Coulombs law for the magnitude of the force F between two pa

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 = -13.0 nC , is located at x1 = -1.685 m ; the second charge, q2 = 35.0 nC , is at the origin (x=0.0000). Part A What is the net force exerted by these two charges on a third charge q3 = 49.5 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

Solution

F = kq1q2 / d^2

when charges are like they repel, when they are opposites they attracts.

so Fnet = due to q1 + due to q2

Fnet = [ kq1q3 / (x1 - x3)^2 ] (-i) + [ kq2q3 / (x3^2) ] (-i)

=- [ (9 x 10^9 x 13 x 10^-9 x 49.5 x 10^-9) / (1.685 - 1.135)^2 ] - [ (9 x 10^9 x 35 x 10^-9 x 49.5 x 10^-9 / 1.135^2 ]

= - 1.91 x 10^-5 - 1.21 x 10^-5 = - 3.12 x 10^-5 N