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=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -12.0 nC , is located at x1 = -1.680 m ; the second charge, q2 = 30.5 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 50.0 nC placed between q1 and q2 at x3 = -1.240 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer 1

Charge Q1 is negative and positioned at the left of Q3.

Hence the force will be to the left

Charge Q2 is positive and positioned at the right of Q3.

Hence the the force will also be to the left

Therefore you can add the forces together (with left being negative)

Calculate the net force.

F = -k*Q1*Q3/r13^2 - k*Q2*Q3/r23^2

=[ -9.0x10^9*12x10^-9*50x10^-9/(1.680 - 1.240)^2 - 9.0x10^9*30.5x10^-9*50x10^-9/1.240^2 ]

F = -3.68 x10^-5. N

negative sign shows it an repulsion force.