Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is: (Take k=frac14piepsilon_0$k=\frac{1}{4\pi\epsilon_{0}}$ and assume kQ^2>Gm^2$kQ^{2}>Gm^{2}$ where G is the Gravitational constant)

### Related Formula K_i + U_i = K_f + U_f$$K_i + U_i = K_f + U_f$$ U = frackq_1q_2r - fracGm_1m_2r$$U = \frac{kq_1q_2}{r} - \frac{Gm_1m_2}{r}$$ ### Core Logic Using energy conservation from the initial state (separation 4R$4R$) to the final state (just touching, so center-to-center separation is 2R$2R$). Both spheres have mass m$m$ and speed u$u$. Initial Energy: E_i = 2left(frac12mu^2right) - fracGm^24R + frackQ^24R$$E_i = 2\left(\frac{1}{2}mu^2\right) - \frac{Gm^2}{4R} + \frac{kQ^2}{4R}$$ Final Energy (just touching implies final velocity is zero): E_f = - fracGm^22R + frackQ^22R$$E_f = - \frac{Gm^2}{2R} + \frac{kQ^2}{2R}$$ ### Step 1: Equating Energies mu^2 - fracGm^24R + frackQ^24R = - fracGm^22R + frackQ^22R$$mu^2 - \frac{Gm^2}{4R} + \frac{kQ^2}{4R} = - \frac{Gm^2}{2R} + \frac{kQ^2}{2R}$$ mu^2 = left(frackQ^22R - frackQ^24Rright) - left(fracGm^22R - fracGm^24Rright)$$mu^2 = \left(\frac{kQ^2}{2R} - \frac{kQ^2}{4R}\right) - \left(\frac{Gm^2}{2R} - \frac{Gm^2}{4R}\right)$$ mu^2 = frackQ^24R - fracGm^24R$$mu^2 = \frac{kQ^2}{4R} - \frac{Gm^2}{4R}$$ u^2 = frac14mR(kQ^2 - Gm^2)$$u^2 = \frac{1}{4mR}(kQ^2 - Gm^2)$$ ### Step 2: Final Conclusion u = sqrtfrackQ^24mRleft(1 - fracGm^2kQ^2right)$$u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$$ ### Pattern Recognition When dealing with two forces (electrostatic repulsion and gravitational attraction), their potentials simply superimpose linearly. The change in total potential energy equals the loss in kinetic energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics Class 11 Physics: Gravitation