Kinetic Energy of Electron Beam.

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Minki
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Homework Statement



Two horizontal plates connected to 600v with the positive at the top.
A positively charged ion between the plates moves downwards under the
force of the field (neglect gravity).
The ion has charge 1.6×10^–19C and mass 2.3×10^–26kg.
Show that the kinetic energy of the ion is 4.8x10-17 J at the instant
that it reaches the bottom plate.

Homework Equations



KE = eV

The Attempt at a Solution



eV = 1.6x10^-19 x 600 = 9.6x10^-17 J.

This is exactly twice the number I am trying to reach and I can't
get any further. The problem is that I don't understand the significance
of the ion reaching the bottom plate.

Thanks.
 
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Hello Minki,

Welcome to Physics Forums!
Minki said:

Homework Statement



Two horizontal plates connected to 600v with the positive at the top.
A positively charged ion between the plates moves downwards under the
force of the field (neglect gravity).
The ion has charge 1.6×10^–19C and mass 2.3×10^–26kg.
Show that the kinetic energy of the ion is 4.8x10-17 J at the instant
that it reaches the bottom plate.

Homework Equations



KE = eV

The Attempt at a Solution



eV = 1.6x10^-19 x 600 = 9.6x10^-17 J.

This is exactly twice the number I am trying to reach and I can't
get any further. The problem is that I don't understand the significance
of the ion reaching the bottom plate.

Thanks.
Your calculation of 9.6x10^-17 J looks correct to me if the ion begins accelerating at the positive (top) plate, and the potential difference between the plates is 600 V. (And of course, if we can ignore gravity, and can assume the plates are in a vacuum.)

Is any information left out of the problem statement? For example is the top plate connected to 600 V and the bottom plate connected to 300 V? Or instead, maybe the ion started accelerating at the midpoint between the plates?

(The part about the ion reaching the bottom plate is significant because it let's you know the ion's potential energy at the end of the experiment. If you know the difference in potential energy between the beginning and the end of the experiment, you can invoke the work-energy theorem to determine the kinetic energy.)
 
That's great Collinsmark. You have answered the question. I did leave out that the ion is midway between the plates.

Many thanks.