SPEAKER 1: Check out this capacitor.
Look at what happens if I hook it up to this light bulb.
CHILDREN (IN UNISON): Boo!
SPEAKER 1: Yeah, nothing happened
because the capacitor is not charged up.
But if we look at up to a battery first,
to charge up the capacitor, and then hook it up
to the light bulb, the light bulb lights up.
CHILDREN (IN UNISON): Ooh!
SPEAKER 1: The reason this happens
is because when a capacitor is charged up, it not only stores
charge, but it stores energy as well.
When we hooked up the capacitor to the battery,
the charges got separated.
These separated charges want to come back together
when given the chance, because opposites attract.
So if you complete the circuit with some wires
and a light bulb, currents going to flow.
And the energy that was stored in the capacitor
turns into light and heat that comes out of the light bulb.
Once the capacitor discharges itself,
and there's no more charges left to transfer,
the process stops and the light goes out.
The type of energy that's stored in capacitors
is electrical potential energy.
So if we want to figure out how much energy
is stored in a capacitor, we need
to remind ourselves what the formula is
for electrical potential energy.
If a charge, Q, moves through a voltage, V,
the change in electrical potential energy of that charge
is just Q times V. Looking at this formula, what
do you think the energy would be of a capacitor that's
been charged up to a charge Q, and a voltage V?
CHILDREN (IN UNISON): Q times V!
SPEAKER 1: Yeah, and that's what I thought it would have been,
too.
But it turns out the energy of a capacitor is 1/2 Q times V.
CHILDREN (IN UNISON): Boo!
SPEAKER 1: Where does this 1/2 come from?
How come the energy is not just Q times V?
Well, the energy of a capacitor would
be Q times V if during discharge, all of the charges
were to drop through the total initial voltage, V.
But during discharge, all of the charges
won't drop through the total voltage, V.
In fact, only the first charge that gets transferred
is going to drop through the total initial voltage, V.
All of the charges that get transferred after that are
going to drop through less and less voltage.
The reason for this is that each time a charge gets
transferred it decreases the total amount of charge
stored on the capacitor.
And as the charge on the capacitor keeps decreasing,
the voltage of the capacitor keeps decreasing.
Remember that the capacitance is defined
to be the charge stored on a capacitor divided
by the voltage across that capacitor.
So as the charge goes down, the voltage goes down.
As more and more charge gets transferred,
there'll be a point where a charge only
drops through 3/4 of the initial voltage.
Wait longer, and there'll come a time
when a charge gets transferred through only a half
of the initial voltage.
Wait even longer, and a charge will only
get transferred through a fourth of the initial voltage.
And the last charge to get transferred drops
through almost no voltage at all,
because there's basically no charge
left that's stored on the capacitor.
If you were to add up all of these drops
in electrical potential energy, you'd
find that the total drop in energy of the capacitor
is just Q, the total charge that was initially
on the capacitor, times 1/2 the initial voltage
of the capacitor.
So basically that 1/2 is there because not all the charge
dropped through the total initial voltage, V. On average,
the charges dropped through only a half the initial voltage.
So if you take the charge stored on a capacitor at any moment,
and multiply by the voltage across the capacitor
at that same moment, divide by 2,
you'll have the energy stored on the capacitor
at that particular moment.
There's another form of this equation that can be useful.
Since capacitance is defined to be charge over voltage,
we can rewrite this as charge equals
capacitance times voltage.
If we substitute the capacitance times voltage
in for the charge, we see that the energy of a capacitor
can also be written as 1/2 times the capacitance times
the voltage across the capacitor squared.
But now we have a problem.
In one of these formulas the V is squared,
and in one of these formulas the V a not squared.
I used to have trouble remembering which is which.
But here's how I remember now.
If you use the formula with the C in it,
then you can see the V squared.
And if you use the formula that doesn't have the C in it,
then you can't see the V squared.
So these are the two formulas for the energy
stored in a capacitor.
But you have to be careful.
The voltage, V, in these formulas
refers to the voltage across the capacitor.
It's not necessarily the voltage of the battery in the problem.
If you're just looking at the simplest
case of one battery that has fully charged up
a single capacitor, then in that case,
the voltage across the capacitor will
be the same as the voltage of the battery.
So if a 9-volt battery has charges up a capacitor
to a maximum charge of four coulombs,
then the energy stored by the capacitor
is going to be 18 joules.
Because the voltage across the capacitor
is going to be the same as the voltage of the battery.
But if you're looking at a case where multiple batteries are
hooked up to multiple capacitors,
then in order to find the energy of a single capacitor,
you've got to use the voltage across
that particular capacitor.
In other words, if you were given this circuit
with these values, you could determine the energy
stored in the middle capacitor by using 1/2 Q-V.
You would just have to be careful to use
the voltage of that capacitor, and not
the voltage of the battery.
Plugging in five coulombs for the charge
lets you figure out that the energy is 7.5 joules.
CHILDREN (IN UNISON): Ooh!
[MUSIC PLAYING]