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Question:

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

Solution:

(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