Magnitude of the force from two charges?

Question

Six charges are placed in a plane at the positions (b,0), (b,b), (0,b), (-b,0), (-b,-b), and (0,-b) as shown. Charges 1, 3, 4, and 6 have charge +6 µC and charges 2 and 5 have charge -2 µC. The distance b has a value of 0.25 m.

QUESTION) What is the magnitude of the force on charge 2 due to charges 1 and 3?

A) 2.44 N

B) 3.03 N

C) 4.19 N

D) 5.52 N

E) 6.08 N

I've got kind of an idea how to do this problem, but I keep getting absurd numbers like 1.73 x 10^12 for my x and y components of the force. I have a feeling I'm missing something in using the force equation F=(kq1q2)/(r^2). Could anybody give me a hand?

Answer 1

one uC=0.000001 C

The two +6uC are the same distance r=b=0.25m

from -2uC which has coordinates (x,y)=(0.25,0.25)

From F=k*q1*q2/r^2

Fx=(9 * (10^9) * (-0.000002) * 0.000006) / (0.25^2) = -1.728

Fy=(9 * (10^9) * (-0.000002) * 0.000006) / (0.25^2) = -1.728

vector force = sqr((-1.72800)^2) + (-1.72800)^2) = 2.44 N

"attractive and points to the origin: (0,0)"