Question

A beam of electrons is shot into a uniform downward electric field of magnitude 1.10 103 N/C. The electrons have an initial velocity of 1.01 107 m/s, directed horizontally. The field acts over a small region, 5.00 cm in the horizontal direction. (a) Find the magnitude and direction of the electric force exerted on each electron. (b) How does the gravitational force on an electron compare with the electric force? (c) How far has each electron moved in the vertical direction by the time it has emerged from the field? (d) What is the electron's vertical component of velocity as it emerges from the field? (Up is the positive y-direction.) (e) The electrons move an additional 19.0 cm after leaving the field. Find the total vertical distance that they have been deflected by the field.

Answer 1

(a)

F = Eq = 1100 ( 1.6 * 10^-19) = 1.76 * 10^-16 N

gravitational force = mg = 9.11 * 10^-31 ( 9.8) = 89.278* 10^-31 N

F/ gravitational force =1.76 * 10^-16 N/89.278* 10^-31 N=1.97 * 10^13

(c)

t = d/v = 5 * 10^-2 / 1.01 * 10^7 = 4.95 * 10^-9 s

a = F/m = 1.76 * 10^-16 N/9.11 * 10^-31 = 1.93 * 10^14

s = 1/2 at^2 = 1/2 * 1.93 * 10^14 ( 4.95 * 10^-9 s)^2 = 2.36 * 10^-3 m

(d)

v=a*t=1.93 * 10^14* 4.95 * 10^-9 s=9.55 * 10^5 m/s

(e)

d=2.36 * 10^-3 m + 9.55 * 10^5*19* 10-2/1.01*10^7=20.33* 10^-3 m