Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1572396 • 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 =1/4pi sum_0, and sum_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, q1 = -15.5 nC, is located at x_1 = -1.685 m; the second charge, q_2 = 37.0 nC, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 54.5 nC placed between q_1 and q_2 at x_3 = 1.215 m? Your answer may be positive or negative, depending on the direction of the force.Explanation / Answer
The force between 1 and 3 will be:
F13 = k q1q3/r13^2
F13 = 9 x 10^9 x (-15.5 x 10^-9) (54.5 x 10^-9)/(1.685 - 1.215)^2 = -3.44 x 10^-5 N
F23 = k q2q3/r23^2
F23 = 9 x 10^9 x (54.5 x 10^-9) (37 x 10^-9)/ 1.215^2 = 1.23 x 10^-5 N
Fnet = -3.44 x 10^-5 N + 1.23 x 10^-5 N = -2.21 x 10^-5 N
Hence, Fnet = -2.21 x 10^-5 N