An electron with a speed of 5.50 108 cm/s enters an electric field of magnitude

Question

An electron with a speed of 5.50 108 cm/s enters an electric field of magnitude 1.00 103 N/C, traveling along the field in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily? 8.60 Incorrect: Your answer is incorrect. Do you realize that the force magnitude is the product of charge magnitude and electric field magnitude? How is force related to acceleration? Do you recall how to use the constant-acceleration equations? Do you recall how kinetic energy is related to the speed? m (b) How much time will have elapsed? s (c) If, instead, the region of electric field is only 2.00 mm long (too small for the electron to stop), what percentage of the electron's initial kinetic energy will be lost in that region? %

Explanation / Answer

q = charge on electron

m = mass of electron

a = retardation caused = qE/m

Vo = initial velocity

Vf = final velocity = 0

d = stopping distance

Using the equation

V2f = Vi2 + 2 a d

02 = (5.5 x 106)2 + 2 (qEd/m)

02 = (5.5 x 106)2 + 2 ((1.6 x 10-19)(1000)d/(9.1 x 10-31))

d = 0.086 m

b)

Vf = Vo + a t

0 = 5.5 x 106 + (qEt/m)

0 = 5.5 x 106 - (1.6 x 10-19)(1000)t/(9.1 x 10-31)

t = 3.13 x 10-8 sec

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