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

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
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