Question

An electron with a speed of 7.61 × 10⁸ cm/s in the positive direction of an x axis enters an electric field of magnitude 1.14 × 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 7.36 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?

Answer 1

here,

initial speed of electron , u = 7.61 * 10^8 cm/s

u = 7.61 * 10^6 m/s

electric feild , E = 1.14 * 10^3 N/C

accelration of electron , a = - E*e/me

a = 1.14 * 10^3 * 1.6 * 10^-19 /(9.1 * 10^-31)

a = - 2.004 * 10^14 m/s^2

(a)

let the distance the electron travel in the field before stopping momentarily be s

using third equation of motion

v^2 - u^2 = 2 * a*s

- (7.61 * 10^6)^2 = - 2 * (2.004 * 10^14 ) * s

s = 0.144 m

the distance the electron travel in the field before stopping momentarily is 0.144 m

(c)

let the time elapsed be t

using first equation of motion

v = u + a*t

0 = 7.61 * 10^6 - 2.004 * 10^14 * t

t = 3.8 * 10^-8 s

the time elapsed is 3.8 * 10^-8 s

(c)

width of regio, w = 7.36 * 10^-3 m

let the velocity at the end of redion be v

using third equation of motion

v^2 - u^2 = 2 * a *s

v^2 - (7.61 * 10^6)^2 = - 2 * (2.004 * 10^14 ) * 7.36 * 10^-3

v = 7.41 * 10^6 m/s

the fraction of kinetic energy lost , f = 0.5 * me * (v^2 - u^2) /(0.5 * me * u^2)

f =   (- (7.41 * 10^6)^2 + (7.61 * 10^6)^2)) / (7.61 * 10^6)^2))

f = 0.0519

f = 5.19 %

the fraction of the electron’s initial kinetic energy will be lost in that region is 5.19%