Using Coulomb's law:
[tex]\begin{gathered} F_{21}=K\frac{Q1\cdot Q2}{r^2} \\ F_{21}=8.988\times10^9\cdot\frac{45\times10^{-6}\cdot(-12\times10^{-6})}{3^2} \\ F_{21}=0.53928N \end{gathered}[/tex][tex]\begin{gathered} F_{31}=K\frac{Q1\cdot Q3}{r^2} \\ F_{31}=8.988\times10^9\cdot\frac{45\times10^{-6}\cdot(18\times10^{-6})}{(\sqrt[]{18})^2^{}} \\ F_{31}=0.40446N \end{gathered}[/tex][tex]\begin{gathered} F_{Tx}=F_{T21}\cos (45) \\ F_{Ty}=F_{T31}\sin (45) \end{gathered}[/tex][tex]\begin{gathered} |F_T|=\sqrt[]{(F_x)^2+(F_y)^2}= \\ |F_T|=\sqrt[]{(-0.25328)^2+(0.285996)^2} \\ |F_T|\approx0.382 \end{gathered}[/tex]
And the angle:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{F_y}{F_x}) \\ \theta\approx-41.52868^{} \end{gathered}[/tex]
