A proton is released from rest in a uniform electric field of magnitude 1.33×105 N/C . Find the speed of the proton after it has traveled (a) 1.40 cm and (b) 13.0 cm .
Uniform electric field = 1.33 * 10^5 N/C
Proton charge = 1.6 * 10^-19 C
Mass of proton = 1.67 * 10^-27 kg
Calculate the force on proton.
F =qE
=1.6*10^-19*1.33*10^5
= 2.128 * 10^-14
Calculate the acceleration.
a = F/m
= 2.128*10^-14/(1.67*10^-27)
Acceleration of the proton = 1.274 * 10^13 m/s^2
s = 1.4 cm
Speed of the proton = sqrt (2 . a. s)
= sqrt (2 * 1.274 * 10^13 * 0.014 )=5.97* 10^5 m/s
when
s = 13 cm
Speed of the proton = sqrt (2 . a. s)
= sqrt (2 * 1.274 * 10^13 * 0.13) = 0.182* 10^7 m/s
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