(a) Three point charges $q,-4q$ and $2q$ are placed at the vertices of an equilateral triangle $ABC$ of side ${}^{\prime }{l}^{\prime }$ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge $q$.
(b) Find out the amount of the work done to separate the charges at infinite distance.

(a) :
Forces acting on the charge $q$ are shown in the figure.
Magnitude of force $F_1$ is given by $F_1=\frac{K2q^2}{l^2}$
Magnitude of force $F_2$ is given by $F_2=\frac{K4q^2}{l^2}$
Resolving these forces in x and y components.
x direction :
Net force $F_x=-(F_1\cos 60+F_2\cos 60)\hat{x}$
$⟹F_x=-(\frac{k2q^2}{l^2}.\frac{1}{2}+\frac{k4q^2}{l^2}.\frac{1}{2})\hat{x}=-\frac{3k{q}^2}{l^2}\hat{x}$
y direction :
Net force $F_y=-(F_2\sin 60-F_2\sin 60)\hat{y}$
$⟹F_y=-(\frac{k4q^2}{l^2}.\frac{\sqrt{3}}{2}-\frac{k2q^2}{l^2}.\frac{\sqrt{3}}{2})\hat{y}=-\frac{\sqrt{3}k{q}^2}{l^2}\hat{y}$
Thus net force on $q$, $F_{net}=F_x+F_y=-\frac{k{q}^2}{l^2}(3\hat{x}+\sqrt{3}\hat{y})$
(b) :
Potential energy of the system $U=U_{q,2q}+U_{q,-4q}+U_{2q,-4q}$
$U=\frac{kq\left(2q\right)}{l}+\frac{kq(-4q)}{l}+\frac{k2q(-4q)}{l}=\frac{-10k{q}^2}{l}$
Thus work done to separate them to infinity $W=U=\frac{-10k{q}^2}{l}$