Consider two identical metallic spheres of radius R each having charge Q and mass m . Their centers have an initial separation of 4 R . Both the spheres are given an initial speed of u towards each other. The minimum

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Correct answer: B. k Q^ 2 4 m R (1- G m^ 2 k Q^ 2 )

Using energy conservation, the system's kinetic energy must overcome the increase in electrostatic potential energy as the spheres approach.
Initial state: separation 4R, speed u each.
Final state: touching (separation 2R), speed zero.
Change in potential energy:

\Delta U = k\frac{Q^2}{2R} - k\frac{Q^2}{4R} = k\frac{Q^2}{4R}

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Initial kinetic energy:

KE_i = mu^2

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Energy equation:

mu^2 = k\frac{Q^2}{4R}\left(1 - \frac{Gm^2}{kQ^2}\right)

.
Therefore,

u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}

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Consider two identical metallic spheres of radius R each having...

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