Question

An electron is released from rest in a uniform electric field. a) If the electron is accelerated to 0.5% of the speed of light after traveling 5 mm, what is the strength of the electric field? b) What speed will the electron have acquired after traveling 15 mm from rest?

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Answer 1

(a) speed of light = 3 times 1068 m/s 0.5% of this speed = 0.015 times 10^8 15 times 10^5 m/s v^2 = u^2 + 2as (15 times 10^5)^2 = 2 times a times (5 times 10^-3) a = (15 times 10^5)^2/2 times 5 times 10^-3 = 2.25 times 10^14 m/s^2 qE = ma E = ma/q = 9.1 times 10^-31 times 2.25 times 10^14/1.6 times 10^-19 = 1279.6 N/C (b) v^2 = u^2 + 2as v = squareroot 2 times 2.25 times 10614 times 15 times 10^-3 = 2598076.21 v = 2.598 times 10^6 m/s