Electron Deflection in Parallel Plate Capacitor

AI Thread Summary
The discussion focuses on calculating the electric field strength required for an electron beam to turn 90 degrees within a parallel plate capacitor. The kinetic energy of the electrons is given as 3.0x10^-17 J, and the problem involves determining the necessary conditions for the electron to exit 1.0 cm away from the entrance hole while traveling perpendicularly. Participants note that the situation resembles projectile motion under constant acceleration, similar to a parabolic trajectory influenced by gravity. They suggest decoupling the x and y motions, treating the x-axis as aligned with the capacitor plates and the y-axis as perpendicular to them. The conversation emphasizes the need for kinematic equations to solve for time and velocity components to derive the required electric field strength.
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Homework Statement


A Problem of Practical interest is to make a beam of electrons turn a 90 degree corner. This can be done wit ha parallel plate capacitor shown. An electric with kinetic energy of 3.0x10^-17J moves up through a small holes in the bottom plate of capacitor.

What is field strength needed if electron is to emerge from an exit hole of 1.0cm away from entrance hole traveling at right angles to its original direction.

http://img4.imageshack.us/img4/6061/platecapacitor.th.jpg


Homework Equations


KE = 1/2mv^2
v1^ = v0^2 +2as

The Attempt at a Solution



I know this somewhat similar to parabola example.
There is constant electric field acceleration caused by plate capacitor.
Initially KE = 1/mv^2 or split with x and y. KE = 1/2mv0x^2 + 1/2mv0y^2 and the y and x-axis will be tilted to adjust to angle of plate capacitor, which I am assuming is 45 degrees.
I know I need to find time with the constant velocity but I am having a bit trouble getting that far. I am not sure if my statements about KE are even correct so if someone could help me through that process and once I derive initial velocities in y and x this problem should be a piece of cake.

Thank You Very Much.
 
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Yes, 1/2mv^2 = 1/2 m (vx^2 + vy^2)
And I agree it's same as a parabola in a constant force field like gravity.
I would orient x-axis along the capacitor plate, and the y-axis perpendicular to the plates, so then is like height in gravity. So then the x and y motions decouple since the force is only along y.
Try it from there using uniform motion along x,
and for y, by the symmetry of the problem, you have to reverse the motion in the alloted time, some kinematics...
 
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