An @lect on with speed of 7.35 × 10 cm/s in the positive direction of an x axis

Question

An @lect on with speed of 7.35 × 10 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.09 x 10s N/C, traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed? (c) If the region containing the electric field is 4.17 mm long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region? (a)N (b) Numb (c) N 735 umberTo

Explanation / Answer

a. The electric field retards the electron's motion. So if the electron travels far enough in the field to stop then the work done by the electric field on the electron has to equal the electron's intial kinetic energy.

K = 1/2mv^2

So W =K

qEd = 1/2 mv^2

d = 1/2 (mv^2/(Eq))

d = 0.734 m = 73.4 cm

b.>

the initial speed and the final speed (0 m/s)

F = qE = ma

a = qE/m

So v = -at + v0 = -qEt/m +v0

v = 0

t = mv0/(qE) where v0 = 7.35x10^6 m/s

t = 1.997x10^-8 sec

c.>

W =qEd

d = 4.17 mm = 0.00417 m

W = 1.396x10^-18 J

This is the kinetic energy lost by the electron. The fraction is:

f = (qEd)/[1/2mv^2]

v= 7.35x10^6 m/s

f = 0.0568 = 5.68%

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