I think the diagram must look like this:
Notice that q3 and q2 are on the same "y-level" in other words the forces between them are only in the x-direction. That means the only charge that is affecting q2, IN THE Y-DIRECTION, is q1. The force of q1 on q2 (at the angle of 45°) is calculated from Coulomb's Law:
where:
k = Coulomb's constant = 8.98756 x 109 kg⋅m3/(s2⋅C2)
q1 = -4.60 x 10-6 C
q2 = 3.75 x 10-5 C
r = 0.283 m
The fact that the charges are opposite tells us that q2 feels an attractive force for q1 meaning the force on q2 is down and to the left. We can now ignore the sign on the charge and just do the math:
F = (8.98756 x 109)(4.60 x 10-6)(3.75 x 10-5)/(0.2832)
F = 19.3579 kg⋅m/s2 = 19.3579 N
But only a portion of that force is in the y-direction. To find out how much, we must break the force vector up into its components. Considering a triangle (as shown in red in the diagram) where the force begins at q2 and points towards q1, that force (19.3579 N) is the hypotenuse of the triangle. The angle at q2 is 45° (because both angles are 45°) and the force in the y-direction (Fy) is adjacent to q2.
So, using trig we can say cos(45°) = adj/hyp = Fy/19.3579 so Fy = 19.3579cos(45) = 10.2 N and the direction is downwards. (so I suppose you could say it is -10.2N)
