A particle A with a charge of 2.0 imes 10⁻⁶ C is held fixed on a horizontal table. A second charged particle of mass 80 g remains in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and the second particle is = 0.2 . Determine the range within which the charge of the second particle may lie.

Step-by-step solution

The maximum force of static friction acting on the second particle is given by f = m g Substituting the given values: f = 0.2 0.08 9.8 = 0.1568 N The electrostatic force between the two charges is F = 1 4 ₀ |q₁ q₂| r^2 Substituting the known values: F = 9 10^9 2.0 10⁻⁶ |q₂| (0.1)^2 F = 1.8 10^6 |q₂| For the second particle to remain in equilibrium, the electrostatic force must be less than or equal to the maximum static friction: F f 1.8 10^6 |q₂| 0.1568 |q₂| 0.1568 1.8 10^6 |q₂| 8.71 10⁻⁸ C Thus, the charge of the