An electron is moving east in a uniform electric field of magnitude 1.48N/C dire

Question

An electron is moving east in a uniform electric field of magnitude 1.48N/C directed to the west. At point A, the velocity of the electron is 4.48×105 pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.385m east of point A?
Use 1.60×10-19C for the magnitude of the charge on an electron and 9.11×10-31kg for the mass of an electron.
2-A proton is moving in the uniform electric field of part (a). At point A, the velocity of the proton is 1.95×104 m/s , again pointed towards the east. What is the speed of the proton at point B?
Use 1.67×10-27 kg for the mass of a proton.

Explanation / Answer

1) force on electron = 1.48*1.6*10-19 = 2.368*10-19 N towards east

acceleration of electron = F/m = 2.6*1011 m/s2

velocity at B = 6.3*105 m/s

2) force on proton = 2.368*10-19 N towards west

acceleration of proton = F/m = 1.41*108 m/s2

velocity at B = 1.6*104 m/s

RATE ME LIFESAVER!!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.