A proton accelerates from rest in a uniform electric field of680 N/C. At one lat

Question

A proton accelerates from rest in a uniform electric field of680 N/C. At one later moment, itsspeed is 1.50 Mm/s (nonrelativisticbecause v is much less than the speed of light). (a) Find the acceleration of the proton.
1 m/s2

(b) Over what time interval does the proton reach this speed?
2 s

(c) How far does it move in this time interval?
3 m

(d) What is its kinetic energy at the end of this interval?
4 J (a) Find the acceleration of the proton.
1 m/s2

(b) Over what time interval does the proton reach this speed?
2 s

(c) How far does it move in this time interval?
3 m

(d) What is its kinetic energy at the end of this interval?
4 J

Explanation / Answer

        Electric field, E = 680N/C         Initial speed, Vi = 0m/s         Final speed, Vf = 1.5 x10 6 m/s         Mass of proton, m = 1.67x 10 -27 kg         Force acting on proton,F = E q                                                = 680 * 1.6 X 10 -19                                                = 1088 X 10 -19 N         (a)         Acceleration, a = F /m                                = 1088 X 10 -19 / 1.67 x 10 -27                                = 651.5 X 10 8 m/s^2         (b)          Time interval, t =( Vf - Vi ) / a                                = 1.5 x 10 6 / 651.5 X 10 8                                = 0.0023 X 10 -2 s                                = 23 s          (c)           Distance S =Vi t + ( 1/2 ) a t 2                           = 0 + 0.5 * 651.5 x 10 8 * 23 2 X 10-12                           = 172321.5 X 10 -4 m                           = 17.23 m          (d)           Kineticenergy, Ke = ( 1/2 ) m Vf 2                                       = 0.5 * 1.67 X 10 -27 * 1.5 2 X 1012                                      = 1.88 x 10 - 15 J

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