An electron with a speed of 6.28 times 10^8 cm/s in the positive direction of an

Question

An electron with a speed of 6.28 times 10^8 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.35 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 7.58 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) Number Units (b) Number Units (c) Number Units

Explanation / Answer

Given, v = speed of electron v= 6.28*10^6 m/s i [ =ve x direction ]

electric field, E = 2.35*10^3 N/C [ +ve x direction, retards motion]

a. force experienced by electron F = e*E

acceleration = eE/m [ m is mass of electron]

so distance travelled before stopping be d, then

2*eE*d/m = v^2

d = v^2*m/2*eE = 4.772 cm

b. time elapsed = t

then

v = eEt/m

t = mv/eE = 1.519*10^-8 s

c. if d' = 7.58 mm

then , potential drop, V = E*d' = 2.35*10^3*7.58*10^-3 = 17.813 V

energy lost = eV = 28.5008 *10^-19 J

initial energy, KE = 0.5mv^2 = 179.44*10^-19 J

so fraction of energy lost = eV/ke = 0.0999.92 percent

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