Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1431173 • 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|/d2, where K = 1/ 4 pi epsilon 0, and epsilon 0 = 8.854 x 10 -12 C2/(N middot m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -19.0 nC , is located at x1 = -1.665 m ; the second charge, q2 = 36.0 nC , is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge qs = 54.0 nC placed between q1 and q2 at x3 = -1.200 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
r13 = 1.665 - 1.200 = 0.465 m
F13 will be along negative x axis
r23 = 1.200 m
F23 will be along negative x axis
Fnet = - |F13| -| F23|
= - K*q3* (q1/r13^2 + q2/r23^2)
= -(9*10^9)*(54*10^-9)*(19*10^-9 / (0.465)^2 + 36*10^-9/ (1.2)^2)
= -(486)*(8.79*10^-8 + 2.5*10^-8)
= - 5.49*10^-5 N
Answer: - 5.49*10^-5 N