An electron with a speed of 4.37 * 10^8 cm/s in the positive direction an times

Question

An electron with a speed of 4.37 * 10^8 cm/s in the positive direction an times axis enters an electric field of magnitude 1.2.3 * 10^3 N/C. travelling along a field line in the direction that retands its motion. How for will electron will the electron travel in the field before stopping momentally, and how much time will have elapsed If the region containing the electric field is 5.30 mm long if (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

a) The electric field retards the electron's motion. So if the electron travels far enough in the field to stop then the work done by the electric field on the electron has to equal the electron's intial kinetic energy.
K = 1/2mv^2
W=work = qEd (where d= distance, q = charge on electron and E = field strength)
So W =K ---> qEd = 1/2 mv^2 and solve for d ---> d = 1/2 (mv^2/(Eq))
d = 0.044 m = 4.4 cm

b) You know the initial speed and the final speed (0 m/s). You cna find the acceleration from:
F = qE = ma ---> a = qE/m
So v = -at + v0 = -qEt/m +v0
solve for t setting v = 0 ---> t = mv0/(qE) where v0 = 4.37x10^6 m/s
t = 2.02 x10^-8 sec

c) Find the work done by the electic field over the 5.3 mm.
W =qEd ---> set d = 5.3 mm = 0.0053 m ---> W = 1.04 x10^-18 J

This is the kinetic energy lost by the electron. The fraction is:
f = (qEd)/[1/2mv^2] ---> v= 4.37x10^6 m/s ---->, f = 0.1197 = 11.97%

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