anyone care for some electrostatic physics?

Question

The force of repulsion that two like charges exert on each other is 3.0 N. What will the force be if the distance between the charges is increased to 6 times its original value?

Answer 1

F1=(k q1 q2 )/ R1
F2=(k q1 q2 )/ R2^2
R2=6R1
F1/F2=(k q1 q2 )/ R1^2/=(k q1 q2 )/ R2^2
F1/F2=R2^2/R1^2
since R2=6R1
F1/F2=36R1^2/R1^2
then
F2=(1/36)F1= 3/36=1/12 N =0.083N