5. Three charges sit on the vertices of an equilateral triangle, the sides of which are 30.0 cm long. If the charges are A = +4.0 UC, B = +5.0 JC and C = +6.0 JC (clockwise from the top vertex), find the force on each charge. [10 marks] As we know the angle of an equilateral triangle is 60, so the half of the angle is 30 FAR FAC 30 cms 9 30 cm B 30 cm For A, FAB=(kqAqB)/12 FAB=((9.00x109 N.me/C2)(4.0x10-C)(5.0x10-C))/(0.3m)2 FAB=2 N [repulsive] FAC=(kqAqc)/12 FAC=((9.00x109 N.m2/C2)(4.0x10-C)(6.0x10-C))/(0.3m)2 FAC=2.4 N[repulsive] FY=FABCOS30+FACCOS30 Fy=(2 N)(cos30)+(2.4 N)(cos30) Fy=3.81 N [upward] Fx=FABSin30 [leftward]+FACsin300 [rightward]