In the figure below, a uniform, upward electric field of magnitude 2.00 x 103 N/
ID: 2132932 • Letter: I
Question
In the figure below, a uniform, upward electric field of magnitude 2.00 x 103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 10.5 cm and separation d = 1.85 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity 0 of the electron makes an angle ? = 49 degree with the lower plate and has a magnitude of 6.00 106 m/s. How far horizontally from the left edge will the electron strike?Explanation / Answer
F= q E where both F and E are vectors
if E points up, and q is negative, then the direction of the force on the electron is down, and the force acts on the electron in the same way gravity acts on a projectile
so we can treat this as a kinematics problem where the acceleration is not due to g, but is given by
F = qE = ma so that a = q E/m where m and q are the mass and charge of an electron
(I get a = 3.5e14 m/s/s but am unsure since you do not give units for the E field, I am assuming MKS units)
now calculate the maximum "height" of the electron from:
vfy^2 = v0y^2 - 2 a d
vfy=final vertical velocity =0 since at apex the vertical velocity is zero
v0y=initial vertical velocity = 6e6 sin 45
a=accleration, just computed from qE/m
d=max height
so d=v0y^2/2a = 0.026m
this means the max height is greater than 2 cm and it is possible the electron will hit the upper plate
now, how long does it take for the electron to reach a height of 0.02m?
y(t)=v0 sin(theta) - 1/2 a t^2
set y=0.02m, v0=6e6m/s, and solve for t;