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SPEAKER 1: Check
out this capacitor.

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Look at what happens if I
hook it up to this light bulb.

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CHILDREN (IN UNISON): Boo!

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SPEAKER 1: Yeah,
nothing happened

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because the capacitor
is not charged up.

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But if we look at up
to a battery first,

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to charge up the capacitor,
and then hook it up

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to the light bulb, the
light bulb lights up.

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CHILDREN (IN UNISON): Ooh!

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SPEAKER 1: The
reason this happens

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is because when a capacitor is
charged up, it not only stores

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charge, but it stores
energy as well.

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When we hooked up the
capacitor to the battery,

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the charges got separated.

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These separated charges
want to come back together

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when given the chance,
because opposites attract.

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So if you complete the
circuit with some wires

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and a light bulb,
currents going to flow.

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And the energy that was
stored in the capacitor

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turns into light and heat that
comes out of the light bulb.

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Once the capacitor
discharges itself,

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and there's no more
charges left to transfer,

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the process stops and
the light goes out.

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The type of energy that's
stored in capacitors

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is electrical potential energy.

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So if we want to figure
out how much energy

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is stored in a
capacitor, we need

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to remind ourselves
what the formula is

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for electrical potential energy.

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If a charge, Q, moves
through a voltage, V,

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the change in electrical
potential energy of that charge

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is just Q times V. Looking
at this formula, what

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do you think the energy would
be of a capacitor that's

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been charged up to a
charge Q, and a voltage V?

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CHILDREN (IN UNISON): Q times V!

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SPEAKER 1: Yeah, and that's what
I thought it would have been,

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too.

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But it turns out the energy of
a capacitor is 1/2 Q times V.

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CHILDREN (IN UNISON): Boo!

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SPEAKER 1: Where does
this 1/2 come from?

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How come the energy
is not just Q times V?

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Well, the energy of
a capacitor would

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be Q times V if during
discharge, all of the charges

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were to drop through the
total initial voltage, V.

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But during discharge,
all of the charges

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won't drop through
the total voltage, V.

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In fact, only the first
charge that gets transferred

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is going to drop through the
total initial voltage, V.

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All of the charges that get
transferred after that are

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going to drop through
less and less voltage.

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The reason for this is that
each time a charge gets

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transferred it decreases
the total amount of charge

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stored on the capacitor.

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And as the charge on the
capacitor keeps decreasing,

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the voltage of the
capacitor keeps decreasing.

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Remember that the
capacitance is defined

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to be the charge stored
on a capacitor divided

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by the voltage across
that capacitor.

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So as the charge goes down,
the voltage goes down.

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As more and more charge
gets transferred,

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there'll be a point
where a charge only

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drops through 3/4 of
the initial voltage.

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Wait longer, and
there'll come a time

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when a charge gets transferred
through only a half

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of the initial voltage.

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Wait even longer, and
a charge will only

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get transferred through a
fourth of the initial voltage.

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And the last charge to
get transferred drops

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through almost no
voltage at all,

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because there's
basically no charge

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left that's stored
on the capacitor.

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If you were to add
up all of these drops

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in electrical
potential energy, you'd

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find that the total drop
in energy of the capacitor

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is just Q, the total
charge that was initially

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on the capacitor, times
1/2 the initial voltage

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of the capacitor.

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So basically that 1/2 is there
because not all the charge

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dropped through the total
initial voltage, V. On average,

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the charges dropped through
only a half the initial voltage.

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So if you take the charge stored
on a capacitor at any moment,

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and multiply by the voltage
across the capacitor

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at that same
moment, divide by 2,

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you'll have the energy
stored on the capacitor

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at that particular moment.

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There's another form of this
equation that can be useful.

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Since capacitance is defined
to be charge over voltage,

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we can rewrite this
as charge equals

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capacitance times voltage.

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If we substitute the
capacitance times voltage

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in for the charge, we see
that the energy of a capacitor

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can also be written as 1/2
times the capacitance times

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the voltage across
the capacitor squared.

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But now we have a problem.

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In one of these formulas
the V is squared,

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and in one of these formulas
the V a not squared.

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I used to have trouble
remembering which is which.

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But here's how I remember now.

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If you use the formula
with the C in it,

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then you can see the V squared.

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And if you use the formula
that doesn't have the C in it,

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then you can't
see the V squared.

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So these are the two
formulas for the energy

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stored in a capacitor.

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But you have to be careful.

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The voltage, V,
in these formulas

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refers to the voltage
across the capacitor.

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It's not necessarily the voltage
of the battery in the problem.

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If you're just looking
at the simplest

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case of one battery that
has fully charged up

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a single capacitor,
then in that case,

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the voltage across
the capacitor will

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be the same as the
voltage of the battery.

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So if a 9-volt battery
has charges up a capacitor

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to a maximum charge
of four coulombs,

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then the energy stored
by the capacitor

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is going to be 18 joules.

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Because the voltage
across the capacitor

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is going to be the same as
the voltage of the battery.

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But if you're looking at a case
where multiple batteries are

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hooked up to
multiple capacitors,

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then in order to find the
energy of a single capacitor,

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you've got to use
the voltage across

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that particular capacitor.

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In other words, if you
were given this circuit

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with these values, you
could determine the energy

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stored in the middle
capacitor by using 1/2 Q-V.

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You would just have
to be careful to use

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the voltage of that
capacitor, and not

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the voltage of the battery.

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Plugging in five
coulombs for the charge

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lets you figure out that
the energy is 7.5 joules.

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CHILDREN (IN UNISON): Ooh!

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[MUSIC PLAYING]