A long cylindrical wire possesses a positive charge with a linear density of 2.0 10⁻⁸ C m ⁻¹ . An electron orbits the wire in a circular path, driven by the attractive electrostatic force. Determine the kinetic energy of the electron. (Note that this energy is independent of the orbital radius.)

Step-by-step solution

The electric field at a distance r from a long cylindrical wire is given by: E = 2 ₀ r The electrostatic force on the electron provides the necessary centripetal force for its circular motion: m v^2 r = eE = e 2 ₀ r m v^2 = e 2 ₀ The kinetic energy of the electron is: K = 1 2 m v^2 = e 4 ₀ Substituting the given values = 2.0 10⁻⁸ C m ⁻¹ , e = 1.6 10⁻¹⁹ C , and 1 4 ₀ = 9 10^9 N m ^2 C ⁻² : K = (9 10^9) (1.6 10⁻¹⁹) (2.0 10⁻⁸) K = 28.8 10⁻¹⁸ J = 2.88 10⁻¹⁷ J