We have the next diagram
r12=0.5
r23=0.95
r13=1.45
Using Columb's law the net force of the particle q3 will be equal to the sum of the forces on particle q3 from particle q1 and q2.
the force between q1 and q3
[tex]\begin{gathered} F_{13}=k\text{ }\frac{q_1\cdot q_3}{r^2_{13}} \\ F_{13}=9\times10^9\text{ }\frac{(-53\times10^{-6}\cdot-88\times10^{-6})}{1.45^2} \\ F_{13}=19.96N \end{gathered}[/tex]then for between q2 and q3
[tex]\begin{gathered} F_{23}=k\text{ }\frac{q_2\cdot q_3}{r^2_{23}} \\ F_{23}=9\times10^9\text{ }\frac{(105\times10^{-6})\cdot(-88\times10^{-6})}{0.95^2} \\ F_{23}=-92.14N \end{gathered}[/tex]Then we analyze
[tex]F_3=F_{13}+F_{23}=19.96-92.14=-72.18[/tex]The force is 72.18N to the left
