Answer

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**Hint:**We need to analyze the charges in the three corners of the equilateral triangle. The charges, the signs of the charges, and the distance between them determine Coulomb's force that will act between two of the charges in the system.

**Complete step-by-step solution**

Let us consider the given system of charges. We know that the direction of force acting on a charge due to another charge is dependent on the sign of the charge of the two under consideration. We know that the like charges repel and the unlike charges attract. So, we can draw the figure marking the forces due to the other charge and the resultant of two forces.

From the figure, we understand the direction in which the forces act on each of the charges. The forces \[{{F}_{1}},{{F}_{2}}\text{ and }{{F}_{3}}\] are the resultant forces acting on the charges \[{{q}_{1}},{{q}_{2}}\text{ and }{{q}_{3}}\] respectively due to each of the other charges.

Now, we can find the magnitude of the force acting on the charges. We know that the distance between all the charges are equal and they have equal magnitudes of charge. Therefore, the Coulomb’s force acting on each charge will be the same, which is given as –

\[\begin{align}

& F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{qQ}{{{r}^{2}}} \\

& \text{here,} \\

& r=l \\

& q=Q=q \\

& \therefore F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}^{2}}}{{{l}^{2}}} \\

\end{align}\]

We understand that each of the charges in the system has a force which is equal to \[F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}^{2}}}{{{l}^{2}}}\] with the directions as described by the figure above.

Now, we can compute the total force acting in the system of charges. We know that the forces \[{{F}_{1}}\text{ and }{{F}_{2}}\] can have a result as shown below.

We can understand from the above figures that the forces due to charges at A and B have a resultant force of F’. The magnitude of F’ is given as –

\[\begin{align}

& F'=\sqrt{{{F}_{1}}^{2}+{{F}_{2}}^{2}+2{{F}_{1}}{{F}_{2}}\cos \theta } \\

& \Rightarrow F'=\sqrt{2{{F}^{2}}+2{{F}^{2}}\cos {{60}^{0}}} \\

& \Rightarrow F'=\sqrt{2{{F}^{2}}+2{{F}^{2}}(\dfrac{1}{2})} \\

& \Rightarrow F'=\sqrt{3{{F}^{2}}} \\

& \therefore F'=\sqrt{3}F \\

\end{align}\]

Now, since the force F’ and \[{{F}_{3}}\] are acting in the opposite directions, the net force acting in the system can be found as –

\[\begin{align}

& {{F}_{net}}=F'-{{F}_{3}} \\

& \Rightarrow {{F}_{net}}=\sqrt{3}F-F \\

& \Rightarrow {{F}_{net}}=(\sqrt{3}-1)F \\

& \text{but,} \\

& F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}^{2}}}{{{l}^{2}}} \\

& \therefore {{F}_{net}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}^{2}}}{{{l}^{2}}}(\sqrt{3}-1)N \\

\end{align}\]

The direction of the force will be in the direction of DC. The resultant force is acting in the system along the direction of DC. This is the required solution.

**Note:**The forces acting in the directions as given by the figure describes that the system can be in equilibrium in this condition even if the charges involved are not the same. The total force acting in the system is zero as the forces between the unlike and like pairs cancel off.

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