Electron Deflection in Parallel Plate Capacitor

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SUMMARY

The discussion centers on calculating the electric field strength required for an electron beam to navigate a 90-degree turn within a parallel plate capacitor. The electron possesses a kinetic energy of 3.0x10^-17 J and must exit 1.0 cm away from its entry point, traveling perpendicular to its initial direction. Key equations utilized include KE = 1/2mv^2 and v1^ = v0^2 + 2as, with the motion analyzed in both x and y components. The problem is approached by decoupling the x and y motions, applying uniform motion along the x-axis and kinematic principles for the y-axis.

PREREQUISITES
  • Understanding of kinetic energy and its relation to motion (KE = 1/2mv^2)
  • Familiarity with electric fields and their effects on charged particles
  • Knowledge of kinematic equations and their application in two-dimensional motion
  • Basic principles of parallel plate capacitors and their electric field characteristics
NEXT STEPS
  • Study the derivation of electric field strength in parallel plate capacitors
  • Learn about the motion of charged particles in electric fields
  • Explore kinematic equations in two dimensions for projectile motion
  • Investigate the effects of varying electric field strengths on particle trajectories
USEFUL FOR

Physics students, electrical engineers, and anyone interested in the dynamics of charged particles in electric fields, particularly in the context of parallel plate capacitors.

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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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