An electron is moving east in a uniform electric field of 1.54 N/C directed to t

Question

An electron is moving east in a uniform electric field of 1.54 N/C directed to the west. At point A, the velocity of the electron is 4.55×105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.350 m east of point A?

Answer:    6.30×105 m/s

Part B

A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.94×104 m/s , again pointed towards the east. What is the speed of the proton at point B?

Explanation / Answer

E = electric field = 1.54 N/C                  towards west

V = 4.55 x 105 m/s       towards east

m = mass of electron = 9.1 x 10-31 kg

q = charge on electron = 1.6 x 10-19 C

Force of Electron due to the electric field is given as

F = qE = (1.6 x 10-19 ) (1.54) = 2.464 x 10-19 N

acceleration of electron is given as

a = F/m = 2.464 x 10-19 / (9.1 x 10-31 ) = 0.271 x 1012 m/s2 towards east

Velocity of electrocn at A = 4.55 x 105 m/s       towards east

d = distance covered to reach poiint B = 0.350 m

Part B :

m = mass of proton = 1.67 x 10-27 kg

q = charge on proton = 1.6 x 10-19 C

Force of Proton due to the electric field is given as

F = qE = (1.6 x 10-19 ) (1.54) = 2.464 x 10-19 N in the west direction   (since positive charges experience elecric force in the same direction as electric field)

acceleration of electron is given as

a = F/m = 2.464 x 10-19 / (1.67 x 10-27 ) = 1.475 x 108 m/s2 towards west

Velocity of electrocn at A = 1.94 x 104 m/s       towards east

d = distance covered to reach poiint B = 0.350 m

using the equation

Vf2 = Vi2 + 2 a d

Vf2 = (1.94 x 104 )2 + 2 (-1.475 x 108 ) (0.350)

Vf = 1.65 x 104 m/s

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