m = mass of electron
q = electron charge
F = m a
q E = m a
a = q E/m
v = Vi - a t
v = 0 at stop
so
t = 5 * 10^5/a
d = Vi t - .5 a t^2
.09 = 5*10^5 t - .5 a t^2
now the easy way :)
if acceleration is constant the average speed during this stop is
2.5 *10^5 m/s
it stops in .09 m
so t = .036 * 10^-5 seconds
a = change in v/change in t
a= 5*10^5/.036*10^-5
a= 139 * 10^10 m/s^2
so
a = q E/m = 139 * 10^10
E = 139*10^10 m/q
= 139 * 10^10 *9.1*10^-31 /1.6*10^-19
An electron with an initial speed of 5.0 ✕ 105 m/s enters a region in which there is an electric field directed along its direction of motion. If the electron travels 9.0 cm in the field before being stopped, what are the magnitude and direction of the electric field?
1 answer
1 answer