An electron with a speed of 7.38 × 10 8 cm/s in the positive direction of an x a

Question

An electron with a speed of 7.38 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.57 × 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 5.78 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

speed of electron ,

u = 7.38 *10^6 m/s

Electric field , E = 2.57 *10^3 N/C

a)

acceleration of electron = q * E/me

acceleration of electron = -1.602 *10^-19 * 2.57 *10^3/(9.1 *10^-31)

acceleration of electron = - 4.52 *10^14 m/s^2

let the distance is d

using third equation of motion

0 - (7.38 *10^6)^2 = -2 * 4.52 *10^14 * d

d = 0.0602 m

d = 6.02 cm

the distance is 6.02 cm

b)

time , t

using first equation of motion

0 = 7.38 *10^6 - 4.52 *10^14 * t

t = 1.63 *10^-8 s

the time taken is 1.63 *10^-8 s

c)

fraction of electron's energy lost = change in potential energy/initial kinetic energy

fraction of electron's energy lost = (1.602 *10^-19 * 2.57 *10^3 * 0.00578)/(0.5 * 9.1 *10^-31 * (7.38*10^6)^2)

fraction of electron's energy lost = 0.096

the fraction of electron's energy lost is 0.096

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