Now that we know what a parallel plate capacitor is
and how to deal with capacitors and how to add them together in series and parallel.
Let´s talk about the energy stored in the capacitor as well as the voltage.
For a given capacitor, we have a different charge on each plate
and so as you can imagine we have an electric field
between those two plates just based on the definition that we had for the electric field.
If our 2 capacitors are a distance d away from each other
and they´re connected by some battery with a voltage V
then we can find the value of this electric field
between the 2 capacitors as simply the voltage applied to the capacitors
divided by the distance between them.
Don´t forget that when we have a charge in the influence of an electric field,
so if we introduce a small positive charge as we have here.
First of all, it will as we said by the definition of the electric field
follow the lines of the electric field, in this case being pushed upwards
and the force on that charge will be equal to that charge´s value q
times the electric field and also don´t forget
that we have Newton´s second law rather in place that F equals ma.
To put charges on a capacitor, when you´re adding a charge
we have a negative charge on the top, in this case, we have a positive charge on the bottom.
If I want to use my battery to keep adding charges in this way,
what we´re forced to do is try to force more positive charge on the bottom plate
and force more negative charge on the top plate.
But since the bottom is already positive and the top is already negative,
trying to force similar charges onto these plates is going to be a process that takes energy
because again like charges repel and so the process of putting these extra charges
on these plates will require energy to add this extra charge
which means that we will have energy stored in our capacitor
if we let that energy come back and be released.
So this energy stored in the capacitor has a value of 1/2
times the charge that´s stored in the capacitor
times the voltage that´s being applied to the capacitor.
What we can do as I said, is if we had this stored charge on the capacitor
store for example a battery, we could then connect our capacitor
to some object that could take and run on electricity
and that energy would be released to the electrons on top of the capacitor
would be repelled by each other, pushed away by each other
and be forced to flow through the circuit maybe lighting up a light bulb.
For example, this is what is used in the flashes of cameras.
We have a capacitor that is charged and when you press the button
to get the flash to turn the camera, the capacitor is what we say discharged
all that energy stored in the capacitor is released very, very quickly
and goes through an object like this light bulb here
lighting it up very briefly as the capacitor discharges.
Because the voltage difference between the plates as it charges,
because the voltage difference between the plates changes rather
as the capacitor discharges, instead of the typical equation for the energy
that you would see from an electric field which we saw was Q times e
for the energy of charge going into an electric field
or Q times the voltage difference rather.
It will only be half the value and the reason that it´s sort of an average.
In other words, as these charges are leaving the plate,
the plates are gaining less and less of an energy difference between the two,
because again charges are leaving and so we have to average the total value for the energy
which would be Q times V using our equation from voltage and instead be half of that value.
So we have that the energy stored on the capacitor is still just 1/2 the charge and the capacitor
times the voltage that´s applied to the capacitor.