WEBVTT

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This is College Physics
Answers with Shaun Dychko.

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The bookshelf of this nervous physicist
is described as a wooden bookshelf

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in order to avoid conducting charges
between the metal shelves.

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And so, the metal shelves can serve as
plates of parallel plate capacitors.

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Now, the capacitance of
these two middle shelves

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is going to be the permittivity of free
space times the area of the shelf

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divided by the separation between them.

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And so, we have 8.85 times ten to
the minus 12 farads per meter,

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times the area of one shelf,
which is 100 square meters,

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divided by 0.200 meters of
separation between them,

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giving a capacitance of 4.43 times ten to
the minus nine farads, or 4.43 nanoFarads.

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And, the next question is: What's
the voltage between the shelves

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if opposite charges of equal magnitude
are placed on adjacent shelves?

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So, that means, the top shelf, say will
have positive two nanoCoulombs,

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and the opposite charge of equal magnitude
will be placed on the other shelf,

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and so we have negative two
nanoCoulombs on the other shelf.

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And, that makes a charge difference between
these shelves of a total of 4 nanoCoulombs

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because two minus negative two makes four.

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And so, in this formula here,

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where we have capacitance equals
the charge divided by the voltage,

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we can solve for <i>V</i> by multiplying
both sides by <i>V</i> over <i>C</i>.

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And, we end up with <i>V</i> equals <i>Q</i> over <i>C</i>,

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and this <i>Q</i> is going to be two times two
nanoCoulombs, for a total of four.

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So, four nanoCoulombs divided by
4.425 nanoFarads gives 0.904 volts.

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And then, the energy stored in
this parallel plate capacitor

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is going to be the charge squared
divided by two times the capacitance.

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I could have used this voltage,
and I could have done

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capacitance times voltage squared
over two, that would have been fine,

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but I prefer not to use some
numbers that I've calculated

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 just in case my calculation was wrong,

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that would make the result
from this formula also wrong.

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And so, I'm using numbers given
to us directly in the question,

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which is the charge.

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And, I have no choice but
to use this capacitance.

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Although, I suppose I could plug
in the formula for capacitance

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if I wanted to use the raw numbers,
but this is good enough for me.

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So, we have charge squared, which
is two times two nanoCoulombs,

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and we square that result and divide by two,

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and also divide by the capacitance
that we calculated in part A,

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and this gives 1.81 times ten
to the minus nine Joules,

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and that is a very harmless
amount of energy,

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and this is a harmless amount of voltage.

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So, the nervous physicist has
no reason to be nervous.