An electron is moving east in a uniform electric field of 1.50 N/C directed to t

Question

An electron is moving east in a uniform electric field of 1.50 N/C directed to the west. At point A, the velocity of the electron is 4.54 105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.365 m east of point A? 0 xm/s Submit Request Answer Part B A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.87x104 m/s, again pointed towards the east. What is the speed of the proton at point B? Up

Explanation / Answer

Fe = q E

Fe = (-1.6 x 10^-19) (-1.50) = 2.4 x 10^-19 N

a = Fe / m = 2.635 x 10^11 m/s^2 to the east.

vf^2 - vi^2 = 2 a d

v^2 - (4.54 x 10^5)^2 = 2(2.635 x 10^11)(0.365)

v = 6.31 x 10^5 m/s ..........Ans

(B) a = - (2.4 x 10^-19)/(1.67 x10^-27)

a = - 1.437 x 10^8 m/s^2 ( to the west)

vp^2 - (1.87 x 10^4)^2 = 2(-1.437 x 10^8)(0.365)

vp = 1.56 x 10^4 m/s ..........Ans

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