Question

A horizontal beam of electrons initially moving at 3.4 times 10^7 m/s is deflected vertically by the vertical electric field between oppositely charged parallel plates as shown in the diagram. The magnitude of the field is 2.20 times 10^4 N/C. What is the direction of the field between the plates? up down right left What is the charge per unit area on the plates? What is the vertical deflection d of the electrons as they leave the plates?

Answer 1

(b)
For parallel plates:
E = σ/(εo)
2.20e4 N/C = σ/(8.85e-12 C^2/(N-m^2))
σ = 1.947e-7 C/m^2

(c)
, x = 0.0200 m

horizontal motion:
x = (vx)(t)
t = x/(vx) [1]

vertical motion:
d = at^2 / 2 [2] (initial vertical velocity is 0)

find acceleration:
F = ma = qE
a = qE/m [3]

Substitute [1] and [3] into [2] and solve for d:
d = (qE/m)(x/(vx))^2 / 2
d = [(1.602e-19 C)(2.20e4 N/C) / (9.109e-31 kg)][(0.0200 m)/(3.4e7 m/s)]^2 / 2
d = 6.69e-4 m = 0.67 mm