At what angle will the electrons in Example 2116 leave the u

At what angle will the electrons in Example 21-16 leave the uniform electric field at the end of the parallel plates (point P in the figure)? Assume the plates are 7.2 cm long, E = 6.0 × 10^3 N/C, and v0 = 1.00x10^7 m/s. Ignore fringing of the field.

Solution

along x axis , there is no force.

hence it x-component of velocity will remian constant.

vx = 1 x 10^7 m/s

time taken to cross plates.

t = Dx / vx = (7.2 x 10^-2 m) / (1 x 10^7 m/s) = 7.2 x 10^-9 sec

In y axis:

iinitial y -component of velocity, uy = 0

electric force, Fe = q E

E is upward and q is negative so Fe will be downwards.

ay = Fe / m = q E / m = (1.6 x 10^-19) (6 x 10^3) / (9.109 x 10^-31)

ay = 1.054 x 10^15 m/s^2

vertical velocity after t,

vy = uy + at

vy = 0 + (1.054 x 10^15) (7.2 x 10^-9) = 7.56 x 10^6 m/s

@ = 360 - tan^-1(vy/vx) = 360 - 37   =323 deg