An electron with a speed of 7.10 × 108 cm/s in the positive direction of an x ax

Question

An electron with a speed of 7.10 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.49 × 103 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 8.19 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?

Explanation / Answer

Given,

v = 7.1 x 10^8 cm/s ; E = 2.49 x 10^3 N/C ;

a)from conservation of energy

qE d = 1/2 m v^2

d = 0.5 m v^2/qE

d = 0.5 x 9.1 x 10^-31 x (7.1 x 10^6)^2/(1.6 x 10^-19 x 2.49 x 10^3) = 0.058 m

Hence, d = 0.058 m = 5.8 cm

b)We know that,

F = qE also F = ma

qE = ma => a = qE/m

a = 1.6 x 10^-19 x 2.49 x 10^3/(9.1 x 10^-31) = 4.38 x 10^14 m/s^2

v = u + at

0 = 7.1 x 10^6 - 4.38 x 10^14 t

t = 7.1 x 10^6/(4.38 x 10^14) = 1.62 x 10^-8 s

Hence, t = 1.62 x 10^-8 s

c)W = qE d

W = 1.6 x 10^-19 x 2.49 x 10^3 x 8.19 x 10^-3 = 3.26 x 10^-18 J

W = 1/2 m v^2 = 0.5 x 9.1 x 10^-31 x (7.1 x 10^6)^2 = 2.29 x 10^-17 J

f = 3.26 x 10^-18/2.29 x 10^-17 = 0.142

Hence, f = 0.142 = 14.2 %

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