The electric field points due east and has a magnitude of 3.0 105 N/C. The mass
ID: 1917809 • Letter: T
Question
The electric field points due east and has a magnitude of 3.0 105 N/C. The mass of the electron is 9.11 10-31 kg and that of the proton is 1.67 10-27 kg. Both the proton and electron have charge magnitudes of 1.60 10-19 C. (d) What is the algebraic expression for the magnitude a of the acceleration that a charge experiences when placed in an electric field? Express your answer in terms of the magnitude |q| and mass m of the charge, and the magnitude E of the electric field. (Answer using q, m, and E.) a = (e) Determine the magnitude ap of the acceleration of the proton. Number Unit ap = (f) Determine the magnitude ae of the acceleration of the electron. Number Unit ae =Explanation / Answer
force on electron = 1.48*1.6*10-19 = 2.368*10-19 N towards eastacceleration of electron = F/m = 2.6*1011 m/s2velocity at B = 6.3*105 m/s2) force on proton = 2.368*10-19 N towards westacceleration of proton = F/m = 1.41*108 m/s2 velocity at B = 1.6*104 m/s