Question

Part A

An electron is moving east in a uniform electric field of 1.48 N/C directed to the west. At point A, the velocity of the electron is 4.47×10⁵ m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.390 m east of point A?

ve =		m/s

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Part B

A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.93×10⁴ m/s , again pointed towards the east. What is the speed of the proton at point B?

vp =		m/s

Answer 1

Part A) v_e = [u_e² + 2(eE/m_e)d_AB]^1/2

=> v_e = {(4.47 * 10⁵)² + [2 * (1.6 * 10^-19 * 1.48 * 0.39 / (9.1 * 10^-31)]}^1/2 = 6.35 * 10⁵ m/s

Part B) v_p = [u_p² - 2(eE/m_p)d_AB]^1/2

=> v_p = {(1.93 * 10⁴)² - [2 * (1.6 * 10^-19 * 1.48 * 0.39 / (1.67 * 10^-27)]}^1/2 = 1.62 * 10⁴ m/s