A proton accelerates from rest in a uniform electric field o

A proton accelerates from rest in a uniform electric field of 638 N/C. At some later time, its speed is 1.02 Times 10^6 m/s. Find the magnitude of the acceleration of the proton. How long does it take the proton to reach this speed? How far has it moved in that interval? Your incorrect answer may have resulted from round off error. Make sure you keep extra significant figures in intermediate steps of your calculation, m What is its kinetic energy at the later time?

Solution

a.

acceleration of the proton is,

a = Eq/m = 638*1.6x10^-19 / 1.67x10^-27 = 610e8 m/s²

b.

the time t is calculated as follows:

   t = v-u/a = 1.02x10⁶ / 610e8 = 16.7e-6 = 16.7 us

c.

the distance is calculated as follows:

    s = 0.5at^2 = 0.5*610e8*16.7e-6 ^2 = 8.5 m

d.

the kinetic energy is,

   K = 0.5mv^2 = 0.5*1.67x10^-27*(1.02x10⁶ )² = 8.68e-16 J = 0.868e-15 J