Electrostatics Part 3

Concept:The minimum speed required for the spheres to just touch is found using conservation of energy, where the initial kinetic energy plus potential energies equals the final potential energies at contact.Explanation:Initial separation between centres: $4R$.Final separation (just touching): $2R$ (each sphere radius $R$).Initial total kinetic energy (both spheres): $\frac{1}{2}m u^2 + \frac{1}{2}m u^2 = m u^2$.Initial electrostatic potential energy: $\frac{k Q^2}{4R}$.Initial gravitational potential energy: $-\frac{G m^2}{4R}$.Final kinetic energy at touching (minimum speed): $0$.Final electrostatic potential energy: $\frac{k Q^2}{2R}$.Final gravitational potential energy: $-\frac{G m^2}{2R}$.Apply conservation of energy:$m u^2 + \frac{k Q^2}{4R} - \frac{G m^2}{4R} = \frac{k Q^2}{2R} - \frac{G m^2}{2R}$.Rearrange: $m u^2 = \frac{k Q^2}{2R} - \frac{G m^2}{2R} - \frac{k Q^2}{4R} + \frac{G m^2}{4R}$.Simplify: $m u^2 = \left( \frac{k Q^2}{4R} \right) - \left( \frac{G m^2}{4R} \right) = \frac{1}{4R} (k Q^2 - G m^2)$.Factor: $u^2 = \frac{k Q^2}{4 m R} \left(1 - \frac{G m^2}{k Q^2} \right)$.Thus $u = \sqrt{ \frac{k Q^2}{4 m R} \left(1 - \frac{G m^2}{k Q^2} \right) }$.Answer:Option D: $\sqrt{\frac{k Q^2}{4 m R} \left(1 - \frac{G m^2}{k Q^2} \right)}$.