Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1543898 • 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 |Qd|/d^2 where K =1/4 pi c_0, and E_o =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, q1=18.0 nC, is located at x_1 = -1.715 m; the second charge, q_2 = 33.5 nC, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3=54.0 nC placed between q_1 and q_2 at x^3 = -1.060m? Your answer may be positive or negative, depending on the direction of the force.Explanation / Answer
From the Columbs law
F 23 = k q2q3/ r1^2 =(9 * 10^9) (33.5 * 10^-9C) (54 * 10^-9)/(-1.060)^2 = 14.49 N (-i)
F 13 = k q1q3/ r2^2 =(9 * 10^9) (-18 * 10^-9C) (54 * 10^-9)/(1.715-1.060)^2 =20.39 N (-i)
net force exerted on the third charge is
F net = 20.39 + 14.49 N = 34.88 N ( along negative x direction)