A particle of mass 'm' and charge 'q', initially at rest, is accelerated by a uniform electric field 'E' through a distance 'D' and is then allowed to approach a fixed static charge 'Q' of the same sign. The distance of the closest approach of the charge q is [ ₀ = permittivity of free space ]

Step-by-step solution

Work done by the uniform electric field on the particle is W = qED . By work-energy theorem, the kinetic energy gained by the particle is K = qED . At the distance of closest approach r , the entire kinetic energy is converted into electrostatic potential energy. The electrostatic potential energy of the system is U = 1 4 ₀ qQ r . Equating kinetic energy and potential energy: qED = 1 4 ₀ qQ r r = Q 4 ₀ ED Answer: Q / 4 ₀ ED