Physics
posted by Jordie on .
an electron is launched at a 30 degree angle with respect to the horizontal and a speed of 3.0*10^6 m/s from the positive plate of the parallel plate capacitor. THe plats of the capacitor are oriented horizontally. The electron lands 5 cm away from the point of release.
a) what is the electric field strength inside the capacitor?
would i use E=(1/4*pi*Eo)(q/r^2)
where q=e and r=5cm
b) what is the minimum separation between the plates?
very confused. any help would be great

a) No don't use that formula. It is the formula for the field strength at distance r from a point source. Your field comes from parallel plates.
This is like a ballistics problem with gravity negligible. The downward acceleration "a" is provided by the field, and equals E*e.
The horizontal range is [V^2/a]sin (2A) = 0.05 m where A is the launch angle of 30 degrees. You may recall that formula from problems in ballistic trajectories. It is easy to derive.
Use that equation to solve for the acceleration a, and then use a = e*E to solve for E.
(b) The plates must be separated by enough distance D to prevent the electrons from hitting the upper plate; otherwise the trajectory is disrupted. That means
(V sin A)^2/2 < a*D
That means the vertical rise distance,
[V sin A/2]*(VsinA/a)
is less than the plate separation, D