Calculate the magnitude and direction of the net electric force on each of the charges.

Three point charges lie along a straight line, where q1 = 6.24 µC, q2 = 1.39 µC, and q3 = -2.06 µC. The separation distances are d1 = 3.00 cm and d2 = 2.00 cm. Calculate the magnitude and direction of the net electric force on each of the charges.



i was using F= Ke |q1||q2|/r^2 but i'm not having any luck getting the answer. 10pts for best help :) thank you

Net force on q1 due to the other two is found as follows.
q1 is pushed to the left by q2 and pulled to the right. by q3
Taking right as positive
F = kq1 (q3 (d1 + d2) ^2 - q2 /d1^2}
F = 9e9* 6.24e-6* 1 e-6 {(2.06/ (0.05) ^2 – (1.39 /0.03^2)}
F = - 40. 46 N
minus indicates that the force is toward left.
=========================

Net force on q2 due to the other two is found as follows
q2 is pushed to the right by q1 and pulled to the right byq3 .
Taking right as positive
F = kq2 (q3 /(d2) ^2 +q1 /d1^2}
F = 9e9*1.39e-6*1e-6 {(2.06 / 0.02^2) + (6.24 /0.03^2)}
F = + 151.2 N
+ indicates that the force is toward right
=================================

Net force on q3 due to the other two is found as follows
q3 is pulled to the left both by q1 and q2
Taking right as positive
F = - kq3 (q1/(d2) ^2+q2 /d1^2}
F = - 9e9*2.06e-6*1e-6*{(6.24 /0.05^2) + (1.39 /0.02^2)}
F = -110.7 N
minus indicates that the force is toward left
============================

Which one is between the other two?