Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1952185 • 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 dis|F|=K rac{|Q Q^prime|}{d^2},
where K= rac{1}{4 pi epsilon_0}. epsilon_0 = 8.854 imes 10^{-12};{ m C^2/(Ncdot m^2)} is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q_1 = -17.0 nC, is located at x_1 = -1.750 m; the second charge, q_2 = 37.5 nC, is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q_3 = 55.0 nC placed between q_1 and q_2 at x_3 = -1.250 m?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
Magnitude = k*[q1/.5^2 -q2/1.25^2]*55*10^-9 =-826.91*55*10^-9 N =-4.548*10^-5 N