Answer:
[tex]F_{net} = -0.7 \hat i + 4.77 \hat j[/tex]
Explanation:
Force due to q2 on q1 is along - X direction due to repulsion between them
so we have
[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]
[tex]F_{12} = \frac{(9\times 10^9)(86 \mu C)(32 \mu C)}{2.5^2}[/tex]
[tex]F_{12} = 3.96 N (-\hat i)[/tex]
Now force between q1 and q3 is given as
[tex]F_{13} = \frac{kq_1q_3}{r^2} \frac{(1.5\hat i + 2.2 \hat j)}{\sqrt{1.5^2 + 2.2^2}}[/tex]
[tex]F_{13} = \frac{(9\times 10^9)(86 \mu C)(53 \mu C)}{(1.5^2 + 2.2^2} \frac{(1.5\hat i + 2.2 \hat j)}{\sqrt{1.5^2 + 2.2^2}}[/tex]
[tex]F_{13} = (5.78)\frac{(1.5\hat i + 2.2 \hat j)}{2.66}[/tex]
[tex]F_{13} = (2.17)(1.5\hat i + 2.2 \hat j)[/tex]
Now net force on q1 is given as
[tex]F_{net} = F_{12} + F_{13}[/tex]
[tex]F_{net} = 3.96(-\hat i ) + (3.26 \hat i + 4.77 \hat j)[/tex]
[tex]F_{net} = -0.7 \hat i + 4.77 \hat j[/tex]
