Answer: magnitude: 1.75 N
direction: 6.2° north of east
The Coulomb force equation is,[tex] F_{12} = \frac{k q_1 q_2}{r^2} [/tex] where, [tex] F_{12} [/tex] is force on 1 due to 2, k is coulomb constant, [tex] q_1 ,q_2 [/tex] are the charges and r is the distance between the charges 1 and 2.
The magnitude of force on 3 from 5 is,[tex] F_{35} = \frac{9 \times 10^9 N m^2/C^2 \times 3 \mu C \frac{10^{-6}C}{1 \mu C} \times 5 \mu C\frac{10^{-6}C}{1 \mu C}}{(0.45 m)^2}=0.67 N [/tex]
The magnitude of force on 3 from 9 is,[tex] F_{39} = \frac{9 \times 10^9 N m^2/C^2 \times 3 \mu C \frac{10^{-6}C}{1 \mu C} \times 9 \mu C\frac{10^{-6}C}{1 \mu C}}{(0.45 m)^2}=1.20 N [/tex]
The resultant magnitude is,[tex] F_R=\sqrt{A^2+B^2+2AB cos \phi}=\sqrt{F_{35}^2+F_{39}^2+2F_{35}F_{39} cos \phi} [/tex][tex] =1.75N [/tex]
The Force from 9 on 3 is large which is in northeast direction so the resultant force will also be in north east direction only. So, 6.2 ° North of east.
