Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1486510 • 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 id |F| = K |QQ'|/d^2 where K = 1/4pisigma_0, and sigma_0 = 8.854 x 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 = -10.0 n^C, is located at x_1 = -1.740 m; the second charge, q_2 = 38.5 n^C, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.070 m Your answer may be positive or negative, depending on the direction of the force.Explanation / Answer
F1 = K*q1*q3/r1^2
r1 = (1.74-1.07) m= 0.67 m
F1 will be towards negative x axis
F1 = K*q1*q3/r1^2
= -(9*10^9)*(10^10^-9)*(47*10^-9) / (0.67)^2
= -9.42*10^-6 N
F2 = K*q2*q3/r2^2
r2 = 1.07 m
F2 will be towards negative x axis
F2 = K*q2*q3/r2^2
= -(9*10^9)*(38.5^10^-9)*(47*10^-9) / (1.07)^2
= -1.42*10^-5 N
Fnet = -9.42*10^-6 -1.42*10^-5 = -2.36*10^-5 N
Answer: -2.36*10^-5 N