Two identical charged particles each having a mass 10 g and charge 2.0 × 10⁻⁷C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms⁻²]

Solution

Physics sir · 5 steps

Limiting equilibrium means the Coulomb push has just reached the maximum static friction, μ mg.

1. Forces on one particle. Like charges repel, so each particle is pushed away from the other along the table. Static friction pulls it back. In limiting equilibrium, friction is at its largest possible value, μ mg.

   kq²/L² = μ m g

2. Friction side. With m = 10 g = 0.01 kg, the limiting friction is μ mg = 0.25 × 0.01 × 10 = 0.025 N.
3. Coulomb side. kq² = 9 × 10⁹ × (2.0 × 10⁻⁷)² = 9 × 10⁹ × 4 × 10⁻¹⁴ = 3.6 × 10⁻⁴ N m².
4. Solve for L. L² = (3.6 × 10⁻⁴)/0.025 = 1.44 × 10⁻² m², so L = 0.12 m.
5. So. The separation is L = 12 cm, option (a).

The idea to keep. Limiting equilibrium is the edge just before slipping. Set the applied force equal to μ N, not to some smaller friction value. Keep every quantity in SI units, then convert at the end.

Answer from the NTA final answer key. Solution by PrepWiser.