An electron is released from rest in a weak electric field g

An electron is released from rest in a weak electric field given by vector E = -2.80X10^-10 N/C j. After the electron has traveled a vertical distance of 1.4 µm, what is its speed? (Do not neglect the gravitational force on the electron.)

Solution

E = - 2.80 x 10^-10 N/C j

electric force, Fe = q E = (-1.6 x 10^-19) (-2.80 x 10^-10 j)

Fe = 4.48 x 10^-29 N j

gravitational force, Fg = m g (-j)

Fg = (9.109 x 10^-31)(9.81)(-j) = - 8.94 x 10^-30 N j

applying Fnet = ma

Fe + Fg = m a

4.48 x 10^-29 - 8.94 x 10^-30 = (9.109 x 10^-31) a

a = 39.37 m/s^2

Applying vf^2 - vi^2 = 2 a d

v^2 - 0^2 = 2 ( 39.37) (1.4 x 10^-6 )

v = 0.0105 m or 10.5 mm