Two spheres carrying charges ( +6 C ) and ( +9 C ). separated by a distance ( d ), experiences a force of repulsion ( F ). When a charge of ( -3 C ) is given to both the sphere and kept at the same distance as before, the new force of repulsion is

Step-by-step solution

Given, charge q₁=+6 C=6 10⁻⁶ C ; charge q₂=+9 C=9 10⁻⁶ C seperation =d Force between charges is given as F= 1 4 ₀ q₁ q₂ d² q₁ q₂ d² = 6 10⁻⁶ 9 10⁻⁶ d² (1) When a charge of +3 C is given to both and separation =d , then q₁=+6 C-3 C=3 C=3 10⁻⁶ Cq₂=+9 C-3 C=6 C=6 10⁻⁶ C The force between charges is given as F^ = 1 4 ₀ q₁^ q₂^ d² q₁^ q₂^ d² = 3 10⁻⁶ 6 10⁻⁶ d² (2) Dividing Eq. (1) by Eq. (2) , we get F F = 6 10⁻⁶ 9 10⁻⁶ 3 10⁻⁶ 6 10⁻⁶ =3 F^ = F 3 Therefore, the new force of repulsion is F 3