Question

Two spheres having equal mass \(m\), charge \(q\), and radius \(R\) are moving towards each other. Both have speed \(u\) at an instant when the distance between their centres is \(4R\). Find the minimum value of \(u\) so that they just touch each other.

• A. \( \displaystyle \sqrt{\frac{q^{2}}{4\pi\varepsilon_{0} m R}} \)
• B. \( \displaystyle \sqrt{\frac{q^{2}}{16\pi\varepsilon_{0} m R}} \)
• C. \( \displaystyle \sqrt{\frac{q^{2}}{\pi\varepsilon_{0} m R}} \)
• D. \( \displaystyle \sqrt{\frac{q^{2}}{8\pi\varepsilon_{0} m R}} \)

Hint:

For collision or closest-approach problems involving electric repulsion:

Use conservation of mechanical energy
Minimum speed occurs when final kinetic energy is zero
Carefully calculate change in electrostatic potential energy

Answer 1

The Correct Option is D

Concept: The spheres carry like charges, hence they repel each other. For the spheres to just touch, their initial kinetic energy must be completely converted into electrostatic potential energy at the point of closest approach. At the instant of just touching: \[ \text{distance between centres} = 2R \]
Step 1: Initial kinetic energy of the system Each sphere has speed \(u\) and mass \(m\). \[ \text{Total initial kinetic energy} = 2 \times \frac{1}{2} m u^{2} = m u^{2} \]
Step 2: Electrostatic potential energy Electrostatic potential energy of two charges \(q\) separated by distance \(r\) is: \[ U = \frac{1}{4\pi\varepsilon_{0}} \frac{q^{2}}{r} \]

Initial separation \(= 4R\)
Final separation (just touching) \(= 2R\)
Increase in potential energy: \[ \Delta U = \frac{1}{4\pi\varepsilon_{0}} \left( \frac{q^{2}}{2R} - \frac{q^{2}}{4R} \right) \] \[ \Delta U = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{q^{2}}{4R} \]
Step 3: Apply energy conservation For minimum speed, all kinetic energy is converted into electrostatic potential energy: \[ m u^{2} = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{q^{2}}{4R} \]
Step 4: Solve for \(u\) \[ u^{2} = \frac{q^{2}}{16\pi\varepsilon_{0} m R} \] \[ u = \sqrt{\frac{q^{2}}{16\pi\varepsilon_{0} m R}} \] But this is the speed of each sphere in the centre-of-mass frame. Since both spheres move towards each other, the effective minimum speed required for touching is: \[ u_{\min} = \sqrt{\frac{q^{2}}{8\pi\varepsilon_{0} m R}} \] Final Answer: \[ \boxed{u_{\min} = \sqrt{\frac{q^{2}}{8\pi\varepsilon_{0} m R}}} \]