Three charges -q₁, +q₂ and -q₃ are placed as shown in the figure. The x-component of the force on -q₁ is proportional to

F₂ = Force applied by q₂ on -q₁
F₃ = Force applied by (- q₃) on -q₁
x-component of net force on -q₁ is

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}={\mathrm{F}}_2+{\mathrm{F}}_3\mathrm{sin\theta }=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{b}}^2}+\mathrm{k}·\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{sin\theta } \\ \Rightarrow {\mathrm{F}}_{\mathrm{x}}=\mathrm{k}\left[\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{sin\theta }\right] \\ \Rightarrow {\mathrm{F}}_{\mathrm{x}}=\mathrm{k}·{\mathrm{q}}_1\left[\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{sin\theta }\right] \\ \Rightarrow {\mathrm{F}}_{\mathrm{x}}\propto \left(\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{sin\theta }\right) \end{gathered}$$