My Notes Ask Your Teac A proton accelerates from rest in a uniform electric fiel

Question

My Notes Ask Your Teac A proton accelerates from rest in a uniform electric field of 630 N/C. At some later time, its speed is 1.02 x 106 m/s. (a) Find the magnitude of the acceleration of the proton. 60310 m/s2 (b) How long does it take the proton to reach this speed? 1.68990-6X Use the acceleration you calculated in part (a) of this problem to find the time required for the proton to reach the specified speed. us (c) How far has it moved in that interval? (d) What is its kinetic energy at the later time? Need Help? Master

Explanation / Answer

Given,

E = 630 N/C ; v = 1.02 x 10^6 m/s

a)We know that the proton will experience a net electric force.

Fnet = ma = q E

a = q E/m = 1.6 x 10^-19 x 630/1.67 x 10^-27 = 6.04 x 10^10 m/s^2

Hence, a = 6.04 x 10^10 m/s^2

b)from eqn of motion we know,

v = u + at => t = (v - u)/a

t = (1.02 x 10^6 - 0)/6.04 x 10^10 = 1.69 x 10^-5 s

Hence, t = 1.67 x 10^-5 s

c)from eqn of motion

v^2 = u^2 + 2 a S

S = (v^2 - u^2)/ 2a

S = [(1.02 x 10^6)^2 - 0^20]/2 x 6.04 x 10^10 = 8.61 m

Hence, S = 8.61 m

d)KE = 1/2 m v^2

KE = 0.5 x 1.67 x 10^-27 x (1.06 x 10^6)^2 = 9.38 x 10^-16 J

Hence, KE = 9.38 x 10^-16 J

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