Three charges $-q_1$, $+q_2$ and $-q_3$ are placed as shown in the figure. ($-q_1$ at origin, $+q_2$ at $(b,0)$, $-q_3$ at $(a\cos \theta ,a\sin \theta )$). The $x$ -component of the force on $-q_1$ is proportional to:

1. Force on $-q_1$ due to $+q_2$: Attractive force along $+x$ axis. $F_{12}=k\frac{q_1{q}_2}{b^2}$. $x$ -component = $+k\frac{q_1{q}_2}{b^2}$.2. Force on $-q_1$ due to $-q_3$: Repulsive force directed away from $-q_3$ towards the origin. The direction is along the line joining $(a\cos \theta ,a\sin \theta )$ to $(0,0)$. The vector component along $x$ is $F_{13}\cos \theta$.$F_{13}=k\frac{q_1{q}_3}{a^2}$. The $x$ -component is $k\frac{q_1{q}_3}{a^2}\cos \theta$.Total $x$ -component $F_x=k{q}_1\left(\frac{q_2}{b^2}+\frac{q_3}{a^2}\cos \theta \right)$.Thus, $F_x\propto \frac{q_2}{b^2}+\frac{q_3}{a^2}\cos \theta$.