Part A An electron is moving east in a uniform electric field of 1.48 N/C direct

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?

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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?

Explanation / Answer

Part A) ve = [ue2 + 2(eE/me)dAB]1/2

=> ve = {(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) vp = [up2 - 2(eE/mp)dAB]1/2

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

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