The force of electrostatic repulsion between two small positively charged objects, A and B, is 3.6 x 10 5 N when AB = 0.12m. What is the force of repulsion if AB is increased to a) 0.24 m b) 0.36 m Share with your friends Share 2 Vipra Mishra answered this Dear Student F=q1q2r2q1q2=Fr2=3.6×105×0.12×0.12=5184 C2When r = 0.24F=51840.242=9×104 NWhen r = 0.36F=51840.362=4×104 N Regards -1 View Full Answer Srishti Kumari answered this Please find this answer 9 Srishti Kumari answered this I hope you can solve now. 1 Srishti Kumari answered this Please find this answer 13