Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1885889 • 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|d2, where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -14.5 nC , is located at x1 = -1.745 m ; the second charge, q2 = 37.0 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 50.5 nC placed between q1 and q2 at x3 = -1.135 m ?
Your answer may be positive or negative, depending on the direction of the force.
Express your answer numerically in newtons to three significant figures.
Force on q3 = _______ N
Explanation / Answer
Net force=Kq3*(q1/0.61^2-q2/1.745^2)
putting q3 ,q2 ,q1 with sign we got
Force=-2.32*10^(-5) N