Coulumb\'s law for the magnitude of the force F between two particles with charg
ID: 2275382 • Letter: C
Question
Coulumb'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= 1/4pie e0 and e0= 8.854*10^-12/ (N*m^2) is the permittivity of free space.
Consider two point charges located on the x axis: one charge,
q1 = -16.0nC , is located atx1 = -1.700m ; the second charge, q2 = 38.5nC ,is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 53.0nC placed between q1 and q2 atx3 = -1.115m ?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
F13 = kq1q3/r^2 = 9*10^9 * (-16)* 53 * 10^-18 / (.585)^2 = -2.23*10^-5 N attractive
F12 = kq2q3/r^2 = 9*10^9 * 38.5 * 53 * 10^-18 /(1.115)^2 = 1.477 * 10^-5 N repulsive
Net force on q3 = F13 +F12 =-3.7 * 10^-5 N toward negative x -axis