Question

An electron is projected with an intial speed of 1.60x10^6 m/s into the uniform field between the parallel plates in the figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates.

A. If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field?

Answer 1

here,

initial horizontal speed , ux = 1.6 * 10^6 m/s

let the time taken to emerge out from the feild be t

t * ux = 0.02

t = 0.02 /( 1.6 * 10^6)

t = 1.25 * 10^-8 s

let the final vertical speed be vy and accelration be ay

s = uy * t + 0.5 * ay * t^2

0.01/2 = 0 + 0.5 * ay *( 1.25 * 10^-8)^2

ay = 6.4 * 10^13 m/s^2

vy = uy + ay * t

vy = 6.4 * 10^13 * 1.25 * 10^-8

vy = 8 * 10^5 m/s

the final speed of the electron , v = sqrt(vx^2 + vy^2)

v = 1.79 * 10^6 m/s