Question

An electron with a speed of 7.81 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.31 × 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 6.64 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

Force acting on electron in electric field is F = m*a = q*E

m*a = q*E = 1.6*10^-19*2.31*10^3 = 3.696*10^-16

mass of the electron m = 9.11*10^-31 kg

then 9.11*10^-31*a = 3.696*10^-16

accelaration a = 4.05*10^14 m/s^2

initial speed is u = 7.81*10^8*10^-2 = 7.81*10^6 m/sec

final speed is v = 0 m/sec

using kinematic equations

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

0^2-(7.81*10^6)^2 = -2*4.05*10^14*S

S = 0.0753 m = 7.53 cm

b) time elapsed is t

v = u+(a*t)

0 = (7.81*10^6) - (4.05*10^14*t)

t = 1.92*10^-8 sec

c) speed at 6.64*10^-3 m

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

v^2-(7.81*10^6)^2 =-2*4.05*10^14*6.64*10^-3

v = 7.45*10^6 m/sec

lost in kinetic energy = Kf-Ki = 0.5*m*(v^-u^2) = 0.5*9.11*10^-31*(7.45^2-7.81^2)*10^12 = -2.502*10^-18 J

initial KE = 0.5*m*u^2 = 0.5*9.11*10^-31*(7.81*10^6)^2 = 2.77*10^-17 J

required fraction is lost in KE/ original KE = 2.508*10^-18/(2.77*10^-17) =0.0905