Solution to Question 44
Physics Principle: Coulomb’s Law, Friction, and Work-Energy Theorem.
Given:
\(m = 50 \text{ g} = 0.05 \text{ kg}\)
\(q = 10 \mu\text{C} = 10^{-5} \text{ C}\)
\(r_0 = 10 \text{ m}\) (Initial separation)
\(\mu = 0.2\)
Step 1: Determine the critical distance for sliding
The second particle will start sliding when the electrostatic repulsion overcomes the maximum static friction. $$F_{elec} = f_{max}$$ $$\frac{k q^2}{r^2} = \mu mg$$ $$\frac{(9 \times 10^9) (10^{-5})^2}{r^2} = 0.2 \times 0.05 \times 10$$ $$\frac{0.9}{r^2} = 0.1 \implies r^2 = 9 \implies r = 3 \text{ m}$$
So, we must push the first particle from \(r_0 = 10 \text{ m}\) to \(r = 3 \text{ m}\).
Step 2: Calculate Minimum Force using Work-Energy
We apply a constant horizontal force \(F_{app}\). To find the minimum force, the particle should reach the critical distance \(r = 3 \text{ m}\) with zero kinetic energy.
Work Done by \(F_{app}\) + Work Done by Friction + Work Done by Electrostatic Force = Change in KE (= 0)
Displacement \(d = 10 – 3 = 7 \text{ m}\). $$W_{app} = F_{app} \cdot d = 7 F_{app}$$ $$W_{fric} = – \mu mg \cdot d = -0.1 \times 7 = -0.7 \text{ J}$$ $$W_{elec} = -\Delta U = – (U_{final} – U_{initial}) = – k q^2 \left( \frac{1}{r} – \frac{1}{r_0} \right)$$ $$W_{elec} = – 0.9 \left( \frac{1}{3} – \frac{1}{10} \right) = -0.9 \left( \frac{7}{30} \right) = -0.21 \text{ J}$$
Summing works to zero: $$7 F_{app} – 0.7 – 0.21 = 0$$ $$7 F_{app} = 0.91$$ $$F_{app} = \frac{0.91}{7} = 0.13 \text{ N}$$