Solution: SUPPOSE that the line from $q_a$ to $q_b$ runs along the $x$-axis. It is clear, from Coulomb's law, that the electrostatic force exerted on any charge placed on this line is parallel to the $x$-axis. Thus, the electric field at any point along this line must also be aligned along the $x$-axis. Let the $x$-coordinates of charges $q_a$ and $q_b$ be $-c/2$ and $+c/2$, respectively. It follows that the ORIGIN ($x=0$) lies halfway between the two charges. The electric field $E_a$ generated by charge $q_a$ at the origin is given by \begin{displaymath}E_a = k_e\,\frac{q_a}{(c/2)^2} = (8.988\times 10^9)\, \frac{(50\times 10^{-6})}{(0.5)^2}= 1.80\times 10^6\,{\rm N\,C}^{-1}.\END{displaymath}The field is positive because it is directed along the $+x$-axis (i.e., from charge $q_a$ towards the origin). The electric field $E_b$ generated by charge $q_b$ at the origin is given by \begin{displaymath}E_b = -k_e\,\frac{q_b}{(c/2)^2} = -(8.988\times 10^9) \,\fra......\times 10^{-6})}{(0.5)^2}= -3.60\times 10^6\,{\rm N\,C}^{-1}.\end{displaymath}The field is NEGATIVE because it is directed along the $-x$-axis (i.e., from charge $q_b$ towards the origin). The resultant field $E$ at the origin is the algebraic sum of $E_a$ and $E_b$ (since all FIELDS are directed along the $x$-axis). Thus, \begin{displaymath}E = E_a + E_b = -1.8 \times 10^6\,{\rm N\,C}^{-1}.\end{displaymath}Since $E$ is negative, the resultant field is directed along the $-x$-axis.The force $f$ acting on a charge $q_c$ placed at the origin is simply \begin{displaymath}f = q_c\,E = (20\times 10^{-6})\,(-1.8\times 10^6)=-36 \,{\rm N}.\end{displaymath}Since $f<0$, the force is directed along the $-x$-axis.Explanation:HOPE IT HELP YOU MARK ME AS BRILLIENT