A uniform electric field exists in the region between two oppositely charged plane-parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.20 cm distant from the first, in a time interval of

a. Find the magnitude of this electric field.

b. Find the speed of the electron when it strikes the second plate

given S = distance from the first = 3.20cm = 0.032m, t = 1.30*10-8 s q = 1.6 x 10_19C using S = at^2/2 acceleration = 0.032 X 2 / (1.30*10-8) ^2

a = 3.79 x 10^14m/s^2

From F = ma F = qE ma = qE

E = ma / q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s