Solution to Question 43
Physics Principle: Conservation of Momentum and Energy in Electrostatics.
Step 1: Initial State Analysis
Initially, the tension \(T\) in the thread counteracts the electrostatic repulsion between the two charges \(q_1\) and \(q_2\) separated by distance \(l\). $$T = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{l^2} \implies \frac{q_1 q_2}{4\pi\epsilon_0} = T l^2$$
Step 2: Conservation of Momentum
When the thread is cut, the balls fly apart. Since the net external force on the system is zero, momentum is conserved.
The first ball (\(m_1\)) has momentum \(p\).
The second ball (\(m_2\)) must have momentum \(-p\) (magnitude \(p\)).
Step 3: Conservation of Energy
The system loses Electrostatic Potential Energy and gains Kinetic Energy. Let the new distance be \(r\). $$\Delta PE = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{l} – \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$ $$KE_{total} = \frac{p^2}{2m_1} + \frac{p^2}{2m_2} = \frac{p^2 (m_1+m_2)}{2 m_1 m_2}$$
Equating \(\Delta PE\) to \(KE_{total}\): $$\frac{q_1 q_2}{4\pi\epsilon_0} \left( \frac{1}{l} – \frac{1}{r} \right) = \frac{p^2 (m_1+m_2)}{2 m_1 m_2}$$ Substitute \(\frac{q_1 q_2}{4\pi\epsilon_0} = T l^2\): $$T l^2 \left( \frac{r-l}{lr} \right) = \frac{p^2 (m_1+m_2)}{2 m_1 m_2}$$ $$T l \left( 1 – \frac{l}{r} \right) = \frac{p^2 (m_1+m_2)}{2 m_1 m_2}$$
Solving for \(r\): $$1 – \frac{l}{r} = \frac{p^2 (m_1+m_2)}{2 m_1 m_2 T l}$$ $$\frac{l}{r} = 1 – \frac{p^2 (m_1+m_2)}{2 m_1 m_2 T l}$$ $$r = \frac{l}{1 – \frac{p^2 (m_1+m_2)}{2 m_1 m_2 T l}} = \frac{2 m_1 m_2 T l^2}{2 m_1 m_2 T l – p^2 (m_1+m_2)}$$