Two identically charged particles are attached to the opposite ends of a spring having a natural length of 10 cm and a spring constant of 100 N m ⁻¹ . The system rests on a smooth horizontal table. Determine the extension in the spring's length if the charge on each particle is 2.0 10⁻⁸ C . Assume the extension is small compared to the natural length.

Step-by-step solution

Let the natural length of the spring be L and the extension be x . The electrostatic repulsive force between the two charges is given by Coulomb's law: F_ e = 1 4 ₀ q² (L+x)² Since the extension x is small compared to the natural length L , L+x L . Thus, F_ e 1 4 ₀ q² L² The restoring force of the spring is: F_ s = kx In equilibrium, the electrostatic repulsive force is balanced by the restoring force of the spring: kx = 1 4 ₀ q² L² Substituting the given values, q = 2.0 10⁻⁸ C , L = 10 cm = 0.1 m , k = 100 N m ⁻¹