This question asks where a positron will leave an electricfield, and at what vel

Question

This question asks where a positron will leave an electricfield, and at what velocity. The particle is moving in thepositive x direction at 1.0 x 10^7 m/s whenit enters the electric field, oriented along theX-axis. I evaluated the first part by looking the the change inelectrical potential energy being in comparison to the kineticpotential energy. This all works out and reveals that the particle leaves thefield at x=0, travelling in the negative x direction. Then we need to explane how fast it is travelling on itsdeparture. I can understand that it would enter the field withkenetic energy, and leave with the same amount of energy. I justdon't know how to adequatley explane this.

Explanation / Answer

(a)    with help of    U = q V    we can translate the graph of volyage into apotential energy graph in eV units    from the information in the problem we cancalculate its kinetic energy in those units    Ki = 284 eV    this is less than the height of thepotrential energy barrier    thus it must reach a turning point and thenreverse its motion (b)    its final velocity then is in negative xdirection with a magnitude equal to that of its initialvelocity    its speed is upon leaving this region 1.0 x107 m / s

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