Question

A beam of electrons traveling horizontally with a velocity of 6.2x10^6 m/s is aimed exactly between two parallel plates a distance 1 cm apart. The plates are 75 meters long. What is the maximum surface charge density on the plates for the electrons to make it through, and what will the electron’s velocity be on the far aside of the plates are charged to that surface density (you may ignore fringe effects)?

Answer 1

time to cross the plates,

t = (75 m) / (6.2 x 10^6 m/s)

t = 1.21 x 10^-5 sec  

perpendicular to the plates,

d = 0.01/2 = 0.005 m  

v0y = 0

d = v0 t + a t^2 / 2

0.005 = 0 + a(1.21 x 10^-5)^2 /2

a = 6.83 x 10^7 m/s^2

and a = q E / m

6.83 x 10^7 = (1.6 x 10^-19)E / (9.109 x 10^-31)

E = 3.89x 10^-4 N/C

and E = sigma / e0

sigma = (8.82 x 10^-12)(3.89 x 10^-4)

= 3.43 x 10^-15 C / m^2 ......Ans

vx = 6.2 x 10^6 m/s

vy = (6.83 x 10^7)(1.21 x 10^-5) = 826 m/s

v = (6.2 x 10^6)i + (826)j m/s