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February 5th, 2019, 05:05 PM   #11
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There is another way to put the data together to check you have the right formula and dimensions without deriving it. Put the formula together so the dimensions are correct.

$\displaystyle \frac{\text{coulombs}}{\text{sec}}=(\frac{\text {electrons}}{L^{3}})( \frac{\text{coulombs}}{\text{electron}})(L^{2})( \frac{L}{\text{sec}})

So all you have to assume is that charge density has the same length units as area.

Last edited by skipjack; February 5th, 2019 at 06:16 PM.
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February 5th, 2019, 06:27 PM   #12
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The denominator that NAC54321 used should be enclosed in parentheses.
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February 6th, 2019, 06:50 AM   #13
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n=1.00x10^{28} electrons/mm^{3}
d=.028 mm
A=Pi(.014)^{2} = 6.16x10^{-4} mm^{2}
i=8.1 coulombs/sec
e=1.60x10^{-19} coulombs/electron
v=drift velocity mm/sec



v= (8.1/(1.6x6.16x10^{5})

v=8.22x10^{-6} mm/sec
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