Please solve this question:- On a rough inclined surface two similar positively charged balls of same mass same charge is placed coefficient of friction is u lower ball is fixed. find ratio of maximum and minimum separation between two balls so that upper ball remains in equilibrium.?

Let two balls be A and B . Let ball B be fixed . Let angle of inclination be Φ. Distance between the balls be `d` and charges on balls be q .Therfore frictional force on ball A is equal to umgcosΦ. The force of repulsion between the charge balls is kq²/d² Component of weight acting along the line of inclination is mgsinΦ . When force of repulsion is greator than mgsinΦ then the separation will be maximum . Therefore kq²/d(max.)²= mgsinΦ+umgcosΦ..........(1) . Now if force of repulsion is less then mgsinΦ the seperation will be minimum . Therfore kq²/d(min)²=mgsinΦ-umgcosΦ............(2) from (1) and (2) you can calculate the desired ratio