Electric fields and mean free path?

Thread starter hobbs125
Start date Oct 14, 2013
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Electric Electric fields Fields Mean Mean free path Path

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The discussion revolves around calculating the required electric field strength for a particle with a mean free path of 50nm to achieve a collision energy of 5eV. A participant suggests that the calculation of 5 divided by 0.00000005 results in 100MV/m, but another advises to double-check the units used in the calculation. The focus is on ensuring accurate unit conversion and proper application of physics principles in determining electric field strength. The conversation highlights the importance of precision in scientific calculations. Accurate unit handling is crucial for correct results in physics problems.

Oct 14, 2013

hobbs125

Just need someone to tell me if I'm doing this right.

If I have accelerated a particle in which the mean free path is 50nm and I need the collision to be 5eV would this be correct in determining the required electric field strength?

5/.00000005 = 100MV/m

Last edited: Oct 14, 2013

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Oct 15, 2013

dauto

No. Double check your units.

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