An electron with a speed of 5.73 times 10^8 cm/s in the positive direction of an

Question

An electron with a speed of 5.73 times 10^8 cm/s in the positive direction of an x axis enters an electric field of magnitude 1.39 times 10^3 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 4.46 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? Number ____ Units ______ Number ____ Units ______ Number ____ Units ______

Explanation / Answer

Initial speed u = 5.73 x10 8 cm / s

= 5.73 x10 6 m/s

Electric field E = 1.39 x10 3 N/C

Accleration a = -Eq/m

Where m = mass of electron = 9.11 x10 -31 kg

q = charge of electron = 1.6 x10 -19 C

Substitute values you get , a = (1.39 x10 3)(1.6x10 -19)/(9.11x10 -31)

= 2.441x10 14 m/s 2

(a). From the relation v 2 - u 2 = 2aS

Where v = fina velocity = 0

- u 2 = 2aS

From this required distance S = -u 2 /2a

S = -( 5.73 x10 6 ) 2 /[2(-2.441x10 14)]

= 67.245x10 -3 m

(b).Required time t = (v-u) / a

= [0-(5.73 x10 6)]/(-2.441x10 14)

= 2.347x10 -8 s

(c).Distance S = 4.46 mm = 4.46x10 -3 m

From the relation v 2 - u 2 = 2aS

v 2 = u 2 + 2aS

=(5.73 x10 6) 2 +[2(-2.441x10 14)(4.46x10 -3)]

=3.065 x10 13

v = 5.536x10 6 m/s

Loss in kinetic energy dK = (1/2) m[u 2- v 2 ]

Required fraction = dK /K

Where K = (1/2) mu 2

So, dK / K = (u 2- v 2 ) / u 2

= 1.137x10 -3

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