Question

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 $k = \frac{1}{4 \pi \epsilon_0}$ and assume $kQ^2>Gm^2$ where $G$ is the Gravitational constant)

• A. \(\sqrt{\frac{kQ^2}{4mR} \left( 1 - \frac{Gm^2}{kQ^2} \right)}\)
• B. \(\sqrt{\frac{kQ^2}{2mR} \left( 1 - \frac{Gm^2}{kQ^2} \right)}\)
• C. \(\sqrt{\frac{kQ^2}{2mR} \left( 1 - \frac{Gm^2}{2kQ^2} \right)}\)
• D. \(\sqrt{\frac{kQ^2}{4mR} \left( 1 + \frac{Gm^2}{kQ^2} \right)}\)

Hint:

When two identical objects move toward each other, remember that the total kinetic energy is $m u^2$. For the "just touch" condition, the final separation is twice the radius.

Answer 1

The Correct Option is A

To solve this problem, we need to consider the forces acting on the two metallic spheres as they move towards each other. The two predominant forces are the electrostatic force and the gravitational force. Here's the step-by-step breakdown:

1. Initially, the separation between the centers of the two metallic spheres is \(4R\).
2. According to the problem, we need to find the minimum initial speed \(u\) such that the spheres just touch each other. This means that finally, their centers would be separated by \(2R\). 
3. The initial potential energy due to electrostatic force when the separation is \(4R\) is given by: \(U_{\text{electrostatic, initial}} = \frac{kQ^2}{4R}\)
4. The final potential energy due to electrostatic force when the separation is \(2R\) is: \(U_{\text{electrostatic, final}} = \frac{kQ^2}{2R}\)
5. Similarly, gravitational potential energy initially is negligible compared to the electrostatic energy since the problem assumes \(kQ^2 > Gm^2\). However, we consider: \(U_{\text{gravitational}} = -\frac{Gm^2}{4R}\) initially and \(U_{\text{gravitational, final}} = -\frac{Gm^2}{2R}\) finally.
6. Using conservation of energy: \(T_{\text{initial}} + U_{\text{initial}} = U_{\text{final}}\), where \(T_{\text{initial}}\) is the initial kinetic energy of the system.
7. The initial kinetic energy is: \(T_{\text{initial}} = 2 \cdot \frac{1}{2} m u^2 = mu^2\)
8. By conservation of energy, we set: \(mu^2 + \frac{kQ^2}{4R} - \frac{Gm^2}{4R} = \frac{kQ^2}{2R} - \frac{Gm^2}{2R}\)
9. Simplify this equation to find \(u^2\): \(mu^2 = \frac{kQ^2}{2R} - \frac{kQ^2}{4R} - \left( \frac{Gm^2}{2R} - \frac{Gm^2}{4R} \right)\)
10. Finally, solving for \(u\), we get: \(u = \sqrt{\frac{kQ^2}{4mR} \left( 1 - \frac{Gm^2}{kQ^2} \right)}\)

Thus, the minimum speed \(u\) with which the spheres should move towards each other is: \(\sqrt{\frac{kQ^2}{4mR} \left( 1 - \frac{Gm^2}{kQ^2} \right)}\).