Electron velocity in Electric Fields

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  • #26
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ok I rechecked it and recalculated and here are my results:

quadratic formula: t= 1.13814 E -8 or t=1.19729 E -8

Using vf = vi - a*t: t= 1.16771 E -8

I keep redoing the math but it turns out the same. :/
 
  • #27
gneill
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ok I rechecked it and recalculated and here are my results:

quadratic formula: t= 1.13814 E -8 or t=1.19729 E -8

Using vf = vi - a*t: t= 1.16771 E -8

I keep redoing the math but it turns out the same. :/

This is an example of where significant figures and rounding can have marked effects on the results. ALWAYS calculate intermediate values using more decimal places than the final result requires, and round only at the end. Quadratics can be notoriously sensitive to parameter values. This particular one should end with a double root... and they should both be close to the 1.16 value. When I calculate the value using high precision values of constants for the charge and mass of the electron, and keeping full precision for all intermediate values during the calculation, I find t = 1.164 x 10-8 s.

However. Let me give you a tip that may help. When you are working on a problem that has some initial velocity vo, a final velocity of zero, and a given distance for the particle to stop, d, then rather than solve the equation:

[itex] d = v_o t - \frac{1}{2} a t^2 [/itex]

Use instead:

[itex] d = \frac{1}{2} a t^2 [/itex]

which provides: [itex] t = \sqrt{\frac{2 d}{a}} [/itex]

and thus only one root to contend with.

The thing is, going from some initial velocity to zero over distance d is just a time-reversed version of going from zero to that speed over the same distance! It's like "running the film backwards" -- the event takes the same amount of time to complete only the formula for the distance is simpler, dropping the initial velocity term.
 
  • #28
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ohhhhhhh now Im getting that, I didnt even think of using that equation. And this problem has taught me a valuable lesson on rounding. lol Thanks again for explaining it. :smile:
 

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