Question

two identical balls having like charges placed at a certain distance apart repel each other with a certain force they are brought in contact and then move the path to a distance equal to how they release initial separation the force of repulsion between them increases 4.5 times in comparison with the initial value the ratio of the initial charges of the balls is

Answer 1

  Let balls have charges Q1 & Q2respectively

 ∴F=[(kQ1Q2)r2]                                                   (1)

when two charges are brought in contact & moved apart then they a charge [(Q1 + Q2) / 2]

∴ in final condition, F' = [k{(Q1+ Q2) / 2}2 / (r/2)2]

 ∴ F' = [{k(Q1 + Q2)2} /r2]                                                      (2)

                    as F' = 4.5F hence from (1) & (2)

                              [{k(Q1 + Q2)2} / r2] = 4.5 [(kQ1Q2) / r2]

                              ∴ (Q1 + Q2)2 = 4.5 Q1Q2

                              ∴ Q12 + Q22 + 2Q1Q2 = 4.5 Q1Q2

                              ∴ Q12 + Q22 – 2.5 Q1Q2 = 0

                              ∴ (Q1 / Q2)2 + 1 – 2.5(Q1Q2) = 0 ------- dividing by Q22

                              Let (Q1 / Q2) = x

                              ∴ x2 – 2.5x + 1 = 0

                                 x2 – 2x – 0.5x + 1 = 0

                              x(x – 2) – 0.5x(x – 2) = 0

                                  x = 2         or       x = 0.5

                              ∴ (Q1 / Q2) = (1/2) or (Q1 / Q2) = 2   

Answer 2

F= K ((+q)×(-2q))/r^2 = -2kq^2/r^2 When the chages +q and –2q comes in contact, than it neutralises the chage+q by –q charge and lefts only –q. These -q charge will be equally distributed both the sphare. If it is seperated the two sphare again than each sphare will aquirea mount of charge of –q/2 and –q/2 .Now the new force between them will be F_n= K ((-q^2/2)(-q^2/2))/r^2 = 1/4 ((kq^2)⁄r^2 ) = (-F)/8So