An electron has a speed of 5.8 x 105 m/s at point A. The voltage at point A is 3700 V and at point B is 5000 V

Question

(a) What is the speed of the electron at point B? Assume forces other than the electric force are negligible.

(b) Repeat the question for a proton instead of an electron. Notice the big difference between accelerating an electron and a proton.

Answer 1

Final speed is given by vf = vi + (2e∆V / m)^(1/2).

(a) vf = (5.8 x 10^5 m/s) + ((2 x (1.6 x 10^-19 C) x (5000 V - 3700 V)) / (9.11 x 10^-31 kg))^(1/2)

= 2.19 x 10^7 m/s

(b) vf = (5.8 x 10^5 m/s) + ((2 x (1.6 x 10^-19 C) x (5000 V - 3700 V)) / (1.67 x 10^-27 kg))^(1/2)

= 1.08 x 10^6 m/s