A problem of practical interest is to make a beam of electrons turn a 90 degree

Question

A problem of practical interest is to make a beam of electrons turn a 90 degree corner. This can be done with the parallel-plate capacitor shown in the figure (Figure 1). An electron with kinetic energy 2.8times10-17J enters through a small hole in the bottom plate of the capacitor. Explain. The bottom plate should be positive. The electron needs to be repelled by the top plate, so the top plate must be negative and the bottom plate positive. In other words, the electric field needs to point away from the bottom plate so the electron's acceleration a is toward the bottom plate. What strength electric field is needed if the electron is to emerge from an exit hole 1.0 cm away from the entrance hole, traveling at right angles to its original direction? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

2. Relevant equations KE = 1/2mv^2 v1^ = v0^2 +2as 3. 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. 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.

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