.205)^2 -38.5e-9/1.205^2]
= -0.0000469 N