Question

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 .

Answer #1

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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