Please answer the following question A proton accelerates fr

Please answer the following question.

A proton accelerates from rest in a uniform electric field of E = 640 N/C. At one later moment, its speed is 1.2 times10^6 m/s. Given that the mass of the proton is m = 1.67 times 10^27 kg, and the charge of the proton is q = 1.6 times 10^19 C (a) Find the acceleration of the proton. (b)Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?

Solution

Electric field E = 640 N/C

Initial speed u = 0 m/s

Charge of proton q = 1.6 x10 ^-19 C

Final speed v = 1.2 x10 ⁶ m/s

mass of proton m = 1.67 x10 ^-27 kg

Charge of proton q = 1.6 x10 ^-19 C

(a). Accleration of the proton a = Eq / m

                                            = (640)(1.6 x10 ^-19 ) /(1.67 x10 ^-27 )

                                            = 6.131 x10 ¹⁰ m/s ²

(b).Required time t = (v- u) / a

                            = [(1.2 x10 ⁶) - 0] /(6.131 x10 ¹⁰)

                            = 1.957 x10 ^-5 s

(c).Distance travel in this time S = ut +(1/2)at ²

                                                = 0 + [0.5 x6.131 x10 ¹⁰ x (1.957x10 ^-5) ² ]

                                                = 11.74 m

(d).Kinetic energy at the end of this interval K.E = (1/2)mv ²

                                                                     = (1/2) (1.67 x10 ^-27 )(1.2 x10 ⁶) ²

                                                                     = 1.2024 x10 ^-15 J