Answer to Question #81328 in Field Theory for aashar

Question

An electron with a speed of 5.00 × 10^8 cm/s enters an electric field of magnitude 1.00 × 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 8.00 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

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 initial kinetic energy.
K =1/2 mv^2
W=work = qEd wher d= distance, q = charge on electron and E = field strength
qEd =1/2 mv^2
d =1/2 ((mv^2)/Eq)
d = 0.712 m = 71.2 cm
b. You know the initial speed and the final speed (0 m/s).
a =qE/m
So
v = -at + v_0 = -qEt/m +v_0
v=0
t =(mv_0)/qE t = 2.85∙〖10〗^(-8) s
c. Find the work done by the electric field over the 8 mm.
W =qEd
W = 1.28∙〖10〗^(-18) J
This is the kinetic energy lost by the electron. The fraction is:
f =qEd/[1/2 mv^2 ]
f = 0.112 = 11.2%.