(a) :Forces acting on the charge q are shown in the figure.Magnitude of force F1 is given by F1=K(2q^2/l^2)Magnitude of force F2 is given by F2=K(4q^2/l^2)Resolving these forces in x and y components.x direction :Net force Fx = -(F1cos60°+F2cos60°) ^x => Fx = -(k(2q^2/l^2)(1/2)+k(4q^2/l^2)(1/2)) ^x = -3kq^2/l^2 ^xy direction :Net force Fy = -(F2sin60° - F1sin60°) ^y => Fy = -(k(4q^2/l^2)√3/2 - k(2q^2/l^2)√3/2) ^y = -√3kq^2/l^2 ^yThus net force on q, Fnet=Fx+Fy = -kq^2/l^2(3^x+√3^y)(b) :Potential energy of the system U=Uq,2q+Uq,-4q+U2q,-4qU=kq(2q)/l+kq(-4q)/l+k(2q)(-4q)/l=-6kq^2/lThus work done to separate them to infinity W=U=-6kq^2/l