A muon is decelerated by an electric field of 2.2 x 103 N/C to rest over a distance of 4.20cm. (a) Find the particle's initial velocity and kinetic energy. (b) Perform the same calculations in (a) for a tau particle

The force on the charged particle q in an electric field is F = qE. The acceleration is a = 48 Here, m is the mass of the particle. a) For a muon, the charge is q = -1.6 x 10-19 C. The mass of muon is m = 1.883 x 10 "kg. The acceleration is a =. -1.6 x 10 1 x 2.2 x 103 1.883 x 10 2 = -1.97 x 1012 _ As this is the deceleration, the relation between the distance and the acceleration is given as v2 = 12 - 2ax. Here, v = 0 is the final velocity of the particle. The initial velocity is, 1 = v2ax = V2 x 1.97 x 102 x (4.20 x 10 ?) = 4.07 x 10- The initial kinetic energy is K =- mxu x 1.883 x 10 2 x (4.07 x 105) = 1.56 x 10 J Explanation: The particle is negative, so the acceleration is also negative here. Step 3/3 A b) For a tau particle, the charge is q = -1.6 x 10 C and the mass is m = 3.168 x 10 -kg. The acceleration will be -1.6 x 10 13 x 2.2 x 103 a = 3.168 x 10-12 = -1.11 x 10 - The initial velocity is, u = V2 x 1.11 x 10" x 4.20 x 10-2 = 9.66 x 101 1 The initial kinetic energy is K =- x 3.168 x 10 2 x (9.66 x 10*)2 = 1.48 x 10 J