Two identical solid metallic spheres, each of mass m , radius R , and charge Q , are initially held such that they are just touching each other. They are released from rest. Assuming the electrostatic repulsion dominates the gravitational attraction, what is the relative speed of separation between the two spheres when the distance between their centers becomes 6R ? (Take k= 1 4 ₀ and G as the universal gravitational

Step-by-step solution

Initial center-to-center distance, r_i = 2R (since the spheres are just touching). Final center-to-center distance, r_f = 6R . From the conservation of mechanical energy, the decrease in potential energy equals the increase in kinetic energy. U = U_i - U_f = ( kQ^2 2R - Gm^2 2R ) - ( kQ^2 6R - Gm^2 6R ) U = (kQ^2 - Gm^2) ( 1 2R - 1 6R ) = kQ^2 - Gm^2 3R Let v be the speed of each sphere when the separation is 6R . The total kinetic energy is: K = 1 2 mv^2 + 1 2 mv^2 = mv^2 Equating U and K : mv^2 = kQ^2 - Gm^2 3R v