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Feynman talking about electricity: http://www.youtube.com/watch?v=qhh32JYkQPk

When brushing your hair, Feynman mentioned that a "few" electrons were transferred to the brush (or vice versa, can't remember off the top of my head how the convention goes). Clearly this will vary depending on how much you comb your hair, so my question is what is the approximate charge of the brush which would be required so that it would be noticeable (i.e. I could make pepper dance with the brush, or jiggle a piece of paper at close range).

Sorry if the question is silly :P

Cheers,

Rofler

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Could whoever gave me a -1 for the question leave a brief comment on how I can improve the quality of the question? –  Rofler Jun 17 '11 at 18:30
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up vote 3 down vote accepted

I'm assuming you just want an order of magnitude answer, because there's no specific numbers that would allow us to be more precise.

According to http://en.wikipedia.org/wiki/Static_electricity#Energies_involved a common voltage for everyday static electricity is 10,000 V, and the amount of energy released in a discharge is less than a joule, usually something like 0.1 J. A simple calculation tells us that the charge involved is $Q = 2 E / V$ or 20 microcoulombs.

How many electrons in 20 microcoulombs? A single electron is $2 \times 10^{-19}$ C, so to make $2 \times 10^{-5}$ C you need about $10^{14}$ electrons, or a hundred trillion. So give or take a few factors of ten, this is the number of electrons transferred from the hair to the brush. It's much more than a billion, but much much less than Avogadro's number.

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That works for me! I was secretly hoping the answer would be some ridiculously tiny number which would blow my mind, like a few thousand, or even something less than a billion, but at least it's a more manageable number than Avagadro's constant. :) –  Rofler Jun 18 '11 at 2:38
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