A particle with a mass of 1 g and a charge of 2.5 10⁻⁴ C is released from rest within an electric field of 1.2 10^4 N C ⁻¹ . Determine the speed of the particle after it has travelled a distance of 40 cm .

Step-by-step solution

Given, mass of the particle, m = 1 g = 10⁻³ kg Charge, q = 2.5 10⁻⁴ C Electric field, E = 1.2 10^4 N C ⁻¹ Distance travelled, s = 40 cm = 0.4 m The force experienced by the particle in the electric field is: F = qE = (2.5 10⁻⁴) (1.2 10^4) = 3.0 N The acceleration of the particle is: a = F m = 3.0 10⁻³ = 3000 m s ⁻² Using the equation of motion v^2 = u^2 + 2as with initial velocity u = 0 : v^2 = 0 + 2 3000 0.4 = 2400 v = 2400 = 20 6 48.99 m s ⁻¹ 49.0 m s ⁻¹