From the geometry of the problem, the force between charges q1 and q2 will only act in the x-direction (left/right), and the force between charges q2 and q3 will only act in the y-direction (up/down).
Since we are only interested in the x-component of the net force, we only need to calculate the magnitude of the force between charges q1 and q2 using the formula for Coulomb's force, namely:
F=k * q1 * q2 / r2,
where k = 1/(4*π*ε0) in SI units. Plugging in π = 3.14, ε0 = 8.85 x 10-12 C2/Nm2, q1 = -4.60 x 10-5 C, q2 = +3.10 x 10-6 C, r = 0.350 m, we find the magnitude of the force to be
F = 1 / [ 4 * (3.14) * (8.85 x 10-12 C2/Nm2) ] * (-4.60 x 10-5 C) * (3.10 x 10-6 C) / (0.350 m)2
= -1.05 x101 N
= -10.5 N.
Like charges repel and opposite charges attract, so since q1 and q2 have opposite charges they will attract. Since q2 is attracted to q1 it will be pulled to the left. The direction of the force is therefore pointed left, which means that if the positive x-direction is to the right, then the net force in the x-direction will be
Fx = -10.5 N,
where the negative sign indicates the direction is to the left.
If we wanted to extend the problem, we could also solve for the y-component of the force on q2.
